I'm Reviewing 10.1
we weren't allowed to use calculators and we had to know the trig chart . . .
In order to do this you must know these formulas:
cos(alpha +/- beta)= cos(alpha)cos(beta) -/+ sin(alpha)sin(beta)
sin(alpha +/- beta)= sin(alpha)cos(beta) +/- cos(alpha)sin(beta)
Alpha and beta will come from the trig chart. You will either add or subtract them to get the angle needed.
Ex: find the exact value of sin15 degrees.
alpha= 45
beta= 30
45-30 = 15
sin(45-30)= (sin 45)(cos 30) - (cos 45)(sin 30)
=(square root of 2/2)(square root of 3/2) - (square root of 2/2)(1/2)
square root of 6/4 - square root of 2/4 = square root of 6 - square root of 2/4
sin 15= square root of 6 - square root of 2/4
Thursday, December 30, 2010
Malorie's Holiday blog #3
I'm reviewing 9.1 right triangles
We learned anout right angles and SOHCAHTOA.
In order to do this, you must know the following formulas:
Sin = opp/hyp
Cos= adj/hyp
tan= opp/adj
cot= adj/opp
sec= hyp/adj
csc= adj/hyp
(*this is only used for right triangles)
Example:
In triangle ABC, Angle A = 90 degrees, Angle B= 30 degrees, and a = 15. Find b and c.
in order to find c you must do cos 30=c/15
c=12.99
in order to find b, you must use the pythagorean theorem.
b= 7.5
We learned anout right angles and SOHCAHTOA.
In order to do this, you must know the following formulas:
Sin = opp/hyp
Cos= adj/hyp
tan= opp/adj
cot= adj/opp
sec= hyp/adj
csc= adj/hyp
(*this is only used for right triangles)
Example:
In triangle ABC, Angle A = 90 degrees, Angle B= 30 degrees, and a = 15. Find b and c.
in order to find c you must do cos 30=c/15
c=12.99
in order to find b, you must use the pythagorean theorem.
b= 7.5
Malorie's Holiday Blog #2
I'm reviewing 8.1 because it was easy.
When solving for theta, you get the trig function by itself and then take the inverse.
Ex: cos(theta)= 36
(theta)=cos-1(36)
Inverses have 2 answers with a few exceptions. Find where the angle is based on the trig function and if the number is tve or -ve.
Steps:
Take the inverse of tve # to find the quadrant one angle.
To find quad 2, ve must be negative and add 180.
To find quad 3, add 180
To find qaud 4, make it negative and add 360.
Example:
Solve for 0 degrees is less than or equal to theta less than 360 degrees.
cos(theta)= square root of 3/2
(on the trig chart)
(Theta)=30
cos is only positive in the first and fourth quadrant, we already know the first so now we need the fourth.
-30 + 360 = 330
(theta)=330
Final answer is (theta)= 30 and 330
When solving for theta, you get the trig function by itself and then take the inverse.
Ex: cos(theta)= 36
(theta)=cos-1(36)
Inverses have 2 answers with a few exceptions. Find where the angle is based on the trig function and if the number is tve or -ve.
Steps:
Take the inverse of tve # to find the quadrant one angle.
To find quad 2, ve must be negative and add 180.
To find quad 3, add 180
To find qaud 4, make it negative and add 360.
Example:
Solve for 0 degrees is less than or equal to theta less than 360 degrees.
cos(theta)= square root of 3/2
(on the trig chart)
(Theta)=30
cos is only positive in the first and fourth quadrant, we already know the first so now we need the fourth.
-30 + 360 = 330
(theta)=330
Final answer is (theta)= 30 and 330
Malorie's Holiday BLog #1
In this blog i'm looking back to the easiest stuff I learned. 7-4 reference angles.
To find a reference angle for sin and cos, you must follow the following:
Step 1: Figure out which quadrant the angle is in.
(if given an angle with pi, use pi as 180 and simplify)
Step 2: determine whether the function is positive or negative.
Step 3: subtract 180 from the angle measure until it is between 0 and 90 degrees.
Step 4: If the angle comes out to be 0, 30, 45, 60, or 90 degrees, use the trig chart to simplify. If not, then leave it as is or plug it in the calculator.
Example:
Find a reference angle for sin 6pi/5
sin (6x180)/5 = 1080/5 = 216
-sin216
this angle would be in the III quadrant because it is between 180 and 270. It would be negative because sin follows the y axsis and y is negative in this quadrant.
216-180 = 36
subtract from 180
your final answer would be -sin36 degrees.
To find a reference angle for sin and cos, you must follow the following:
Step 1: Figure out which quadrant the angle is in.
(if given an angle with pi, use pi as 180 and simplify)
Step 2: determine whether the function is positive or negative.
Step 3: subtract 180 from the angle measure until it is between 0 and 90 degrees.
Step 4: If the angle comes out to be 0, 30, 45, 60, or 90 degrees, use the trig chart to simplify. If not, then leave it as is or plug it in the calculator.
Example:
Find a reference angle for sin 6pi/5
sin (6x180)/5 = 1080/5 = 216
-sin216
this angle would be in the III quadrant because it is between 180 and 270. It would be negative because sin follows the y axsis and y is negative in this quadrant.
216-180 = 36
subtract from 180
your final answer would be -sin36 degrees.
Charlie's Christmas Holiday's Blog #4
one recent thing we did was arithmetic sequences. where you use the formula to find the certain number of the sequence that you want to find or just find the real formula to find that certain number of the sequence. the formula you use is t*n = t*1 + (n - 1) d; where t*n is the sequence anser you want to find, t*1 is the firse numbe of the sequence, and d is the number being added.
For example:
6, 12, 18, 24
t*n = 6 + (n - 1) 6
= 6 + 6n - 6
= 6n
~~>all you do is find what is being added, then plug into the formula.
For example:
t*1 = 3 and t*2 = 7, find t*6
t*n = 3 + (n-1) 4
= 3 + 4n - 4
= -1 + 4n
t*6 = -1 + 4(6)
= -1 + 24
= 23
~~> you find what is being added between the first 2, then plug into the formula. then change the t*n to t*6 and replace the n in the new formula with 6.
For example:
6, 12, 18, 24
t*n = 6 + (n - 1) 6
= 6 + 6n - 6
= 6n
~~>all you do is find what is being added, then plug into the formula.
For example:
t*1 = 3 and t*2 = 7, find t*6
t*n = 3 + (n-1) 4
= 3 + 4n - 4
= -1 + 4n
t*6 = -1 + 4(6)
= -1 + 24
= 23
~~> you find what is being added between the first 2, then plug into the formula. then change the t*n to t*6 and replace the n in the new formula with 6.
Charlie's Christmas Holiday's Blog #3
this first semester we also did law of sines. the formula (sinA/opposite leg = sinB/opposite leg) is used when you have a nonright triangle and have atleast one pair of an angle and oposite leg with either a different leg or another angle. to work this you just plug into the formula and cross mulitply like we use to do in like prealgebra back in the gap.
For example:
angle Z = 69
angle Y = 22
side y = 4
side z = ?
sin69/z = sin22/4
4sin69 = z sin22
4sin69/sin22 = z
z = 9.97
~~> you're suppose to draw a triangle for this too, but i'm not that talented to do it on here. for this you take angle Y and put it over side y and set it equal to angle Z over side z, then cross mulitply to solve for z. so side z equals 9.97.
For example:
angle Z = 69
angle Y = 22
side y = 4
side z = ?
sin69/z = sin22/4
4sin69 = z sin22
4sin69/sin22 = z
z = 9.97
~~> you're suppose to draw a triangle for this too, but i'm not that talented to do it on here. for this you take angle Y and put it over side y and set it equal to angle Z over side z, then cross mulitply to solve for z. so side z equals 9.97.
Charlie's Christmas Holiday's Blog #2
in another chapter we went through sine, cosine, and tangent for right triangles. for this we use the 'word' SOHCAHTOA. this means that sine equals opposite over hypotenuse, cosine equals adjacent oer hypotenuse, and tangent equals opposite over adjacent. to do this you are given the information that the triangle is a right triangle and either a side length and an angle other than the 90 degree one or two side lengths. in order to solve, though, a triangle must be drawn.
For example:
angle C = 90 degrees
angle A = 12 degrees
side b = 4
side a = ?
tan 12 = a/4
a = 4tan12
a = .9
~~> obviously i can't draw a triangle on here. but i took the opposite side length of angle A which was unknow and put it over the adjacent side length which was 4. since tangent is opposite over adjacent.
For example:
angle C = 90 degrees
angle A = 12 degrees
side b = 4
side a = ?
tan 12 = a/4
a = 4tan12
a = .9
~~> obviously i can't draw a triangle on here. but i took the opposite side length of angle A which was unknow and put it over the adjacent side length which was 4. since tangent is opposite over adjacent.
Charlie's Christmas Holiday's Blog #1
we went through captial sigma one section in the first semester, where there is a big E looking thing called captial sigma, a limit at the top of it [#], a limit with an index at the bottom of it [x = #], and the summand on the right of the big E thing [f(x)]. to solve you plug the bottom limit into the summand that's on the right of the captil sigma and add it with the same formula but the next numberical number until you get to the limitation that is on the top of the big E thing.
For example:
4
E p+3
p=2
(2+3) + (3+3) + (4+3)
= 5 + 6 + 7
= 18
~~>for this you pluged the 2 into the formula p + 3, then added it with the same formula but with 3 instead of 2, then again with 4 instead of 3, after that you couldn't go any further because the limit at the top was 4. then you solve to get 18.
For example:
4
E p+3
p=2
(2+3) + (3+3) + (4+3)
= 5 + 6 + 7
= 18
~~>for this you pluged the 2 into the formula p + 3, then added it with the same formula but with 3 instead of 2, then again with 4 instead of 3, after that you couldn't go any further because the limit at the top was 4. then you solve to get 18.
Wednesday, December 29, 2010
lawrence's holiday blog 2
we did chapter 9-1. It talked about solving for right triangles and law of sines.
Right triangle:
1. hypotenuse is opposite of the right angle
2. area= 1/2 base times height
3. SOH CAH TOA
-sin theta =opposite/hypotenuse
-csc theta=hypotenuse/opposite
-cos theta=adjacent/hypotenuse
-sec theta=hypotenuse/adjacent
-tan theta=opposite/adjacent
-cot theta=adjacent/opposite
Ex: In triangle ABC, angle A=28 degrees, angle C=90 degrees, side a=49, what is side b?
tan28degrees=40/b
btan28degrees=40
b=75.229
Ex: In triangle ABC, angle A=50 degrees, angle C= 90 degrees, side c=10, what is side a?
sin 50 degrees= a/10
=10 sin 50 degrees= a
a= 7.660
Law of Sines:
sinA/a = sinB/b = sinC/c
Law of sines is used when you know pairs in non-right triangles
To use the law of sines, you are setting up a proportion
Ex: in triangle ABC, angle A=115 degrees, side a=123, side b=16, what is angle B?
sin115 degrees/123 = sin theta/16
123sin theta=16sin115 degrees
sin theta=16sin115 degrees/123
theta =sin inverse (16sin115 degrees/123)
theta =6.771 degrees
Ex: in triangle ABC, angle A=30 degrees, angle B=15 degrees, side a=4, what is side B?
sin 30 degrees/4 = sin 15 degrees/x
x sin 30 degrees = 4 sin 15 degrees
x= 4 sin 15 degrees/ sin 30 degrees
=4 sin 15 degrees/ 1/2
x= 2.071
Right triangle:
1. hypotenuse is opposite of the right angle
2. area= 1/2 base times height
3. SOH CAH TOA
-sin theta =opposite/hypotenuse
-csc theta=hypotenuse/opposite
-cos theta=adjacent/hypotenuse
-sec theta=hypotenuse/adjacent
-tan theta=opposite/adjacent
-cot theta=adjacent/opposite
Ex: In triangle ABC, angle A=28 degrees, angle C=90 degrees, side a=49, what is side b?
tan28degrees=40/b
btan28degrees=40
b=75.229
Ex: In triangle ABC, angle A=50 degrees, angle C= 90 degrees, side c=10, what is side a?
sin 50 degrees= a/10
=10 sin 50 degrees= a
a= 7.660
Law of Sines:
sinA/a = sinB/b = sinC/c
Law of sines is used when you know pairs in non-right triangles
To use the law of sines, you are setting up a proportion
Ex: in triangle ABC, angle A=115 degrees, side a=123, side b=16, what is angle B?
sin115 degrees/123 = sin theta/16
123sin theta=16sin115 degrees
sin theta=16sin115 degrees/123
theta =sin inverse (16sin115 degrees/123)
theta =6.771 degrees
Ex: in triangle ABC, angle A=30 degrees, angle B=15 degrees, side a=4, what is side B?
sin 30 degrees/4 = sin 15 degrees/x
x sin 30 degrees = 4 sin 15 degrees
x= 4 sin 15 degrees/ sin 30 degrees
=4 sin 15 degrees/ 1/2
x= 2.071
lawrence's holiday blog 1
we learned the first part of chapter 10. it is really easy. just know the formula and plug in.
formulas:
cos(alpha + or - beta) =sine alpha cos beta - or + sin alpha cos beta
sin (alpha + or - beta)= sin alpha cos beta + or - cos alpha sin beta
* aplha and beta come from the trig chart and add or subtract to get the angle you are looking for.
ex. find the exact value of sin 15
alpha= 45
beta= 30
sin(45-30)= sin45 cos30- cos45 sin 30
square root of2/2 square root of3/2 - square root of2/2 1/2
square root of6/4- square root of2/4 = square root of 6-2/f
ex.2
suppose that sin alpha=4/5 and sin beta= 5/13. where 0
cos alpha cos beta- sinalpha sin beta
(3/5)(-12/13)- (4/5)(5/13)
(-36/65)-(20/65)
(-56/65)
that is all that i learned in chapter 10 section 1. i hope this helped out the people who dont understand even though i dont know how you wont get this section when it is the easiest thing we probably will cover.
formulas:
cos(alpha + or - beta) =sine alpha cos beta - or + sin alpha cos beta
sin (alpha + or - beta)= sin alpha cos beta + or - cos alpha sin beta
* aplha and beta come from the trig chart and add or subtract to get the angle you are looking for.
ex. find the exact value of sin 15
alpha= 45
beta= 30
sin(45-30)= sin45 cos30- cos45 sin 30
square root of2/2 square root of3/2 - square root of2/2 1/2
square root of6/4- square root of2/4 = square root of 6-2/f
ex.2
suppose that sin alpha=4/5 and sin beta= 5/13. where 0
cos alpha cos beta- sinalpha sin beta
(3/5)(-12/13)- (4/5)(5/13)
(-36/65)-(20/65)
(-56/65)
that is all that i learned in chapter 10 section 1. i hope this helped out the people who dont understand even though i dont know how you wont get this section when it is the easiest thing we probably will cover.
Tuesday, December 28, 2010
Helen's Holiday Blog #4
9-1
Examples:
1) The safety instructions for a 20ft ladder indicate that the ladder should not be inclined at more than 70 degrees with the ground. Suppose the ladder is leaned against a house at this angle, as shown. Find the distance from the base of the house to the foot of the ladder and the height reached by the ladder.
cos70=a/20
a=20cos70=6.840ft
sin70=b/20
b=20sin70=18.794ft
2) The highest tower in the world is in Toronto, Canada and is 553m high. An observer at point A, 100m from the center of the tower's base, sights the top of the tower, the angle of elevation is angle A. Find the measure of this angle to the nearest tenth of degrees.
tan A=553/100
A=tan inverse (553/100)=79.750 degrees
9-2
Formulas: 1/2 bh for right angles only or 1/2 ab sin(angle b/w) for non right angles
Examples:
1) Two sides of a triangle have lengths 7cm and 4cm. The angle between the sides measures 73 degrees. Find the area of the triangle.
A=1/2ab sin (theta)
A=1/2(7)(4)sin(73)
A=1/2(28)sin73
=14 sin 73
=13.388
Examples:
1) The safety instructions for a 20ft ladder indicate that the ladder should not be inclined at more than 70 degrees with the ground. Suppose the ladder is leaned against a house at this angle, as shown. Find the distance from the base of the house to the foot of the ladder and the height reached by the ladder.
cos70=a/20
a=20cos70=6.840ft
sin70=b/20
b=20sin70=18.794ft
2) The highest tower in the world is in Toronto, Canada and is 553m high. An observer at point A, 100m from the center of the tower's base, sights the top of the tower, the angle of elevation is angle A. Find the measure of this angle to the nearest tenth of degrees.
tan A=553/100
A=tan inverse (553/100)=79.750 degrees
9-2
Formulas: 1/2 bh for right angles only or 1/2 ab sin(angle b/w) for non right angles
Examples:
1) Two sides of a triangle have lengths 7cm and 4cm. The angle between the sides measures 73 degrees. Find the area of the triangle.
A=1/2ab sin (theta)
A=1/2(7)(4)sin(73)
A=1/2(28)sin73
=14 sin 73
=13.388
Helen's Holiday Blog #3
8-5
Things you CAN'T do!!!!!!!!!!!
2.Cancel from the inside of a trig function.
Ex: sin(2x)/2 = you CAN'T cancel the 2
Examples:
1)2sin^2 theta-1=0
2sin^2 theta=1
sin^2 theta=1/2
sin theta = +/- sq.root of 1/2
theta= sin inverse (sq. root of 1/2)
2)sin^2x-sinx=cos^2 x
sin^2x-sinx=1-sin^2x
2sin^2x-sinx-1=0
(2sin^2 x -2sinx)+(sinx-1)=0
2sinx(sinx-1)+(sinx-1)=0
(2sinx+1)(sinx-1)=0
2sinx+1=0 sinx-1=0
sinx=-1/2 sinx=1
x=sin inverse (1/2) x=sin inverse (1)
x=7Ï€/6,11Ï€/6,Ï€/2 x=Ï€/2
3)sinxtanx=3sinx
sinxtanx-3sinx=0
sinx(tanx-3)=0
sinx=0 tanx-3=0
x=sin inverse(0) x=tan inverse (3)
Answer: 0,71.565,180,251.565,360
- You would use the identities to get to the same trig function. If you can't try to make it tan or cot by sin/cos or cos/sin
- Follow the inverse rules
Things you CAN'T do!!!!!!!!!!!
- Divide by a trig function when solving to cancel.
2.Cancel from the inside of a trig function.
Ex: sin(2x)/2 = you CAN'T cancel the 2
Examples:
1)2sin^2 theta-1=0
2sin^2 theta=1
sin^2 theta=1/2
sin theta = +/- sq.root of 1/2
theta= sin inverse (sq. root of 1/2)
2)sin^2x-sinx=cos^2 x
sin^2x-sinx=1-sin^2x
2sin^2x-sinx-1=0
(2sin^2 x -2sinx)+(sinx-1)=0
2sinx(sinx-1)+(sinx-1)=0
(2sinx+1)(sinx-1)=0
2sinx+1=0 sinx-1=0
sinx=-1/2 sinx=1
x=sin inverse (1/2) x=sin inverse (1)
x=7Ï€/6,11Ï€/6,Ï€/2 x=Ï€/2
3)sinxtanx=3sinx
sinxtanx-3sinx=0
sinx(tanx-3)=0
sinx=0 tanx-3=0
x=sin inverse(0) x=tan inverse (3)
Answer: 0,71.565,180,251.565,360
Helen's Holiday Blog #1
8-1
QIII: add 180 degrees
QIV: make negative, add 360 degrees
Examples:
1) 3cos(theta) +9=7
3cos(theta)=-2
cos(theta)=-2/3
theta=cos inverse (-2/3)
=48.1897 degrees
Answer: theta= 131.810,228.190
2) 5 sec (theta)+6=0
5 sec (theta)=-6
sec(theta)= -6/5
theta= sec inverse (6/5)
theta= cos inverse (5/6)
Answer: theta= 146.443,213.557
- To solve for theta you get the trig function by itself and then take an inverse.
- An inverse has 2 answers, find where the angle is based on trig function and if the number is positive or negative.
- Steps: Take the inverse of the postive # to find the QI angle.
QIII: add 180 degrees
QIV: make negative, add 360 degrees
Examples:
1) 3cos(theta) +9=7
3cos(theta)=-2
cos(theta)=-2/3
theta=cos inverse (-2/3)
=48.1897 degrees
Answer: theta= 131.810,228.190
2) 5 sec (theta)+6=0
5 sec (theta)=-6
sec(theta)= -6/5
theta= sec inverse (6/5)
theta= cos inverse (5/6)
Answer: theta= 146.443,213.557
Helen's Holiday Blog #2
8-3
y=A sin(Bx +/- C) +/- D
or
y=A cos( Bx +/-C) +/-D
Steps:
1) Follow previous steps for period change with 5 points.
3)move ally values up or down.
Examples:
1) y= 3cos(Ï€x + 1/2)+2
amp: 3 period: 2Ï€/Ï€=2
1. o/Ï€=0 1. 0-1/2=-1/2
2. π/2/π=1/2 2. 1/2-1/2=0
3 .Ï€/Ï€=1 3. 1-1/2=1/2
4. 3Ï€/2/Ï€=3/2 4. 3/2-1/2=1
5. 2Ï€/2/Ï€=2 5. 2-1/2=3/2
Then you would graph it.
And your answer would be in a different color
y=A sin(Bx +/- C) +/- D
or
y=A cos( Bx +/-C) +/-D
Steps:
1) Follow previous steps for period change with 5 points.
- take all 5 points and divide by B
3)move ally values up or down.
Examples:
1) y= 3cos(Ï€x + 1/2)+2
amp: 3 period: 2Ï€/Ï€=2
1. o/Ï€=0 1. 0-1/2=-1/2
2. π/2/π=1/2 2. 1/2-1/2=0
3 .Ï€/Ï€=1 3. 1-1/2=1/2
4. 3Ï€/2/Ï€=3/2 4. 3/2-1/2=1
5. 2Ï€/2/Ï€=2 5. 2-1/2=3/2
Then you would graph it.
And your answer would be in a different color
Nicala's Blog
in chapter nine section 3
law of sine
you used the law of sine with non-right triangles and only when you know the angle and opposite leg's value.
formula
sin(angle)over its opposite leg= sine(angle two) over its opposite leg
you cross multiply to solve
A engineer wants to determine the distance between point A and point C. She knows that AB=25 angle A=110 and angle B=20. Find AC and BC.
first you HAVE to draw the triangle so you can solve it otherwise it will be very hard to visual what the triangle looks like.
you need to subtract the two angles away from 180.
180-110-20=50
then you but it in the formula
sin50 over 25=sin120 over AC
ACsin50=25sin20
AC=25sin20 overs sin 50
AC=11.2
law of sine
you used the law of sine with non-right triangles and only when you know the angle and opposite leg's value.
formula
sin(angle)over its opposite leg= sine(angle two) over its opposite leg
you cross multiply to solve
A engineer wants to determine the distance between point A and point C. She knows that AB=25 angle A=110 and angle B=20. Find AC and BC.
first you HAVE to draw the triangle so you can solve it otherwise it will be very hard to visual what the triangle looks like.
you need to subtract the two angles away from 180.
180-110-20=50
then you but it in the formula
sin50 over 25=sin120 over AC
ACsin50=25sin20
AC=25sin20 overs sin 50
AC=11.2
Nicala's Blog
in chapter nine section one
its the right triangle that we did in geometry.
when doing the right triangles you must remember what all the trig functions symbolize on the triangle. the easiest way to remember is to use SOHCAHTOA.
the S stands for sine and the OH- stands for what sine is on the triangle- Opposite over Hypothenuse.
its the same for the rest of it too.
CAH- cosine=adjecent over hypothenuse
TOA-tangent=opposite over adjacent
its the right triangle that we did in geometry.
when doing the right triangles you must remember what all the trig functions symbolize on the triangle. the easiest way to remember is to use SOHCAHTOA.
the S stands for sine and the OH- stands for what sine is on the triangle- Opposite over Hypothenuse.
its the same for the rest of it too.
CAH- cosine=adjecent over hypothenuse
TOA-tangent=opposite over adjacent
For the rest of the trig function, cotangent, secant, and cosecant, they are the inverse of other three trig functions i mentioned.
Cotangent is the opposite of tangent= adjacent over opposite
Secant is the opposite of cosine= hypothenuse over adjacent
and Cosecant is the opposite of sine=hypothenuse over adjacent
Hint: the hypothenuse is always the opposite of the ninety
Example
Solve for each part of the triangle.
A=35 and a=4
B=55 and b=?
C=90 and c=5
first need to draw the triangle with this information on it.
the you need to look and see what you need
since you know that this is a right triangle you should the pythagorean theorm which is a(squared)+ b(squared)=c(squared)
4(squared)+b(squared)=5(squared)
16+b=25
b=3
Nicala's Blog
in chapter seven section two
all you have to do is remember the formulas and the hints to pass this section
formulas
s=r(theta)
k-one half(r)squaredtheta
k=onehalf (r)(s)
s=arc length
r=radius
theta=central angle
k=area
word problem hints
theta=apparent size
s=diameter
r=distance between objects
Also your theta must be in radians
Example
A sector of a circle has an arc length of 8 cm and the area of 75 cm. Find its radius and the measure of its central angle.
s=8cm
k=75 cm
r=?
theta=?
using one of the formulas for area
k=one half(r)(s)
75=(1/2)(r)(8)
75=4r
then you divide four by both sides
r=18.75cm
next you use the arclength formula
s=(r)theta
8=(18.75)(theta)
theta=2.34375 rads
all you have to do is remember the formulas and the hints to pass this section
formulas
s=r(theta)
k-one half(r)squaredtheta
k=onehalf (r)(s)
s=arc length
r=radius
theta=central angle
k=area
word problem hints
theta=apparent size
s=diameter
r=distance between objects
Also your theta must be in radians
Example
A sector of a circle has an arc length of 8 cm and the area of 75 cm. Find its radius and the measure of its central angle.
s=8cm
k=75 cm
r=?
theta=?
using one of the formulas for area
k=one half(r)(s)
75=(1/2)(r)(8)
75=4r
then you divide four by both sides
r=18.75cm
next you use the arclength formula
s=(r)theta
8=(18.75)(theta)
theta=2.34375 rads
Sunday, December 26, 2010
Nathan's blog
Nothing says Merry Christmas like doing your math blogs. The first sentence of this blog is, of course, a lie. I will probably review chapter 13 for this blog.
Chapter 13-1 involved arithmetic and geometric sequences. In an arithmetic sequence, you add or subtract in order to get to the next number. In a geometric sequence you multiply or divide to get the next number.
Ex. Is the following sequnce arithmetic, geometric, or neither.
1.) 2,4,8,16,..........Geometric because you are multiplying by 2.
2.) 15,20,25,30,.......Arithmetic because you are adding by 5.
3.) 3,7,12,18,......Neither because their is not common difference.
You will need to know these formulas in order to get the next number in the sequence.
Arithmetic- tn=t1(n-1)d
Geometric- tn=t1 x r^n-1
Find the first four terms for tn=15n+7
t1=15(1)+7=22
t2=15(2)+7=37
t3=15(3)+7=52
t4=15(4)+7=67
That's all for this blog. I hope that someone learned a thing or two about chapter 13. Happy New Year (2011)!
Chapter 13-1 involved arithmetic and geometric sequences. In an arithmetic sequence, you add or subtract in order to get to the next number. In a geometric sequence you multiply or divide to get the next number.
Ex. Is the following sequnce arithmetic, geometric, or neither.
1.) 2,4,8,16,..........Geometric because you are multiplying by 2.
2.) 15,20,25,30,.......Arithmetic because you are adding by 5.
3.) 3,7,12,18,......Neither because their is not common difference.
You will need to know these formulas in order to get the next number in the sequence.
Arithmetic- tn=t1(n-1)d
Geometric- tn=t1 x r^n-1
Find the first four terms for tn=15n+7
t1=15(1)+7=22
t2=15(2)+7=37
t3=15(3)+7=52
t4=15(4)+7=67
That's all for this blog. I hope that someone learned a thing or two about chapter 13. Happy New Year (2011)!
Sunday, December 19, 2010
Nathan's week before Christmas blog
This is the first, I repeat first, of the three blogs that B-Rob demands that we do over the holidays.
In this blog, I will review chapter 7, because that was probably the easiest chapter we did all year.
The first thing that I will go over is converting degrees to radians and radians to degrees.
Convert 57 degrees to radians.
-All you do is multiply 57*Ï€/180=57Ï€/180; This reduces down to 19Ï€/60
Convert 4Ï€ to degrees.
-All you do is multiply 4Ï€*180/Ï€=720 degrees.
Now I will go over how to get co-terminal angles.
Ex. Give a positive co-terminal angle for 55.
-Now you just add 360 to 55=415 degrees
Give a positive co-terminal angle for π/2.
-You can type this into your calculator by putting 1/2+2 and then convert it to a fraction.
You are adding 2Ï€ and the final answer is:
=5Ï€/2
That is about it for this blog. Still got blog prompts to answer, so................
In this blog, I will review chapter 7, because that was probably the easiest chapter we did all year.
The first thing that I will go over is converting degrees to radians and radians to degrees.
Convert 57 degrees to radians.
-All you do is multiply 57*Ï€/180=57Ï€/180; This reduces down to 19Ï€/60
Convert 4Ï€ to degrees.
-All you do is multiply 4Ï€*180/Ï€=720 degrees.
Now I will go over how to get co-terminal angles.
Ex. Give a positive co-terminal angle for 55.
-Now you just add 360 to 55=415 degrees
Give a positive co-terminal angle for π/2.
-You can type this into your calculator by putting 1/2+2 and then convert it to a fraction.
You are adding 2Ï€ and the final answer is:
=5Ï€/2
That is about it for this blog. Still got blog prompts to answer, so................
Monday, December 13, 2010
TAYLOR'S BLOG REVIEW!!! 3 MERRY CHRISTMAS AND HAPPY NEW YEAR!!
REVIEW TIME!!
The following is part one of my notes on lesson 9.1.
For this lesson we will only be using right triangles . We will be using Sin for these problems as well and only Sin for part 1 of this lesson.
To find Sin remember this:
Sin Theta = opposite/ adjacent
(REMEMBER I DO NOT KNOW HOW TO PUT PICTURES ON THIS BLOG SO JUST DO AS I SAY AND REMEMBER THAT THE LENGTH OPPOSITE OF THE ANGLE IS THE LOWER CASE LETTER OF THAT ANGLE)
Triangle ABC is a right triangle. Use the following information to solve for it.
A=90 degrees a= 13
B= 80 degrees b=16
C= 10 degrees c=?
First, draw your triangle with the information that I gave you includes.
Sin 10= ?/13 Then, write out your equation like the formula I gave you.
c= 2.257 Finally, just multiply five on both sides, round to the third place after the decimal and you get this as your answer.
Triangle ABC is a right triangle. Use the following information to solve for it.
A=90 degrees a= 12
B= ? degrees b=2
C= ? degrees c=6
First, draw your triangle with the information that I gave you includes.
Sin C= 6/12 Then, write out your equation like the formula I gave you.
C= Sin^-1 (6/12) Then, you need to find the inverse.
C= 30 degrees Then, solve. Then, you need to make sure that the other possible angle cannot be used so turn the answer into a negative and add 180 degrees. You should get 150 degrees. Then, add 90 degrees to it and you get 240. Since this is over 180 degrees we cannot use it for triangles only have 180 degrees in them.
180 degrees -90 degrees -30 degrees= 60 degrees Finally, you just have to subtract 90 degrees and 30 degrees from 180 degrees and you will get 60 degrees for angle B.
So your answers will be C= 30 degrees and B= 60 degrees.
THAT IS ALL FOR THIS PART OF MY LESSON 9.1 PART 1 NOTES !! SO UNTIL NEXT TIME JA NE!!!! (ja ne is goodbye in Japanese)
The following is part one of my notes on lesson 9.1.
For this lesson we will only be using right triangles . We will be using Sin for these problems as well and only Sin for part 1 of this lesson.
To find Sin remember this:
Sin Theta = opposite/ adjacent
(REMEMBER I DO NOT KNOW HOW TO PUT PICTURES ON THIS BLOG SO JUST DO AS I SAY AND REMEMBER THAT THE LENGTH OPPOSITE OF THE ANGLE IS THE LOWER CASE LETTER OF THAT ANGLE)
Triangle ABC is a right triangle. Use the following information to solve for it.
A=90 degrees a= 13
B= 80 degrees b=16
C= 10 degrees c=?
First, draw your triangle with the information that I gave you includes.
Sin 10= ?/13 Then, write out your equation like the formula I gave you.
c= 2.257 Finally, just multiply five on both sides, round to the third place after the decimal and you get this as your answer.
Triangle ABC is a right triangle. Use the following information to solve for it.
A=90 degrees a= 12
B= ? degrees b=2
C= ? degrees c=6
First, draw your triangle with the information that I gave you includes.
Sin C= 6/12 Then, write out your equation like the formula I gave you.
C= Sin^-1 (6/12) Then, you need to find the inverse.
C= 30 degrees Then, solve. Then, you need to make sure that the other possible angle cannot be used so turn the answer into a negative and add 180 degrees. You should get 150 degrees. Then, add 90 degrees to it and you get 240. Since this is over 180 degrees we cannot use it for triangles only have 180 degrees in them.
180 degrees -90 degrees -30 degrees= 60 degrees Finally, you just have to subtract 90 degrees and 30 degrees from 180 degrees and you will get 60 degrees for angle B.
So your answers will be C= 30 degrees and B= 60 degrees.
THAT IS ALL FOR THIS PART OF MY LESSON 9.1 PART 1 NOTES !! SO UNTIL NEXT TIME JA NE!!!! (ja ne is goodbye in Japanese)
TAYLOR'S BLOG REVIEW!!! 2
REVIEW TIME!!
The following is part two of my notes on lesson 9.1.
For this lesson we will only be using right triangles . We will be using Cos for these problems as well and only Cos for part 2 of this lesson.
To find Cos remember this:
Cos Theta = adjacent/ hypotenuse
(REMEMBER I DO NOT KNOW HOW TO PUT PICTURES ON THIS BLOG SO JUST DO AS I SAY AND REMEMBER THAT THE LENGTH OPPOSITE OF THE ANGLE IS THE LOWER CASE LETTER OF THAT ANGLE)
Triangle ABC is a right triangle. Use the following information to solve for it.
A=90 degrees a= 7
B= 50 degrees b=?
C= 40 degrees c=2
First, draw your triangle with the information that I gave you includes.
Cos 40= ?/7 Then, write out your equation like the formula I gave you.
b= 5.362 Finally, just multiply five on both sides, round to the third place after the decimal and you get this as your answer.
Triangle ABC is a right triangle. Use the following information to solve for it.
A=90 degrees a= 12
B= ? degrees b=8
C= ? degrees c=6
First, draw your triangle with the information that I gave you includes.
Cos C= 8/12 Then, write out your equation like the formula I gave you.
C= Cos^-1 (8/12) Then, you need to find the inverse.
C= 48.190 degrees Then, solve and round to the third place after the decimal. Then, you would normally need to make sure that there is no other possible angle but since the positive form of
Cos is on the first and forth quadrant, you cannot do it for the forth quadrant is up to 360 degrees and triangles can only go up to 180 degrees so this is your only answer.
180 degrees -90 degrees - 48.190 = 41.81 degrees Finally, you just have to subtract 90 degrees and 48.190 degrees from 180 degrees and you will get 41.81 degrees for angle B.
So your answers will be C= 48.190 degrees and B= 41.81 degrees.
THAT IS ALL FOR THIS PART OF MY LESSON 9.1 PART 2 NOTES !! SO UNTIL NEXT TIME JA NE!!!! (ja ne is goodbye in Japanese)
The following is part two of my notes on lesson 9.1.
For this lesson we will only be using right triangles . We will be using Cos for these problems as well and only Cos for part 2 of this lesson.
To find Cos remember this:
Cos Theta = adjacent/ hypotenuse
(REMEMBER I DO NOT KNOW HOW TO PUT PICTURES ON THIS BLOG SO JUST DO AS I SAY AND REMEMBER THAT THE LENGTH OPPOSITE OF THE ANGLE IS THE LOWER CASE LETTER OF THAT ANGLE)
Triangle ABC is a right triangle. Use the following information to solve for it.
A=90 degrees a= 7
B= 50 degrees b=?
C= 40 degrees c=2
First, draw your triangle with the information that I gave you includes.
Cos 40= ?/7 Then, write out your equation like the formula I gave you.
b= 5.362 Finally, just multiply five on both sides, round to the third place after the decimal and you get this as your answer.
Triangle ABC is a right triangle. Use the following information to solve for it.
A=90 degrees a= 12
B= ? degrees b=8
C= ? degrees c=6
First, draw your triangle with the information that I gave you includes.
Cos C= 8/12 Then, write out your equation like the formula I gave you.
C= Cos^-1 (8/12) Then, you need to find the inverse.
C= 48.190 degrees Then, solve and round to the third place after the decimal. Then, you would normally need to make sure that there is no other possible angle but since the positive form of
Cos is on the first and forth quadrant, you cannot do it for the forth quadrant is up to 360 degrees and triangles can only go up to 180 degrees so this is your only answer.
180 degrees -90 degrees - 48.190 = 41.81 degrees Finally, you just have to subtract 90 degrees and 48.190 degrees from 180 degrees and you will get 41.81 degrees for angle B.
So your answers will be C= 48.190 degrees and B= 41.81 degrees.
THAT IS ALL FOR THIS PART OF MY LESSON 9.1 PART 2 NOTES !! SO UNTIL NEXT TIME JA NE!!!! (ja ne is goodbye in Japanese)
TAYLOR'S BLOG REVIEW!! 1
REVIEW TIME!!!!
The following is part three of my notes on lesson 9.1.
For this lesson we will only be using right triangles . We will be using Tan for these problems as well and only Tan for part 3 of this lesson.
To find Tan remember this:
Tan Theta = opposite/adjacent
(REMEMBER I DO NOT KNOW HOW TO PUT PICTURES ON THIS BLOG SO JUST DO AS I SAY AND REMEMBER THAT THE LENGTH OPPOSITE OF THE ANGLE IS THE LOWER CASE LETTER OF THAT ANGLE)
Triangle ABC is a right triangle. Use the following information to solve for it.
A=90 degrees a= 20
B= 30 degrees b=50
C= 60 degrees c=?
First, draw your triangle with the information that I gave you includes.
Tan 30= 50/? Then, write out your equation like the formula I gave you.
Tan 30 x ?= 50 Then you would multiply by ? And get this as your answer.
c= 86.603 Finally, just divide by Tan 30 both sides, round to the third place after the decimal and you get this as your answer.
Triangle ABC is a right triangle. Use the following information to solve for it.
A=90 degrees a= 7
B= ? degrees b=6
C= ? degrees c=8
First, draw your triangle with the information that I gave you includes.
Tan C= 8/6 Then, write out your equation like the formula I gave you.
C= Tan^-1 (8/6) Then, you need to find the inverse.
C= 53.130 degrees Then, solve and round to the third place after the decimal. Then, you would normally need to make sure that there is no other possible angle but since the positive form of
Tan is on the first and third quadrant, you cannot do it for the third quadrant is up to 270 degrees and triangles can only go up to 180 degrees so this is your only answer.
180 degrees -90 degrees -53.130 degrees= 86.97 degrees Finally, you just have to subtract 90 degrees and 53.130 degrees from 180 degrees and you will get 86.87 degrees for angle B.
So your answers will be C= 53.130 degrees and B= 86.87 degrees.
THAT IS ALL FOR THIS PART OF MY LESSON 9.1 PART 2 NOTES !! SO UNTIL NEXT TIME JA NE!!!! (ja ne is goodbye in Japanese)
The following is part three of my notes on lesson 9.1.
For this lesson we will only be using right triangles . We will be using Tan for these problems as well and only Tan for part 3 of this lesson.
To find Tan remember this:
Tan Theta = opposite/adjacent
(REMEMBER I DO NOT KNOW HOW TO PUT PICTURES ON THIS BLOG SO JUST DO AS I SAY AND REMEMBER THAT THE LENGTH OPPOSITE OF THE ANGLE IS THE LOWER CASE LETTER OF THAT ANGLE)
Triangle ABC is a right triangle. Use the following information to solve for it.
A=90 degrees a= 20
B= 30 degrees b=50
C= 60 degrees c=?
First, draw your triangle with the information that I gave you includes.
Tan 30= 50/? Then, write out your equation like the formula I gave you.
Tan 30 x ?= 50 Then you would multiply by ? And get this as your answer.
c= 86.603 Finally, just divide by Tan 30 both sides, round to the third place after the decimal and you get this as your answer.
Triangle ABC is a right triangle. Use the following information to solve for it.
A=90 degrees a= 7
B= ? degrees b=6
C= ? degrees c=8
First, draw your triangle with the information that I gave you includes.
Tan C= 8/6 Then, write out your equation like the formula I gave you.
C= Tan^-1 (8/6) Then, you need to find the inverse.
C= 53.130 degrees Then, solve and round to the third place after the decimal. Then, you would normally need to make sure that there is no other possible angle but since the positive form of
Tan is on the first and third quadrant, you cannot do it for the third quadrant is up to 270 degrees and triangles can only go up to 180 degrees so this is your only answer.
180 degrees -90 degrees -53.130 degrees= 86.97 degrees Finally, you just have to subtract 90 degrees and 53.130 degrees from 180 degrees and you will get 86.87 degrees for angle B.
So your answers will be C= 53.130 degrees and B= 86.87 degrees.
THAT IS ALL FOR THIS PART OF MY LESSON 9.1 PART 2 NOTES !! SO UNTIL NEXT TIME JA NE!!!! (ja ne is goodbye in Japanese)
Sunday, December 12, 2010
Dom's blog
Chapter 13-1
formulas:
Arithmetic = tn = t1+(n-1)d
Geometric = tn = t1 x r^n-1
Arithmetic = tn = t1+(n-1)d
Geometric = tn = t1 x r^n-1
Example 1:
3, 7, 11, 15, 19 has a1 = 3, d = 4,
and n = 5. The explicit formula is
an = 3 + (n – 1)·4 = 4n – 1
Example 2:
These two terms are 12 – 5 = 7 places apart, so, from the definition of a geometric sequence, I know that a12 = ( a5 )( r7 ). I can use this to solve for the value of the common ratio r:
160 = (5/4)(r7)
128 = r7
2 = r
Since a5 = ar4, then I can solve for the value of the first term a:
5/4 = a(24) = 16a
5/64 = a
well thats it
Feroz's Blog
So. tired. here is. blog. 13 -1.
formulas:
Arithmetic = tn = t1+(n-1)d
Geometric = tn = t1 x r^n-1
ex. find first 3 terms for tn = 201n + 3
t1 = 201(1) +3
201 + 3 = 204
t2 = 201(2) +3
402 + 3 = 405
t3 = 201(3) +3
603 + 3 = 606
ex. find formula for n^m term :3,7,11,15 (Arithmetic)
tn = t1 + (n-1)d
tn = 3 + (n-1)(4)
tn = 3 + 4n - 4
tn = 4n-1
*insert witty comment*
formulas:
Arithmetic = tn = t1+(n-1)d
Geometric = tn = t1 x r^n-1
ex. find first 3 terms for tn = 201n + 3
t1 = 201(1) +3
201 + 3 = 204
t2 = 201(2) +3
402 + 3 = 405
t3 = 201(3) +3
603 + 3 = 606
ex. find formula for n^m term :3,7,11,15 (Arithmetic)
tn = t1 + (n-1)d
tn = 3 + (n-1)(4)
tn = 3 + 4n - 4
tn = 4n-1
*insert witty comment*
Kaitlyn's bloggg
In this blogg, i am going to go over 10-1. This section was probably one of the easiest ones we have done so far this year because all you have to do is know the formulas and you will know how to solve the problems. you also have to make sure you know the trig chart.
Formulas: sin(A+-B)=sinAcosB+-cosAsinB
cos(A+-B)=cosAcosB-+sinAsinB
Ex: find the exact value of sin75
-sin(45+30)=sin45cos30+cos45sin30
-sin(75)=(squarerootof 2/2)(squarerootof 3/2)+(squarerootof 2/2)(1/2)
-sin(75)=(squarerootof6/4)+(squarerootof2/4)
-sin(75)=(squarerootof 6+squarerootof 2)/4 <----answerrrrrrrr!
ex: simplifyyy cos40cos20-sin40sin20
-cos(40+20)
-cos(60)
-1/2 <----answerrr!!
Formulas: sin(A+-B)=sinAcosB+-cosAsinB
cos(A+-B)=cosAcosB-+sinAsinB
Ex: find the exact value of sin75
-sin(45+30)=sin45cos30+cos45sin30
-sin(75)=(squarerootof 2/2)(squarerootof 3/2)+(squarerootof 2/2)(1/2)
-sin(75)=(squarerootof6/4)+(squarerootof2/4)
-sin(75)=(squarerootof 6+squarerootof 2)/4 <----answerrrrrrrr!
ex: simplifyyy cos40cos20-sin40sin20
-cos(40+20)
-cos(60)
-1/2 <----answerrr!!
Nathan's Blog
For this blog, I will be going over section 9-3 and 9-4, which deals with the Law of Sines and the Law of Cosines.
Law of Sines:
-Only used with non-right triangles.
-Only use when you have an angle and opposite leg value you know.
Formula:
sin(angle)/opposite leg=sin(angle 2)/opposite leg 2
*Cross multiply to solve.
Ex. A civil engineer wants to determine the distance from points A and B to an inaccessible point C. From direct measurement, the engineer knows that AB=25m, angle A=110 degrees and angle B=20 degrees. Find AC and BC.
sin50/25=sin110/BC
BC=25sin110/sin50
BC=30.667 m
sin50/25=sin20/AC
AC=25sin20/sin50
AC=11.162 m
Law of Cosines:
-Used when you don't have a pair in a non-right triangle.
Ex. Suppose that two sides of a triangle have lengths of 3 cm and 7 cm and that the angle between them measures 130 degrees. Find the third side of the triangle using the Law of Cosines.
p^2=3^2+7^2-2(3)(7)cos130
p=√3^2+7^2-2(3)(7)cos130
p=9.22 cm
Law of Sines:
-Only used with non-right triangles.
-Only use when you have an angle and opposite leg value you know.
Formula:
sin(angle)/opposite leg=sin(angle 2)/opposite leg 2
*Cross multiply to solve.
Ex. A civil engineer wants to determine the distance from points A and B to an inaccessible point C. From direct measurement, the engineer knows that AB=25m, angle A=110 degrees and angle B=20 degrees. Find AC and BC.
sin50/25=sin110/BC
BC=25sin110/sin50
BC=30.667 m
sin50/25=sin20/AC
AC=25sin20/sin50
AC=11.162 m
Law of Cosines:
-Used when you don't have a pair in a non-right triangle.
Ex. Suppose that two sides of a triangle have lengths of 3 cm and 7 cm and that the angle between them measures 130 degrees. Find the third side of the triangle using the Law of Cosines.
p^2=3^2+7^2-2(3)(7)cos130
p=√3^2+7^2-2(3)(7)cos130
p=9.22 cm
lawrence's blog
on this blog i will be going over chapter 9-1. it is one of the easiest chapters. you just have to know the formulas and you will be fine.
Right triangle:
1. hypotenuse is opposite of the right angle
2. area= 1/2 base times height
3. SOH CAH TOA
-sin theta =opposite/hypotenuse
-csc theta=hypotenuse/opposite
-cos theta=adjacent/hypotenuse
-sec theta=hypotenuse/adjacent
-tan theta=opposite/adjacent
-cot theta=adjacent/opposite
Ex: In triangle ABC, angle A=28 degrees, angle C=90 degrees, side a=49, what is side b?
tan28degrees=40/b
btan28degrees=40
b=75.229
Ex: In triangle ABC, angle A=50 degrees, angle C= 90 degrees, side c=10, what is side a?
sin 50 degrees= a/10
=10 sin 50 degrees= a
a= 7.660
Law of Sines:
sinA/a = sinB/b = sinC/c
Law of sines is used when you know pairs in non-right triangles
To use the law of sines, you are setting up a proportion
Ex: in triangle ABC, angle A=115 degrees, side a=123, side b=16, what is angle B?
sin115 degrees/123 = sin theta/16
123sin theta=16sin115 degrees
sin theta=16sin115 degrees/123
theta =sin inverse (16sin115 degrees/123)
theta =6.771 degrees
Ex: in triangle ABC, angle A=30 degrees, angle B=15 degrees, side a=4, what is side B?
sin 30 degrees/4 = sin 15 degrees/x
x sin 30 degrees = 4 sin 15 degrees
x= 4 sin 15 degrees/ sin 30 degrees
=4 sin 15 degrees/ 1/2
x= 2.071
Right triangle:
1. hypotenuse is opposite of the right angle
2. area= 1/2 base times height
3. SOH CAH TOA
-sin theta =opposite/hypotenuse
-csc theta=hypotenuse/opposite
-cos theta=adjacent/hypotenuse
-sec theta=hypotenuse/adjacent
-tan theta=opposite/adjacent
-cot theta=adjacent/opposite
Ex: In triangle ABC, angle A=28 degrees, angle C=90 degrees, side a=49, what is side b?
tan28degrees=40/b
btan28degrees=40
b=75.229
Ex: In triangle ABC, angle A=50 degrees, angle C= 90 degrees, side c=10, what is side a?
sin 50 degrees= a/10
=10 sin 50 degrees= a
a= 7.660
Law of Sines:
sinA/a = sinB/b = sinC/c
Law of sines is used when you know pairs in non-right triangles
To use the law of sines, you are setting up a proportion
Ex: in triangle ABC, angle A=115 degrees, side a=123, side b=16, what is angle B?
sin115 degrees/123 = sin theta/16
123sin theta=16sin115 degrees
sin theta=16sin115 degrees/123
theta =sin inverse (16sin115 degrees/123)
theta =6.771 degrees
Ex: in triangle ABC, angle A=30 degrees, angle B=15 degrees, side a=4, what is side B?
sin 30 degrees/4 = sin 15 degrees/x
x sin 30 degrees = 4 sin 15 degrees
x= 4 sin 15 degrees/ sin 30 degrees
=4 sin 15 degrees/ 1/2
x= 2.071
Holiday Blog Prompt 3
What concept(s) from Algebra II did you most struggle with? Why? We will be revisiting many Algebra II concepts this semester, what do you plan to do differently to master them?
Holiday Blog Prompt 2
Explain the different types of polar graphs and their equations. Find sites with images of each. **Hint use google and search the images tab. Images can be pasted into blogger and include the link. You should have different sites.
Holiday Blog Prompt 1
Come up with your own trig graphing problem and walk someone through it step by step explaining each step and formula as though they have never taken Advanced Math.
Friday, December 10, 2010
Taylor's 15th blog
These are my notes from Ch. 13.2.
A recursive means define in terms of what came before.
This is what you need to remember:
tn-1 previous term
tn-2 2 terms back
tn-3 ....
Ok let's do some examples.
Find the third, fourth, and fifth terms given. t1=9 tn=5tn-1+6
t1=5 (9)+6= 51 First replace tn-1 with the previous terms. You will keep doing this until you get all three answers. Remember to use your answers and not t1!!
t2= 5 (51) + 6 = 261
t3= 5 (261) + 6= 1311
t4= 5 (1311) + 6 = 6561
t5= 5 (6561) + 6 = 32811
So your answers will be 1311, 6561, and 32811.
Now let's try finding the recursive definition for 10, 12, 14, 16...
First, you need to figure out how much is in-between each number and how to get to the next number with the previous numbers help.
Since there is 2 in between each number your answer or equation will be tn-1+2.
These are the rest of my notes on Ch. 13.2. So UNTIL MONDAY SEE YALL!!!
A recursive means define in terms of what came before.
This is what you need to remember:
tn-1 previous term
tn-2 2 terms back
tn-3 ....
Ok let's do some examples.
Find the third, fourth, and fifth terms given. t1=9 tn=5tn-1+6
t1=5 (9)+6= 51 First replace tn-1 with the previous terms. You will keep doing this until you get all three answers. Remember to use your answers and not t1!!
t2= 5 (51) + 6 = 261
t3= 5 (261) + 6= 1311
t4= 5 (1311) + 6 = 6561
t5= 5 (6561) + 6 = 32811
So your answers will be 1311, 6561, and 32811.
Now let's try finding the recursive definition for 10, 12, 14, 16...
First, you need to figure out how much is in-between each number and how to get to the next number with the previous numbers help.
Since there is 2 in between each number your answer or equation will be tn-1+2.
These are the rest of my notes on Ch. 13.2. So UNTIL MONDAY SEE YALL!!!
Nicala's Blog
chapter 11 section 1
convert to rectangular use the following formulas
x=r cos theta
y=r sin theta
y=r sin theta
and convert to polar use the formulas
r=squared root x squared plus y squared
Example: Give the polar point for (3,5)
r=squared root 3 squared plus 5 squared= squared root of 34
tan theta=three over 5
theta= tan inverse(three over five)=31
put it on the coordinating plane
theta=211
so the answer is squared root of 34, 31 and the negative squared root of 34,211
r=squared root x squared plus y squared
Example: Give the polar point for (3,5)
r=squared root 3 squared plus 5 squared= squared root of 34
tan theta=three over 5
theta= tan inverse(three over five)=31
put it on the coordinating plane
theta=211
so the answer is squared root of 34, 31 and the negative squared root of 34,211
Monday, December 6, 2010
Week 7 Prompt
What have you personally accomplished in math this semester? Do you feel more confident? What will you do differently next semester?
Sunday, December 5, 2010
Feroz's ASDJEOIAHFOIADS;ADIFSJ
JUST FINISHED THAT MATH PROJECT FOR BROB AND I CAN'T STAY AWAKE IF MY LIFE DEPENDED ON IT.
Section 10-1
When everyone and their mother went over this section I chose not to, just in case I had no idea what to do a blog on, and what do ya know.
Formulas:
Sin: (alpha +/- beta) = sinAcosB +/- cosAsinB
Cos: (alpha -/+ beta) = cosAcosB +/- sinAsinB
ex. Find the exact value of sin 75
sin45cos30 + cos45sin30
= sqrt(2/2)(1/2) + sqrt(2/2)sqrt(3/2)
= sqrt(6) + sqrt(2)
----------------------
4
*yawn*
Section 10-1
When everyone and their mother went over this section I chose not to, just in case I had no idea what to do a blog on, and what do ya know.
Formulas:
Sin: (alpha +/- beta) = sinAcosB +/- cosAsinB
Cos: (alpha -/+ beta) = cosAcosB +/- sinAsinB
ex. Find the exact value of sin 75
sin45cos30 + cos45sin30
= sqrt(2/2)(1/2) + sqrt(2/2)sqrt(3/2)
= sqrt(6) + sqrt(2)
----------------------
4
*yawn*
Ch. 13-1
Sequences
arithmetic-add or subtract
geometric-multiply or divide
the variable "d" stands for the +/- term in arithmetic sequences
the variable "r" stands for the # of term in geometric sequences
ex: a. 4,8,12,16,... arithmetic (adding 4)
b. 9, 27/2, 81/4,...geometric (multiply by 3/2)
c. 4,7,11,16,22,...neither
Arithmetic: tn=t1+(n-1)d
Geometric: tn=t1*r^n-1
ex. find formula for n^m term :3,7,11,15
tn=t1+(n-1)d
tn=3+(n-1)(4)
tn=3+4n-4
tn=4n-1
arithmetic-add or subtract
geometric-multiply or divide
the variable "d" stands for the +/- term in arithmetic sequences
the variable "r" stands for the # of term in geometric sequences
ex: a. 4,8,12,16,... arithmetic (adding 4)
b. 9, 27/2, 81/4,...geometric (multiply by 3/2)
c. 4,7,11,16,22,...neither
Arithmetic: tn=t1+(n-1)d
Geometric: tn=t1*r^n-1
ex. find formula for n^m term :3,7,11,15
tn=t1+(n-1)d
tn=3+(n-1)(4)
tn=3+4n-4
tn=4n-1
What did we learn this week?
Problems we have to face,
and problems we'll overcome.
What ever choice you make, you'll have to live with it,
because what's done is done.
This week we went over things,
that some eyes have already seen.
Yet, those eyes still learned,
many many things.
Holiday time is here,
and love in all in the air.
Finals our coming soon,
and in time its going to be chaos everywhere.
But minus the hard thinking,
I'm blessed to be here.
Senor year is almost over,
so adulthood is near.
Problems we'll have to face,
and problems we'll overcome.
Just think before you react,
and that problem will be over with and done.
Enjoy
Good luck on testing everyone.
and problems we'll overcome.
What ever choice you make, you'll have to live with it,
because what's done is done.
This week we went over things,
that some eyes have already seen.
Yet, those eyes still learned,
many many things.
Holiday time is here,
and love in all in the air.
Finals our coming soon,
and in time its going to be chaos everywhere.
But minus the hard thinking,
I'm blessed to be here.
Senor year is almost over,
so adulthood is near.
Problems we'll have to face,
and problems we'll overcome.
Just think before you react,
and that problem will be over with and done.
Enjoy
Good luck on testing everyone.
Dylan's blog... :/
chp. 13-1 Sequences
-a list of numbers
arithmetic-add same # to every term
geometric-multiply by same # every term
the variable "d" stands for the +/- term in arithmetic sequences
the variable "r" stands for the *// term in geometric sequences
Ex. of sequences
a. 4,8,12,16,... arithmetic (adding 4)
b. 9, 27/2, 81/4,...geometric (multiply by 3/2)
c. 4,7,11,16,22,...neither
To find the n^m term of a sequence
Arithmetic: tn=t1+(n-1)d
Geometric: tn=t1*r^n-1
ex. find first 3 terms for tn=4n-3
t1=4(1)-3=1
t2=4(2)-5=5
t3=4(3)-3=9
ex. find formula for n^m term :3,7,11,15
tn=t1+(n-1)d
tn=3+(n-1)(4)
tn=3+4n-4
tn=4n-1
-a list of numbers
arithmetic-add same # to every term
geometric-multiply by same # every term
the variable "d" stands for the +/- term in arithmetic sequences
the variable "r" stands for the *// term in geometric sequences
Ex. of sequences
a. 4,8,12,16,... arithmetic (adding 4)
b. 9, 27/2, 81/4,...geometric (multiply by 3/2)
c. 4,7,11,16,22,...neither
To find the n^m term of a sequence
Arithmetic: tn=t1+(n-1)d
Geometric: tn=t1*r^n-1
ex. find first 3 terms for tn=4n-3
t1=4(1)-3=1
t2=4(2)-5=5
t3=4(3)-3=9
ex. find formula for n^m term :3,7,11,15
tn=t1+(n-1)d
tn=3+(n-1)(4)
tn=3+4n-4
tn=4n-1
toriss.
Tori here for the last blog before exams. I am extremely worried about this exam because i dont remember anyyyythingg from chapter 8. its gonna be really hard to pass but i hope i wont be the only one in the room taking the exam.
I'm gonna go over polar/rectangular: i wish the entire exam would be this stufff!
here are your formulas:
polar:(r, theta)r =+/-squareroot of x^ 2 + y^2::tan theta=inverse of y/x
rectangular:(x,y)x=rcosthetay=rsintheta
give the polar point for (3,4)
since the second number is not a degree, you know its in rectangular
so to convert to polar, you look at your formulas
r = squareroot of 3^2 +4^2=square root of 25 = +/-5
so r=+/-5theta=tan inverse of 4/3 =53.130degrees
so your point in polar would be (5, 53.130) and (-5, 233.130) because there are two quadrants where tan is positive
To plot, you just take your coordinate plane, go the regular for the x and for the degree, you just imagine your unit circle and go the approximate amount of degrees around. and for the other point, you just mirror it.
blah blah blahh. notes notes noteess and there ya go.
I'm gonna go over polar/rectangular: i wish the entire exam would be this stufff!
here are your formulas:
polar:(r, theta)r =+/-squareroot of x^ 2 + y^2::tan theta=inverse of y/x
rectangular:(x,y)x=rcosthetay=rsintheta
give the polar point for (3,4)
since the second number is not a degree, you know its in rectangular
so to convert to polar, you look at your formulas
r = squareroot of 3^2 +4^2=square root of 25 = +/-5
so r=+/-5theta=tan inverse of 4/3 =53.130degrees
so your point in polar would be (5, 53.130) and (-5, 233.130) because there are two quadrants where tan is positive
To plot, you just take your coordinate plane, go the regular for the x and for the degree, you just imagine your unit circle and go the approximate amount of degrees around. and for the other point, you just mirror it.
blah blah blahh. notes notes noteess and there ya go.
Kaitlynnnnnn's blogggg
In this blog, i am going to go overr what we learned in chapterr 13.
Arithmetic: a sequence of numbers in which you have to add or subtract to each number to get to the next number.
Geometric: a sequence of numbers in which you have to multiply or divide each number to get to the next number.
Ex: 4,6,8,10... find the next 2 terms.
-12,14 ---because you have to add 2 to each numberrr, this sequence is arithmatic.
Ex: 3,9,27....
-81,243 ---because you have to multiply by 3 to each number this sequence is geometric.
Limitssss!
1)if top exponent is greater than the bottom, the limit is infinity.
2)if top exponent is less than the bottom, the limit is 0.
3)if top exponent is equal to the bottom, the limit is the coefficient of the top and bottom.
Ex: 2x^2-1/2x^3
The limit is 0 because the top exponent is less than the bottom.
Arithmetic: a sequence of numbers in which you have to add or subtract to each number to get to the next number.
Geometric: a sequence of numbers in which you have to multiply or divide each number to get to the next number.
Ex: 4,6,8,10... find the next 2 terms.
-12,14 ---because you have to add 2 to each numberrr, this sequence is arithmatic.
Ex: 3,9,27....
-81,243 ---because you have to multiply by 3 to each number this sequence is geometric.
Limitssss!
1)if top exponent is greater than the bottom, the limit is infinity.
2)if top exponent is less than the bottom, the limit is 0.
3)if top exponent is equal to the bottom, the limit is the coefficient of the top and bottom.
Ex: 2x^2-1/2x^3
The limit is 0 because the top exponent is less than the bottom.
Helen's Bog
Chapter 13
Definitions:
Arithmetic Sequence- when you add or subtract to get the next term in the pattern.
Geometric Sequence- when you divide or multiply to get the next term in the pattern.
Examples:
1) Find the next four term 5,11,17,23,....
The sequence is arithmetic because you would add 6 to get the next term.
5+6=11
11+6=17
17+6=23
23+6=29
29+6=35
35+6=41
41+6=47
Answer : 29,35,41,47
2) Find the next four terms 2,6,18,54
The sequence is geometric because you would multiply by 5 to get the next four terms.
2x3=6
6x3=18
18x3=54
54x3=162
162x3=486
486x3=1458
1458x3=4374
Answer: 162,486,1458,4374
Limits
1) If the top and bottom exponent is equal , then the answer would be the coefficients.
2)If the top exponent is greater than the bottom, then the answer would be infinite.
3)If the top exponent is lesser than the bottom, then the answer would be zero.
Definitions:
Arithmetic Sequence- when you add or subtract to get the next term in the pattern.
Geometric Sequence- when you divide or multiply to get the next term in the pattern.
Examples:
1) Find the next four term 5,11,17,23,....
The sequence is arithmetic because you would add 6 to get the next term.
5+6=11
11+6=17
17+6=23
23+6=29
29+6=35
35+6=41
41+6=47
Answer : 29,35,41,47
2) Find the next four terms 2,6,18,54
The sequence is geometric because you would multiply by 5 to get the next four terms.
2x3=6
6x3=18
18x3=54
54x3=162
162x3=486
486x3=1458
1458x3=4374
Answer: 162,486,1458,4374
Limits
1) If the top and bottom exponent is equal , then the answer would be the coefficients.
2)If the top exponent is greater than the bottom, then the answer would be infinite.
3)If the top exponent is lesser than the bottom, then the answer would be zero.
Lawrence's blog
in this blog imma talk about section 9-2. this section is trying to get you to find the area of a right triangle or a non right triangle and it is two very simple formulas.
first formula
1/2 bh only use this for a right triangle
second formula
1/2 bh sin(angleb/w) only use this formula for a non right triangle.
those are the two formulas now you pretty much just plug the numbers in.
ex. two sides of a triangle have lengths 7cm & 4cm. the angle between the sides measure 73 degrees. find the area of the triangle.
you use a=1/2 ab sin theta
then you just plug in the numbers to that formula
a=1/2 (7)(4) sin73 degrees
a=14sin73 degrees
a=13.399 cm squared. (final answer)
that is what i learned in section 9-2. it is very easy as long as you know the formulas and you just plug the numbers into them. that is pretty much this whole chapter. thought i would go over this since we are in the exam review and this will be our next packet to do.
first formula
1/2 bh only use this for a right triangle
second formula
1/2 bh sin(angleb/w) only use this formula for a non right triangle.
those are the two formulas now you pretty much just plug the numbers in.
ex. two sides of a triangle have lengths 7cm & 4cm. the angle between the sides measure 73 degrees. find the area of the triangle.
you use a=1/2 ab sin theta
then you just plug in the numbers to that formula
a=1/2 (7)(4) sin73 degrees
a=14sin73 degrees
a=13.399 cm squared. (final answer)
that is what i learned in section 9-2. it is very easy as long as you know the formulas and you just plug the numbers into them. that is pretty much this whole chapter. thought i would go over this since we are in the exam review and this will be our next packet to do.
Nathan's Blog
I am just going to give a review for chapter 13 for this blog.
Section 1 dealt with sequences. An arithmetic sequence is when you add or subtract to get to the next number. A geometric sequence is when you multiply or divide to get to the next number.
Ex. 5,7,9,11.... This sequence is arithmetic because you added 2.
3,6,12,24.... This sequence is geometric because you multiplied by 2.
Now, on to limits. This was a very easy section to do.
1.) If the top exponent is equal to the bottom exponent, then you take the coefficients.
2.) If the top exponent is greater than the bottom exponent, the answer is +- infinite.
3.) If the top exponent is less than the bottom exponent, the answer is 0.
n^2/10n^3=0 (2 is less than 3)
2n^4/n^4=2 (same exponent; coefficient is 2/1)
5n^7/3n^2=Infinite (7 is greater than 2)
Now we are done with chapter 13. This chapter was easier than some of the previous ones.
Section 1 dealt with sequences. An arithmetic sequence is when you add or subtract to get to the next number. A geometric sequence is when you multiply or divide to get to the next number.
Ex. 5,7,9,11.... This sequence is arithmetic because you added 2.
3,6,12,24.... This sequence is geometric because you multiplied by 2.
Now, on to limits. This was a very easy section to do.
1.) If the top exponent is equal to the bottom exponent, then you take the coefficients.
2.) If the top exponent is greater than the bottom exponent, the answer is +- infinite.
3.) If the top exponent is less than the bottom exponent, the answer is 0.
n^2/10n^3=0 (2 is less than 3)
2n^4/n^4=2 (same exponent; coefficient is 2/1)
5n^7/3n^2=Infinite (7 is greater than 2)
Now we are done with chapter 13. This chapter was easier than some of the previous ones.
Friday, December 3, 2010
Taylor's 14th(I Think XD) Blog
This is the rest of my notes from Ch. 13.1.
A sequence is a set of numbers. Today we will learn a bit about geometric sequence which is when you multiply the same number to every term.
This is the formula:
tn= t1 x r ^n-1
r is the number that is being multiplied or divided. The n and one by the t are small to.
Ok let's do some examples.
Find the first 4 terms of this equation tn= 6 x 5^ n-1
t1=6 x 5^ 1-1 First replace n with 1,2,3,and 4 and solve the equation.
t2=6 x 5^ 2-1
t3=6 x 5^ 3-1
t4=6 x 5^ 4-1
t1= 6 Then you have your answers.
t2= 30
t3= 150
t4= 750
4,8,,16,32.... find the formula for the nth term.
4,8,,16,32.... First you need to figure out what is being multiplied or divided from each number. In this case it is multiplying by 2.
tn= t1 x r ^n-1Then write out your equation.
tn= 4 x2 ^n-1 Then fill out the parts that you know and since you cannot do anything else to this equation this is your answer.
These are the rest of my notes on Ch. 13.1. So UN TIL MONDAY SEE YALL!!!
A sequence is a set of numbers. Today we will learn a bit about geometric sequence which is when you multiply the same number to every term.
This is the formula:
tn= t1 x r ^n-1
r is the number that is being multiplied or divided. The n and one by the t are small to.
Ok let's do some examples.
Find the first 4 terms of this equation tn= 6 x 5^ n-1
t1=6 x 5^ 1-1 First replace n with 1,2,3,and 4 and solve the equation.
t2=6 x 5^ 2-1
t3=6 x 5^ 3-1
t4=6 x 5^ 4-1
t1= 6 Then you have your answers.
t2= 30
t3= 150
t4= 750
4,8,,16,32.... find the formula for the nth term.
4,8,,16,32.... First you need to figure out what is being multiplied or divided from each number. In this case it is multiplying by 2.
tn= t1 x r ^n-1Then write out your equation.
tn= 4 x2 ^n-1 Then fill out the parts that you know and since you cannot do anything else to this equation this is your answer.
These are the rest of my notes on Ch. 13.1. So UN TIL MONDAY SEE YALL!!!
Nicala's Blog
In chapter 10 section1
we can not use calculators, and we needed to know these two formulas:
cos(alpha+-beta)=cos(alpha)cos(beta)-+sin(alpha)sin(beta)
sin(alpha+-beta)=sin(alpha)cos(beta)+-cos(alpha)sin(beta)
we can not use calculators, and we needed to know these two formulas:
cos(alpha+-beta)=cos(alpha)cos(beta)-+sin(alpha)sin(beta)
sin(alpha+-beta)=sin(alpha)cos(beta)+-cos(alpha)sin(beta)
Ex. Find the exact value of sin 75.
sin(45+30)=sin45cos30+cos45sin30
squared root of 2 over 2 times one half plus squared root of 2 over 2 times squared root of 3 over 2=squared root of two plus six over four
Charlie's..
This week in advanced math we did chapter 13 (i think) review.
we also started to review for our mid - term exam.
To do this we did our old chapter 7 multiple choice test, our old chapter 7 free response test, our old chapter 8 multiple choice test, and our old chapter 8 free response test. Next week we'll do the rest of our old test.
For my blog, though, i"m going to go over chapter 13 stuff.
chapter 13 had to do with sequences and such.
one was arithmetic sequences.
you use the formula:
t*n = t*1 + (n-1) d
example:
#1:
4, 12, 20, 28
t*n = 4 + (n-1) 8
= 4 + 8n -8
= -4 + 8n
#2:
3, 7, 11, 15, 19
t*n = 3 + (n-1) 4
= 3 + 4n-4
= -1 + 4n
we also started to review for our mid - term exam.
To do this we did our old chapter 7 multiple choice test, our old chapter 7 free response test, our old chapter 8 multiple choice test, and our old chapter 8 free response test. Next week we'll do the rest of our old test.
For my blog, though, i"m going to go over chapter 13 stuff.
chapter 13 had to do with sequences and such.
one was arithmetic sequences.
you use the formula:
t*n = t*1 + (n-1) d
example:
#1:
4, 12, 20, 28
t*n = 4 + (n-1) 8
= 4 + 8n -8
= -4 + 8n
#2:
3, 7, 11, 15, 19
t*n = 3 + (n-1) 4
= 3 + 4n-4
= -1 + 4n
Monday, November 29, 2010
Week 6 Prompt
How do you determine if a sequence is arithmetic or geometric? What are the rules for finding limits? Give examples of each.
Sunday, November 28, 2010
another freebie!
another freebie! haha, what else do i remeber that was easy....
these formula's, i'm pretty sure they are for when you have an angle that they want the sin or cos of but its not on the trig chart but its the sum or difference of two angles that ARE on the trig chart.
its all about sin and cos
sin(alpha +/- beta) = sinalphacosbeta+/-cosalphasinbeta
cos(alpha +/-beta) = cosalphacosbeta -/+sinalphasinbeta
they could as you to find the exact value of cos 75
and B-rob told us that you use the cos formula above and use two trig chart angles that add or subtract to give you the anlge they want.
so you know that 45 and 30 are both trig chart angles and they ADD to give you 75, so you go to your cos formula and plug in the first part for addition, then you can look further down the formula and see that if its addtion in the front, that it'll be subtraction in the back.
you plug it in and get
cos45cos30-sin45sin30
so you use your brain the remeber the trig chart and you know this
square root of two over 2 * 1/2 - square root of 3 over 2 * square root of 2 over 2which simplifies to square root of 2 + square root of 6 all over 4
but if they give you something like this
sin15cos15-cos15sin15
you can know from the formula and from the sign inbetween that you have to ADD these two angles, which is the other side of the formula.
so you get sin 30
which is 1/2
waaaaaahoooooooo
holidays are overrr:(
these formula's, i'm pretty sure they are for when you have an angle that they want the sin or cos of but its not on the trig chart but its the sum or difference of two angles that ARE on the trig chart.
its all about sin and cos
sin(alpha +/- beta) = sinalphacosbeta+/-cosalphasinbeta
cos(alpha +/-beta) = cosalphacosbeta -/+sinalphasinbeta
they could as you to find the exact value of cos 75
and B-rob told us that you use the cos formula above and use two trig chart angles that add or subtract to give you the anlge they want.
so you know that 45 and 30 are both trig chart angles and they ADD to give you 75, so you go to your cos formula and plug in the first part for addition, then you can look further down the formula and see that if its addtion in the front, that it'll be subtraction in the back.
you plug it in and get
cos45cos30-sin45sin30
so you use your brain the remeber the trig chart and you know this
square root of two over 2 * 1/2 - square root of 3 over 2 * square root of 2 over 2which simplifies to square root of 2 + square root of 6 all over 4
but if they give you something like this
sin15cos15-cos15sin15
you can know from the formula and from the sign inbetween that you have to ADD these two angles, which is the other side of the formula.
so you get sin 30
which is 1/2
waaaaaahoooooooo
holidays are overrr:(
hmm maryy.
so we do get two freebie blogs to go over whhhaatever we feel like correct?
well i'm gonna go over the easiest stuff i can think of.
sooo SOHCAHTOA everyone!
well i'm gonna go over the easiest stuff i can think of.
sooo SOHCAHTOA everyone!
sin(theta)=opposite/hypotenuse
cos(theta)=adjacent/hypotenuse
tan(theta)=opposite/adjacent
you can find all the parts of the triangle with just the ones on top, but if they ask for something specific like csc, sec, or cot,you go by this
csc(theta)=hypotenuse/opposite
sec(theta)=hypotenuse/adjacent
cot(theta)=adjacent/opposite
this will only work if you are given a RIGHT triangle, where the hypotenuse will be opposite of the right angle.
When you are given an angle and the right angle, you will go and find the other angle inside the triangle, but you should still always use the one you are given incase you have made a mistake, because if you made a mistake on that, your whole problem will be WRONG!
Example.
A=90 C=65 and a=18
you draw your triangle and use 180-90-65, to get your B angle.
then you use cos to find c
and you use sin, to get b
makeup blogggggg
this blog is from the week i missed because of swimming
section 13-2
this section is just about recorsive numbers in a sequence
tn-1 means previous term
tn-2 means two terms back
tn-3 means three terms back
recorsive means define in terms of what came before
find the 2nd and 3rd terms
they give you a starting number and a sequence
t1=7 tn=tn-1+1
brob said you look at tn-1 as the term before and not as anything minus 1
so you take your first number and you plug it in
you will get tn=4(7)+1
then you get 29
so then you plug that number in to get the next one
so you get tn=4(29)+1 which equals 117
you do that to find infinite numbers in a recorsive sequence
example number 2
they could ask you what was asked above or they could give you the numbers and ask you to write a recorsive definition. which means a formula!
so if they gave you 4,7,10,13
you can see that they are adding three
soo tn=tn-1=3
and you use that formula to find other numbers
Dom's blog
Chapter 13- 4
Limits
limits have rules:
If the top exponent is equal to the bottom exponent the answer is coefficient
If the top exponent is greater than the bottom exponent answer is +/- infinite
If the top exponent is less than the bottom exponent answer is 0
limits have rules:
If the top exponent is equal to the bottom exponent the answer is coefficient
If the top exponent is greater than the bottom exponent answer is +/- infinite
If the top exponent is less than the bottom exponent answer is 0
The examples are not cooperating so im moving on.
Chapter 7-1
Converting to Radians
This is very easy
This is very easy
formula to radians: pi/180
formula to degrees: 180/pi
ex. 2pi degrees
2 x 180 = 360 degrees (just get rid of the pi)
ex 2. 540 degrees to radians
540/180 = 3pi
formula to degrees: 180/pi
ex. 2pi degrees
2 x 180 = 360 degrees (just get rid of the pi)
ex 2. 540 degrees to radians
540/180 = 3pi
dylan's back to school blog...
I'm going to go over 13-2 Recursive Definitions
An example of recursive definitions(meaning pre-term):
tn-1
tn-2
tn-3
...
ex. Find the 3rd, 4th, and 5th terms given:
t1=7, tn=4tn-1+1
t2=4(7)+1
=29
t3=4(29)+1
=117
t4=4(117)+1
=469
t5=4(469)+1
=1877
ex. Give a recursive def. for
4,7,10,13
tn=tn-1+3
An example of recursive definitions(meaning pre-term):
tn-1
tn-2
tn-3
...
ex. Find the 3rd, 4th, and 5th terms given:
t1=7, tn=4tn-1+1
t2=4(7)+1
=29
t3=4(29)+1
=117
t4=4(117)+1
=469
t5=4(469)+1
=1877
ex. Give a recursive def. for
4,7,10,13
tn=tn-1+3
Helen's Blog
Section 7-1
Converting:
convert degrees to rads: x pi/180
convert rads to degrees: x 180/pi
Section 10-1
Formulas:
sin(A+/-B)=sinA cosB+/ -cosA sinB
cos(A+/-B)=cosA cosB-/+sinA sinB
Examples:
1) Find the exact value of cos 75
cos(45+30)=cos 45 cos 30+ sin 45 sin 30
= (sq. root 2/2)(sq. root 3/2) + (sq. root 2/2)(1/2)
= Sq. root 6 + sq.root 2/4
2)simplify: cos 90 cos 45+sin 90 sin 45
= cos(90-45)
=cos 45
=sq. root 2/2
Converting:
convert degrees to rads: x pi/180
convert rads to degrees: x 180/pi
Section 10-1
Formulas:
sin(A+/-B)=sinA cosB+/ -cosA sinB
cos(A+/-B)=cosA cosB-/+sinA sinB
Examples:
1) Find the exact value of cos 75
cos(45+30)=cos 45 cos 30+ sin 45 sin 30
= (sq. root 2/2)(sq. root 3/2) + (sq. root 2/2)(1/2)
= Sq. root 6 + sq.root 2/4
2)simplify: cos 90 cos 45+sin 90 sin 45
= cos(90-45)
=cos 45
=sq. root 2/2
kaitlyn's bloggggg
In this blogg i will go overrrr chapter 10 section 1 since it was realllyyyyyyy easyy and i actually understood it prettyy well. All we really needed to know in this section was our two formulas and you also had to make sure you knew your trig chart prettyy well toooo.
Formulas: sin(A+-B)=sinAcosB+-cosAsinB
cos(A+-B)=cosAcosB-+sinAcosB
Exampleeeee: Find the exact value of sin75
-sin(45+30)=sin45cos30+cos45sin30
-sin(75)=(squarerootof 2/2)(squarerootof 3/2)+(squarerootof 2/2)(1/2)
-sin(75)=(squarrootof6/4)+(squarerootof 2/4)
-sin(75)=(squarerootof 6)+(squarerootof 2)/4 <-----answerrrrr!
Exampleeeeee: Simplifyyy cos60cos30-sin60sin30
-cos(60+30)
-cos(90)
- 1 <---answerrrr!!!
Well thats myy bloggg for this weekendddddd, hope everyone had a greatt thanksgivingg holidayy!!!! (: (:
Formulas: sin(A+-B)=sinAcosB+-cosAsinB
cos(A+-B)=cosAcosB-+sinAcosB
Exampleeeee: Find the exact value of sin75
-sin(45+30)=sin45cos30+cos45sin30
-sin(75)=(squarerootof 2/2)(squarerootof 3/2)+(squarerootof 2/2)(1/2)
-sin(75)=(squarrootof6/4)+(squarerootof 2/4)
-sin(75)=(squarerootof 6)+(squarerootof 2)/4 <-----answerrrrr!
Exampleeeeee: Simplifyyy cos60cos30-sin60sin30
-cos(60+30)
-cos(90)
- 1 <---answerrrr!!!
Well thats myy bloggg for this weekendddddd, hope everyone had a greatt thanksgivingg holidayy!!!! (: (:
tori michelleee.
Im gonna go over section 11.1 because thats the only one i rememberrr.
In this section we learned how to convert polar to rectangular and how to convert polar to rectangular. To do this, you must know the following formulas:to convert to rectangular--x=rcos(theta) y=rsin(theta)to convert to polar--r=square root of x^2 + y^2 tan(theta)= y/x(r,(theta)) - polar form(x,y) - rectangular
Example 1:Give the polar point for (1,1)r=square root of 1^2 + 1^2 = 1
*first you use the polar formulatan(theta)= 1/1 = 1
*use tan(theta) y/xtheta=tan^-1(1) = 45
*find the inverse*
because tan is positive, find where tan is positive on the trig chart.
it is positive in quadrant 1 and 3 so your answers are 45 and 225
so you gett, (1, 45) ; (-1, 225)
In this section we learned how to convert polar to rectangular and how to convert polar to rectangular. To do this, you must know the following formulas:to convert to rectangular--x=rcos(theta) y=rsin(theta)to convert to polar--r=square root of x^2 + y^2 tan(theta)= y/x(r,(theta)) - polar form(x,y) - rectangular
Example 1:Give the polar point for (1,1)r=square root of 1^2 + 1^2 = 1
*first you use the polar formulatan(theta)= 1/1 = 1
*use tan(theta) y/xtheta=tan^-1(1) = 45
*find the inverse*
because tan is positive, find where tan is positive on the trig chart.
it is positive in quadrant 1 and 3 so your answers are 45 and 225
so you gett, (1, 45) ; (-1, 225)
Over the break post
10.1 formulas for cos(X+B) and sin(X+B)
cos(X+-B)= cosXcosB -+ sinXsinB
sin(X+-B)= sinXcosB+- cosXsinB
Example #1
Find the exact value
Cos75
cos(45+30) X=45 B=30
cos45cos30 - sin45sin30
cos75= the sqaure root of 6 minus the square root of 2 all over 4.
Example #2
Sin15
sin(45-30) X=45 B=30
sin45cos30 - cos45sin30
sin15= the sqaure root of 6 minus the square root of 2 all over 4
cos(X+-B)= cosXcosB -+ sinXsinB
sin(X+-B)= sinXcosB+- cosXsinB
Example #1
Find the exact value
Cos75
cos(45+30) X=45 B=30
cos45cos30 - sin45sin30
cos75= the sqaure root of 6 minus the square root of 2 all over 4.
Example #2
Sin15
sin(45-30) X=45 B=30
sin45cos30 - cos45sin30
sin15= the sqaure root of 6 minus the square root of 2 all over 4
lawrence's blog
in this part i will go over chapter 7 sections 1 and 2. they are very simple but you need to know these formulas because you do use them in all the other chapters. if you havent learned by now when b rob said you need to know all of this stuff im sure everyone knows now. anyway i will start go over these to sections
SECTION 1
two units of measure: degrees and radians
convert degrees to rads is x pi/180
convert rads to degrees is x 180/pi
coterminal angles +/- 360 or +/- 2pi if in rads
to convert to minutes and seconds.
to convert to minutes and seconds you just multiply by 60
to get them out of minutes and seconds its
x/60 + y/3600
examples:
1) 315 degrees convert to rads
315 x pi/180
=7pi/4
2) 25.335 degrees convert to minutes and seconds
.335 x60= 20.1
.1 x60= 6
25 degrees 20' 6"
SECTION 2
these are the formulas
s=r theta
k=1/2 r squared theta
k= 1/2 rs
know s=arc length r=radius theta= central angles and k=area of a sector
theta must also be in radians!!!!!!!
example
1) sector of a circle has arc length 6cm& 75cm squared. find its radius and the measure of its central angle
s= 6cm k=75
r=? theta=?
75= 1/2r(6)
75=3r now you divide by 3
r=25cm dont forget the units
now you use s=r theta
6=25(theta)
theta= 6/25 rads
i hope this helped some people who may not totally get this and i hope those who didnt finally have it and i hope everyone had a great holiday break.
SECTION 1
two units of measure: degrees and radians
convert degrees to rads is x pi/180
convert rads to degrees is x 180/pi
coterminal angles +/- 360 or +/- 2pi if in rads
to convert to minutes and seconds.
to convert to minutes and seconds you just multiply by 60
to get them out of minutes and seconds its
x/60 + y/3600
examples:
1) 315 degrees convert to rads
315 x pi/180
=7pi/4
2) 25.335 degrees convert to minutes and seconds
.335 x60= 20.1
.1 x60= 6
25 degrees 20' 6"
SECTION 2
these are the formulas
s=r theta
k=1/2 r squared theta
k= 1/2 rs
know s=arc length r=radius theta= central angles and k=area of a sector
theta must also be in radians!!!!!!!
example
1) sector of a circle has arc length 6cm& 75cm squared. find its radius and the measure of its central angle
s= 6cm k=75
r=? theta=?
75= 1/2r(6)
75=3r now you divide by 3
r=25cm dont forget the units
now you use s=r theta
6=25(theta)
theta= 6/25 rads
i hope this helped some people who may not totally get this and i hope those who didnt finally have it and i hope everyone had a great holiday break.
Feroz's Double Blog
Oh my damn. I haven't done any of these and it's like almost midnight.
Feroz's Blog: Part 1: Chapter 13: Section 4: Limits
Like with everything in math, limits have rules:
1.) If the top exponent is equal to the bottom exponent the answer is coefficient
2.) If the top exponent is greater than the bottom exponent answer is +/- infinite
3.) If the top exponent is less than the bottom exponent answer is 0
Not that hard to remember actually.
*So yeah, showing an example for this is not working out for me, like it looks really bad. So um, all you really need to know is the 3 rules above, and if the exponents are the same on the top and bottom use the coefficients, and you should be able to figure out which side is greater or if they're the same.
Part 2: Chapter 7: Section 1: Converting to Radians
Converting something to degrees or radians is extremely simple. If I could marry a lesson, it would probably be this one.
To radians: pi/180
To degrees: 180/pi
ex. 2pi to degrees
2 x 180 = 360 degrees (just get rid of the pi)
ex 2. 540 degrees to radians
540/180 = 3pi
There. I would go into more detail but my laptops dying. Not that you needed more detail. Yeah.
Feroz's Blog: Part 1: Chapter 13: Section 4: Limits
Like with everything in math, limits have rules:
1.) If the top exponent is equal to the bottom exponent the answer is coefficient
2.) If the top exponent is greater than the bottom exponent answer is +/- infinite
3.) If the top exponent is less than the bottom exponent answer is 0
Not that hard to remember actually.
*So yeah, showing an example for this is not working out for me, like it looks really bad. So um, all you really need to know is the 3 rules above, and if the exponents are the same on the top and bottom use the coefficients, and you should be able to figure out which side is greater or if they're the same.
Part 2: Chapter 7: Section 1: Converting to Radians
Converting something to degrees or radians is extremely simple. If I could marry a lesson, it would probably be this one.
To radians: pi/180
To degrees: 180/pi
ex. 2pi to degrees
2 x 180 = 360 degrees (just get rid of the pi)
ex 2. 540 degrees to radians
540/180 = 3pi
There. I would go into more detail but my laptops dying. Not that you needed more detail. Yeah.
Saturday, November 27, 2010
Nathan's turkey blog
It's Saturday, LSU just lost, im mad and bored, so I might as well get this blog over with so that I can do nothing tommorrow.
I am going to review chapter ten, sections one and two. These were probably the easiest sections since chapter 7.
In section 10-1, we couldn't use calculators, and we needed to know these two formulas:
cos(alpha+-beta)=cos(alpha)cos(beta)-+sin(alpha)sin(beta)
sin(alpha+-beta)=sin(alpha)cos(beta)+-cos(alpha)sin(beta)
Ex. Find the exact value of sin 15.
sin(45-30)=sin45cos30-cos45sin30
=√2/2(√3/2)-√2/2(1/2)
=√6-√2/4
cosπ/4 cosπ/4-sinπ/4 sinπ/4
=cos(π/4+π/4)=cos 2π/4=cosπ/2=0
Section 10-2 deals with the same kind of problem solving; the formulas you need to know is:
tan(alpha+beta)=tan(alpha)+tan(beta)/1-tan(alpha)tan(beta)
tan(alpha-beta)=tan(alpha)-tan(beta)/1+tan(alpha)tan(beta)
tan110-tan50/1+tan110tan50
=tan(110-50)
=tan60
=√3
tan27+tan18/1-tan27tan18
=tan(27+18)
=tan45
=1
Another easy section was 13-1. This section dealt with arithmetic and geometric sequences. An arithmetic sequence is when you add or subtract the numbers. A geometric sequence is when you multiply or divide to get the next number.
Ex. In a geometric sequence, t3=12 and t6=96, find t11
96/12=8
r^3=8
r=2
tn=3*2^11-1
t11=3*2^10
t11=3*1024
t11=3072
I am going to review chapter ten, sections one and two. These were probably the easiest sections since chapter 7.
In section 10-1, we couldn't use calculators, and we needed to know these two formulas:
cos(alpha+-beta)=cos(alpha)cos(beta)-+sin(alpha)sin(beta)
sin(alpha+-beta)=sin(alpha)cos(beta)+-cos(alpha)sin(beta)
Ex. Find the exact value of sin 15.
sin(45-30)=sin45cos30-cos45sin30
=√2/2(√3/2)-√2/2(1/2)
=√6-√2/4
cosπ/4 cosπ/4-sinπ/4 sinπ/4
=cos(π/4+π/4)=cos 2π/4=cosπ/2=0
Section 10-2 deals with the same kind of problem solving; the formulas you need to know is:
tan(alpha+beta)=tan(alpha)+tan(beta)/1-tan(alpha)tan(beta)
tan(alpha-beta)=tan(alpha)-tan(beta)/1+tan(alpha)tan(beta)
tan110-tan50/1+tan110tan50
=tan(110-50)
=tan60
=√3
tan27+tan18/1-tan27tan18
=tan(27+18)
=tan45
=1
Another easy section was 13-1. This section dealt with arithmetic and geometric sequences. An arithmetic sequence is when you add or subtract the numbers. A geometric sequence is when you multiply or divide to get the next number.
Ex. In a geometric sequence, t3=12 and t6=96, find t11
96/12=8
r^3=8
r=2
tn=3*2^11-1
t11=3*2^10
t11=3*1024
t11=3072
Nicala's Post
chapter 13 section 1
geometric and arithmetic sequences
arithmetic sequence is a sequence where the same number is added each time.
Formula: tn=t1+(n-1)d
geometric sequence is a sequence where the same number is multiplied each time.
Formula: tn=t1xr^n-1
Sample Problem
Find thed first four terms and state whether the sequences is arithmetic, geometric, or neither.
To find the first four terms, replace the n with 1.
Then 2, then 3, and 4.
tn=3n+2
tn=3(1)+2
tn=3+2
tn=5
tn=3n+2
tn=3(2)+2
tn=6+2
tn=8
tn=3n+2
tn=3(3)+2
tn=9+2
tn=11
tn=3n+2
tn=3(4)+2
tn=12+2
tn=14
Answer: 5,8,11,14
the sequence is arithmetic and d=3
geometric and arithmetic sequences
arithmetic sequence is a sequence where the same number is added each time.
Formula: tn=t1+(n-1)d
geometric sequence is a sequence where the same number is multiplied each time.
Formula: tn=t1xr^n-1
Sample Problem
Find thed first four terms and state whether the sequences is arithmetic, geometric, or neither.
To find the first four terms, replace the n with 1.
Then 2, then 3, and 4.
tn=3n+2
tn=3(1)+2
tn=3+2
tn=5
tn=3n+2
tn=3(2)+2
tn=6+2
tn=8
tn=3n+2
tn=3(3)+2
tn=9+2
tn=11
tn=3n+2
tn=3(4)+2
tn=12+2
tn=14
Answer: 5,8,11,14
the sequence is arithmetic and d=3
Friday, November 26, 2010
Taylor"s 12(I think) Blog
This is some of my notes from Ch. 13.1.
A sequence is a set of numbers. Today we will learn a bit about arithmetic sequence which is when you add the same number to every term.
This is the formula:
tn= t1 +(n-1)d
d is the number that is being added or subtracted. The n and one by the t are small to.
Ok let's do some examples.
Find the first 5 terms of this equation tn+ 6n+2
t1= 6(1)+2 First replace n with 1,2,3, and for and solve the equation.
t2=6(2)+2
t3=6(3)+2
t4=6(4)+2
t1= 8 Then you have your answers.
t2= 14
t3= 20
t4= 26
8,13,18,23.... find the formula for the nth term.
8,13,18,23.... First you need to figure out what is being added or subtracted from
each number. In this case it is adding 5.
tn=t1+(n-1)d Then write out your equation.
tn=8+(n-1)5 Then fill out the parts that you know2.
tn=8+5n-5 Then multiply 5 times n and -1.
tn= 3+ 5n Then you just add up your like terms and this is your answer.
These are some of my notes on Ch. 13.1. So UN TIL SCHOOL STARTS SEE YALL!!!
A sequence is a set of numbers. Today we will learn a bit about arithmetic sequence which is when you add the same number to every term.
This is the formula:
tn= t1 +(n-1)d
d is the number that is being added or subtracted. The n and one by the t are small to.
Ok let's do some examples.
Find the first 5 terms of this equation tn+ 6n+2
t1= 6(1)+2 First replace n with 1,2,3, and for and solve the equation.
t2=6(2)+2
t3=6(3)+2
t4=6(4)+2
t1= 8 Then you have your answers.
t2= 14
t3= 20
t4= 26
8,13,18,23.... find the formula for the nth term.
8,13,18,23.... First you need to figure out what is being added or subtracted from
each number. In this case it is adding 5.
tn=t1+(n-1)d Then write out your equation.
tn=8+(n-1)5 Then fill out the parts that you know2.
tn=8+5n-5 Then multiply 5 times n and -1.
tn= 3+ 5n Then you just add up your like terms and this is your answer.
These are some of my notes on Ch. 13.1. So UN TIL SCHOOL STARTS SEE YALL!!!
Wednesday, November 24, 2010
charlie's 2nd thanksgiving bloggy thinggy.
Okay, so a while back we learned how to turn polar form into rectangular form.
i forgot what chapter and section it's in though.
so.. the definition of polar is to graph using angles
polar form is (r, theta) and rectangular form is (x, y)
the formulas we used for this was...
r = `the square root of` x^2 + y^2
and
tantheta = (y/x)
[ this easily changes to theta = tangent inverse of (y/x) ]
so to do this you basically just plug in the (x, y) rectangular form numbers into the two formulas to get the (r, theta) polar form.
EXAMPLE:
#1. turn rectangular form (-1, 2) into polar form
[ first, you plug the -1 and the 2 into the formulas ]
r = `the square root of` -1^2 + 2^2
= `the square root of` 1 + 4
= +/- `the square root of` 5
theta = tangent inverse of (2/-1)
= 63.565
[ tangent is negative on the unit circle in quadrant 2 &4, so now you make 63 negative and add 180 & make 63 negative and add 360 to get to those quadrants ]
.... `the square root of` 5, 116. 565
.... `the square root of` -5, 296.565
i forgot what chapter and section it's in though.
so.. the definition of polar is to graph using angles
polar form is (r, theta) and rectangular form is (x, y)
the formulas we used for this was...
r = `the square root of` x^2 + y^2
and
tantheta = (y/x)
[ this easily changes to theta = tangent inverse of (y/x) ]
so to do this you basically just plug in the (x, y) rectangular form numbers into the two formulas to get the (r, theta) polar form.
EXAMPLE:
#1. turn rectangular form (-1, 2) into polar form
[ first, you plug the -1 and the 2 into the formulas ]
r = `the square root of` -1^2 + 2^2
= `the square root of` 1 + 4
= +/- `the square root of` 5
theta = tangent inverse of (2/-1)
= 63.565
[ tangent is negative on the unit circle in quadrant 2 &4, so now you make 63 negative and add 180 & make 63 negative and add 360 to get to those quadrants ]
.... `the square root of` 5, 116. 565
.... `the square root of` -5, 296.565
Week 6 Prompt
Due to the fact that the internet was not dependable on my laptop I was unable to post the blog prompt for the holidays. Therefore it is a freebie for everyone. Just make sure to post your 2 regular blogs. They can come from any review topic.
Sunday, November 21, 2010
What I learned this week.
This week was on chapter 13,
and there are many more sections and chapters to come.
This chapter had six sections,
and none of the six were fun.
But I learned something new,
and true knowledge I received.
More problems and more formulas,
seems to always keep my mind off ease.
This is the time of the year,
were high school kids always seems to slack off.
This is how teens go down the wrong path,
and the knowledge that was gained, ends up lost.
I tend to not follow the crowd,
and get lost in the struggle.
I plan to keep my grades up,
and stay away from trouble.
The week was long and tough,
but now we have a break.
Now I have time to catch up on sleep,
from those long nights of staying up late.
Happy holidays to everyone,
and have of mind full of love instead of hate.
And don't take this time for granted,
go to sleep instead of staying up eating thanksgiving cake.
#1: Find the first 5 terms of t*n = 2n - 9
t*1 = 2(1) - 9 = 2 - 9 = - 7
t*2 = 2(2) - 9 = 4 - 9 = -5
t*3 = 2(3) - 9 = 6 - 9 = -3
t*4 = 2(4) - 9 = 8 - 9 = -1
t*5 = 2(5) - 9 = 10 - 9 = 1
#2: 2, 4, 6, 8, 10
t*n = t*1 + (n-1)d
t*n = 2 + (n-1)2
t*n = 2 + 2n -2
t*n = 2n
t*1 = 2(1) - 9 = 2 - 9 = - 7
t*2 = 2(2) - 9 = 4 - 9 = -5
t*3 = 2(3) - 9 = 6 - 9 = -3
t*4 = 2(4) - 9 = 8 - 9 = -1
t*5 = 2(5) - 9 = 10 - 9 = 1
#2: 2, 4, 6, 8, 10
t*n = t*1 + (n-1)d
t*n = 2 + (n-1)2
t*n = 2 + 2n -2
t*n = 2n
torisss.
I don't know what to do this blog on so im just gonna do it on the same one i did last week.
This week we did chapter 13.
the chapter is about sequences, which are a listing of numbers.
Each sequence is either arithmetic (adding the same number every term) or geometric (multipling by the same number for every term).
For this we have two formulas:
arithmetic formula: t*n = t*1 + (n-1)d
geometric formula: t*n = t*1 X r^n-1
For these formulas d is the number being added or subtracted and r is the number being multiplied by.
Or you may have to find certain terms.
Fo this you get a formula like t*n = #n -# and you make n the term you're looking for.
Some Examples:
#1: Find the first 5 terms of t*n = 2n - 9
t*1 = 2(1) - 9 = 2 - 9 = - 7
t*2 = 2(2) - 9 = 4 - 9 = -5
t*3 = 2(3) - 9 = 6 - 9 = -3
t*4 = 2(4) - 9 = 8 - 9 = -1
t*5 = 2(5) - 9 = 10 - 9 = 1
#2: 2, 4, 6, 8, 10
t*n = t*1 + (n-1)d
t*n = 2 + (n-1)2
t*n = 2 + 2n -2
t*n = 2n
This week we did chapter 13.
the chapter is about sequences, which are a listing of numbers.
Each sequence is either arithmetic (adding the same number every term) or geometric (multipling by the same number for every term).
For this we have two formulas:
arithmetic formula: t*n = t*1 + (n-1)d
geometric formula: t*n = t*1 X r^n-1
For these formulas d is the number being added or subtracted and r is the number being multiplied by.
Or you may have to find certain terms.
Fo this you get a formula like t*n = #n -# and you make n the term you're looking for.
Some Examples:
#1: Find the first 5 terms of t*n = 2n - 9
t*1 = 2(1) - 9 = 2 - 9 = - 7
t*2 = 2(2) - 9 = 4 - 9 = -5
t*3 = 2(3) - 9 = 6 - 9 = -3
t*4 = 2(4) - 9 = 8 - 9 = -1
t*5 = 2(5) - 9 = 10 - 9 = 1
#2: 2, 4, 6, 8, 10
t*n = t*1 + (n-1)d
t*n = 2 + (n-1)2
t*n = 2 + 2n -2
t*n = 2n
Nathan's Blog
I am going to do this blog on section 13-4. These are the rules for fractions.
1.) If the top exponent is equal to the bottom exponent the answer is coefficient
2.) If the top exponent is greater than the bottom exponent answer is +- infinite
3.) If the top exponent is less than the bottom exponent answer is 0
Examples: All equations have lim/n→(infinite)
sin(1/n)=0
n^2+1/2n^2-3n=1/2 (exponents are equal; take coefficients)
7n^3/4n^2-5=infinite (top exponent is greater)
-n^2/n+1= -infinite
n+5/n=1
2n^4/6n^5+7=0
Now I will go over section 13-5 which deals with the sum of an infinite geometric series. The formula is: sn=t1/1-r
To write a repeating decimal as a fraction: number repeating/last place-1
ex. .4646=46/100-1=46/99
Find the sum of the infinite geometrice series.
9-6+4-....
r=-6/9=-2/3; 4/-6=-2/3
sn=9/1-(-2/3)=27/5
Write .5(repeating) as a fraction.
5/10-1=5/9
Overall, these sections were pretty easy. If you don't know your formulas, then you might have a little trouble with the sections in chapter 13. Happy Turkey Day.
1.) If the top exponent is equal to the bottom exponent the answer is coefficient
2.) If the top exponent is greater than the bottom exponent answer is +- infinite
3.) If the top exponent is less than the bottom exponent answer is 0
Examples: All equations have lim/n→(infinite)
sin(1/n)=0
n^2+1/2n^2-3n=1/2 (exponents are equal; take coefficients)
7n^3/4n^2-5=infinite (top exponent is greater)
-n^2/n+1= -infinite
n+5/n=1
2n^4/6n^5+7=0
Now I will go over section 13-5 which deals with the sum of an infinite geometric series. The formula is: sn=t1/1-r
To write a repeating decimal as a fraction: number repeating/last place-1
ex. .4646=46/100-1=46/99
Find the sum of the infinite geometrice series.
9-6+4-....
r=-6/9=-2/3; 4/-6=-2/3
sn=9/1-(-2/3)=27/5
Write .5(repeating) as a fraction.
5/10-1=5/9
Overall, these sections were pretty easy. If you don't know your formulas, then you might have a little trouble with the sections in chapter 13. Happy Turkey Day.
lawrences blog
im doing this blog on chapter 10 section 1. this chapter took me a little while to get but now i know how to do it and it is easy. it is really simple. you have to know the formulas and you have to know the trig chart. everything from chapter 7 and on you better know or you will be beat down by b-rob with her tests.
formulas:
cos(alpha +/- beta) cos alpha cos beta -/+ sin alpha sin beta
sin(alpha +/- beta) sin alpha cos beta +/- cos alspha sin beta
aplha and beta come from the trig chart and add or subtract to get the angle you are looking for
examples:
1)
cos(75)
alpha 45
cos 30
sin(45+30)= sin45 cos30- cos45 sin30
=square root of2/2 (square root of 2/2)- square root of 2/2(1/2)
square root of6/4- square root of 2/4
= square root of 6-2 all over 4 (final answer)
this is what i learned from chapter 10 section 1. hope this helps some people
formulas:
cos(alpha +/- beta) cos alpha cos beta -/+ sin alpha sin beta
sin(alpha +/- beta) sin alpha cos beta +/- cos alspha sin beta
aplha and beta come from the trig chart and add or subtract to get the angle you are looking for
examples:
1)
cos(75)
alpha 45
cos 30
sin(45+30)= sin45 cos30- cos45 sin30
=square root of2/2 (square root of 2/2)- square root of 2/2(1/2)
square root of6/4- square root of 2/4
= square root of 6-2 all over 4 (final answer)
this is what i learned from chapter 10 section 1. hope this helps some people
Friday, November 19, 2010
Charlie's.
Today we took notes on the 'captial sigma'.
This is in chapter 13, i think section 6.
She gave us the definition of a sigma; which is, a sigma is a series written in condensed form.
For this you have the little capital sigma sign (it looks like a weird E)
On top of it is #, at the bottom is n=#, and on the right side is f(x).
Where # is the limits of summation, n is the index, and f(x) is the summand.
Examples:
#1:
6
E 2(x)+4
x=2
So you just plug this into the summand [which is 2(x) +4] starting with 2 and ending at 6, adding them inbetween.
2(2)+4 + 2(3)+4 + 2(4)+4 + 2(5)+4 + 2(6)+4
= 8 + 10 + 12 + 14 + 16
= 60
#2:
3
E (3-p)^2
p=-1
(3+1)^2 + (3-0)^2 + (3-1)^2 + (3-2)^2 + (3-3)^2
= 4^2 + 3^2 + 2^2 + 1^2 + 0^2
= 16 + 9 + 4 + 1 + 0
=30
This is in chapter 13, i think section 6.
She gave us the definition of a sigma; which is, a sigma is a series written in condensed form.
For this you have the little capital sigma sign (it looks like a weird E)
On top of it is #, at the bottom is n=#, and on the right side is f(x).
Where # is the limits of summation, n is the index, and f(x) is the summand.
Examples:
#1:
6
E 2(x)+4
x=2
So you just plug this into the summand [which is 2(x) +4] starting with 2 and ending at 6, adding them inbetween.
2(2)+4 + 2(3)+4 + 2(4)+4 + 2(5)+4 + 2(6)+4
= 8 + 10 + 12 + 14 + 16
= 60
#2:
3
E (3-p)^2
p=-1
(3+1)^2 + (3-0)^2 + (3-1)^2 + (3-2)^2 + (3-3)^2
= 4^2 + 3^2 + 2^2 + 1^2 + 0^2
= 16 + 9 + 4 + 1 + 0
=30
Taylor's 12th (I think XD) Blog
These are the other half of my notes on Chapter 11.1.
We will be learning how to convert rectangular to polar.
Here are the formulas.
r= square root of x^2 + y^2 tan theta= y/x
polar is in (r,theta) form and rectangular is in (x,y) form.
Now let's work on some problems.
C
onvert (8,9) to polar.
r= square root of 8^2 + 9^2 tan theta= 9/8 First, write out your problem so that
it will match the formula that I gave you.
r= square root of 64 + 81 theta= tan^-1 (9/8) Then for the first part of the equation you square the two numbers. Then for the second part set up the tan so that you are finding the inverse of the fraction.
r= square root of 145 theta= 48.366 degrees Then for the first part of the equation you add the square roots and for the second part you will find the inverse and you will get what I have after you round to the 3rd place after the decimal.
Since you cannot find the square root of 145 you will leave this as your answer and since I cannot use a graph to show you, you need to visualize the quadrant plane for tan and since it is positive we will be using the first and third quadrant. So to get 48.366 degrees to quadrant three you would add 180 degrees and get 228.366 degrees as the answer.
So your answer for this question should look like this (square root of 145 , 48.366 degrees ) and ( - square root of 145 , 228.366 degrees).
Convert (10,12) to polar.
r= square root of 10^2 + 12^2 tan theta= 12/10 First, write out your problem so that it will match the formula that I gave you.
r= square root of 100 + 144 theta= tan^-1 (12/10) Then for the first part of the equation you square the two numbers. Then for the second part set up the tan so that you are finding the inverse of the fraction.
r= square root of 244 theta= 50.194 degrees Then for the first part of the equation you add the square roots and for the second part you will find the inverse and you will get what I have after you round to the 3rd place after the decimal.
Since you cannot find the square root of 145 you will factor out the four and get 2 square root of 61 and since I cannot use a graph to show you, you need to visualize the quadrant plane for tan and since it is positive we will be using the first and third quadrant. So to get 50.194 degrees to quadrant three you would add 180 degrees and get 230.194 degrees as the answer.
So your answer for this question should look like this ( 2 square root of 61, 50.194 degrees ) and ( - 2 square root of 61, 230.194 degrees).
That is the last half of my notes from Chapter 11.1.
SO UNTIL NEXT TIME JA NE!!!!! (Japanese for goodbye.)
We will be learning how to convert rectangular to polar.
Here are the formulas.
r= square root of x^2 + y^2 tan theta= y/x
polar is in (r,theta) form and rectangular is in (x,y) form.
Now let's work on some problems.
C
onvert (8,9) to polar.
r= square root of 8^2 + 9^2 tan theta= 9/8 First, write out your problem so that
it will match the formula that I gave you.
r= square root of 64 + 81 theta= tan^-1 (9/8) Then for the first part of the equation you square the two numbers. Then for the second part set up the tan so that you are finding the inverse of the fraction.
r= square root of 145 theta= 48.366 degrees Then for the first part of the equation you add the square roots and for the second part you will find the inverse and you will get what I have after you round to the 3rd place after the decimal.
Since you cannot find the square root of 145 you will leave this as your answer and since I cannot use a graph to show you, you need to visualize the quadrant plane for tan and since it is positive we will be using the first and third quadrant. So to get 48.366 degrees to quadrant three you would add 180 degrees and get 228.366 degrees as the answer.
So your answer for this question should look like this (square root of 145 , 48.366 degrees ) and ( - square root of 145 , 228.366 degrees).
Convert (10,12) to polar.
r= square root of 10^2 + 12^2 tan theta= 12/10 First, write out your problem so that it will match the formula that I gave you.
r= square root of 100 + 144 theta= tan^-1 (12/10) Then for the first part of the equation you square the two numbers. Then for the second part set up the tan so that you are finding the inverse of the fraction.
r= square root of 244 theta= 50.194 degrees Then for the first part of the equation you add the square roots and for the second part you will find the inverse and you will get what I have after you round to the 3rd place after the decimal.
Since you cannot find the square root of 145 you will factor out the four and get 2 square root of 61 and since I cannot use a graph to show you, you need to visualize the quadrant plane for tan and since it is positive we will be using the first and third quadrant. So to get 50.194 degrees to quadrant three you would add 180 degrees and get 230.194 degrees as the answer.
So your answer for this question should look like this ( 2 square root of 61, 50.194 degrees ) and ( - 2 square root of 61, 230.194 degrees).
That is the last half of my notes from Chapter 11.1.
SO UNTIL NEXT TIME JA NE!!!!! (Japanese for goodbye.)
Nicala's Blog
chapter 7 section 4
finding a reference angle
finding a reference angle
- find the quadrant the angle is in.
- determine if the trig function is postive or negative
- subtract 180 degrees from the angle until theta is between 0 and angle.
- if it is a trig chart angle plug in, if not leav it or plug in calculator.
Hint: evaluate means # answer unlesas otherwise specified.
Find tan inverse (1) without a calculator.
- put it on the coordinating plane
- subtract from 180
- keep subtracting until you get a answer between zero and ninety.
the answer is forty five degrees.
Monday, November 15, 2010
Week 5 Blog Prompt
What is a famous sequences and series? What is it used for and who discovered it? Everyone should find a different type.
Sunday, November 14, 2010
Dom's blog
Chapter 13-1
tsub2=4(2)-3=8-3=5
tsub3=4(3)-3=12-3=9
tsub4=4(4)-3=16-3=12
arithmetic +4
Sequences-
is a list of numbers. There are two kinds of sequences:
1) Arithmetic- +/- by the same number every term
2) Geometric- *// by the same number every term
is a list of numbers. There are two kinds of sequences:
1) Arithmetic- +/- by the same number every term
2) Geometric- *// by the same number every term
Ex: if tsubnl=4n-3, find the first 4 terms:
tsub1=4(1)-3=4-3=1tsub2=4(2)-3=8-3=5
tsub3=4(3)-3=12-3=9
tsub4=4(4)-3=16-3=12
arithmetic +4
Well, I'm done.
Malorie's Blog
13.1 Sequences
In this lesson, we learned the two different types of sequencing, Arithmetic and Geometric. Arithmetic is when you add the same number to every term. Geometric is when you multiply by the same number for every term.
The formula to find the nth term for Arithmetic is tn= t1 + (n - 1)d
Geometric is: tn= t1 x r^(n-1)
t= term n= last number of sequence r= what is being added or multipled d= what is being added or subtracted
Example 1:
If tn=4n-3 find the first 4 terms and say whether it is arithmetic, geometric, or neither.
t1= 4(1)-3= 1
t2=4(2)-3=5
t3=4(3)-3=9
t4=4(4)-3=13
Arithmetic because 4 is being added to every term
Example 2:
If tn=2n+3 find the first 4 terms and say whether it is arithmetic, geometric, or neither.
t1=2(1)+3=5
t2=2(2)+3=7
t3=2(3)+3=9
t4+2(4)+3=11
Arithmetic because 2 is being added to every term
In this lesson, we learned the two different types of sequencing, Arithmetic and Geometric. Arithmetic is when you add the same number to every term. Geometric is when you multiply by the same number for every term.
The formula to find the nth term for Arithmetic is tn= t1 + (n - 1)d
Geometric is: tn= t1 x r^(n-1)
t= term n= last number of sequence r= what is being added or multipled d= what is being added or subtracted
Example 1:
If tn=4n-3 find the first 4 terms and say whether it is arithmetic, geometric, or neither.
t1= 4(1)-3= 1
t2=4(2)-3=5
t3=4(3)-3=9
t4=4(4)-3=13
Arithmetic because 4 is being added to every term
Example 2:
If tn=2n+3 find the first 4 terms and say whether it is arithmetic, geometric, or neither.
t1=2(1)+3=5
t2=2(2)+3=7
t3=2(3)+3=9
t4+2(4)+3=11
Arithmetic because 2 is being added to every term
kldfjalkjdv
this week i wish i didn't skip thursday because i had soo much trouble on my homework that it wasn't even funny..
this week, we learned about chapter 13
this chapter is about sequences, which is a list of numbers, there are two kinds
Arithmetic- add the same number to every term
Geometric- multiply by the same number for each number
if you divide the numbers in a sequences and you find a like ratio, then its geometric
here are your formulas
arithmetic- t(n)=t1 +(n-1)d
geometric= t(n)=t1 x r ^n-1
where d is what is being added or subtracted and where r is what you are multiplying by
example=t(n)=4(n)-3 find the four first terms
t1=4(1)-3=1
t2=4(2)-3=5
t3=4(3))-3=9
t4=4(4)-3=13
and from this you can determine that it is an aritmetic sequence
example 2
3,7,11,15... find a formula for the nth term
you take your fomula(arithmetic)
you know that tn is the total number
you know that t1 is the first number in the sequence
and you know that d is the number being added or subtracted
you can tell you are adding by the information given
now you just plug in your formula
tn=3 +(n-1)4
then simplify
tn=3+4n-4
tn=4n-1
and there ya have it
i only have one example of this same process but with geometric and i can't do geometric,i needd some help tomorrow B-Robbbb
Dylan's Bloggity Blog
Sec. 13-1 Sequences
A sequence is a list of numbers. There are two kinds:
Arithmetic- +/- by the same number every term
Geometric- *// by the same number every term
Ex 6,4,2,0,... is Arithmetic
4,7,11,16,22,... is neither
To find the n^m term, you use these formulas:
Arithmetic- tn=t1+(n-1)d
Geometric- tn=t1*r
( d= the number that is used for A and G, the number being //+/-/*)
A sequence is a list of numbers. There are two kinds:
Arithmetic- +/- by the same number every term
Geometric- *// by the same number every term
Ex 6,4,2,0,... is Arithmetic
4,7,11,16,22,... is neither
To find the n^m term, you use these formulas:
Arithmetic- tn=t1+(n-1)d
Geometric- tn=t1*r
( d= the number that is used for A and G, the number being //+/-/*)
What did we learn this week?
These chapters we've learned,
made us think a great deal.
We have to go into deep thinking
to sought out whats real.
Formulas are hard
and questions seem impossible to do.
But this is a good test,
to see what kind of person are u.
You learn more than you think,
and even more when you don't complain,
your mind is your thoughts,
so your mind is going to change.
Make the right decisions
even if the right thing is too hard to do
because the wrong thing is easier
So do most challenging, and his plan will go through.
Continue to seek change
and let your mind feel at ease
keep your faith in God
and always believe.
This week's material was all on chapter 13.
Arithmetic= tn= t1(n-1)d
Geometric= t1 x r^n-1
#1: Find the first 5 terms of t*n = 2n - 9
t*1 = 2(1) - 9 = 2 - 9 = - 7
t*2 = 2(2) - 9 = 4 - 9 = -5
t*3 = 2(3) - 9 = 6 - 9 = -3
t*4 = 2(4) - 9 = 8 - 9 = -1
t*5 = 2(5) - 9 = 10 - 9 = 1
#2: 2, 4, 6, 8, 10
t*n = t*1 + (n-1)d
t*n = 2 + (n-1)2
t*n = 2 + 2n -2
t*n = 2n
I'm sleepy goodnight bloggers!
made us think a great deal.
We have to go into deep thinking
to sought out whats real.
Formulas are hard
and questions seem impossible to do.
But this is a good test,
to see what kind of person are u.
You learn more than you think,
and even more when you don't complain,
your mind is your thoughts,
so your mind is going to change.
Make the right decisions
even if the right thing is too hard to do
because the wrong thing is easier
So do most challenging, and his plan will go through.
Continue to seek change
and let your mind feel at ease
keep your faith in God
and always believe.
This week's material was all on chapter 13.
Arithmetic= tn= t1(n-1)d
Geometric= t1 x r^n-1
#1: Find the first 5 terms of t*n = 2n - 9
t*1 = 2(1) - 9 = 2 - 9 = - 7
t*2 = 2(2) - 9 = 4 - 9 = -5
t*3 = 2(3) - 9 = 6 - 9 = -3
t*4 = 2(4) - 9 = 8 - 9 = -1
t*5 = 2(5) - 9 = 10 - 9 = 1
#2: 2, 4, 6, 8, 10
t*n = t*1 + (n-1)d
t*n = 2 + (n-1)2
t*n = 2 + 2n -2
t*n = 2n
I'm sleepy goodnight bloggers!
tori's
this week, we learned chapter 13 section 1.
it was extremely easy in my opinion!
Sequences-a list of numbers
arithmetic- when you add/subtract the same number to every term
geometric- when you multiply/divide the same number for every term
ex. 1
are the following sequences arithmetic, geometric, or neither?
a. 4, 8, 12, 16,...------> arithmetic +4
b. 9, 27/2, 81/4,...----> geometric x3/2
to find the nth term of a sequence
FORMULAS:
arithmetic- tsubn=tsub1+(n-1)d
geometric- tsubn=tsub1xr^(n-1)
where d is what is being added or subtracted and r is what you are multiplying or dividing by.
example 2:
if tsubnl=4n-3, find the first 4 terms:
tsub1=4(1)-3=4-3=1
tsub2=4(2)-3=8-3=5
tsub3=4(3)-3=12-3=9
tsub4=4(4)-3=16-3=12
arithmetic +4
and there we gooo!
it was extremely easy in my opinion!
Sequences-a list of numbers
arithmetic- when you add/subtract the same number to every term
geometric- when you multiply/divide the same number for every term
ex. 1
are the following sequences arithmetic, geometric, or neither?
a. 4, 8, 12, 16,...------> arithmetic +4
b. 9, 27/2, 81/4,...----> geometric x3/2
to find the nth term of a sequence
FORMULAS:
arithmetic- tsubn=tsub1+(n-1)d
geometric- tsubn=tsub1xr^(n-1)
where d is what is being added or subtracted and r is what you are multiplying or dividing by.
example 2:
if tsubnl=4n-3, find the first 4 terms:
tsub1=4(1)-3=4-3=1
tsub2=4(2)-3=8-3=5
tsub3=4(3)-3=12-3=9
tsub4=4(4)-3=16-3=12
arithmetic +4
and there we gooo!
Charlie's.
This week we started chapter 13.
the chapter is about sequences, which are a listing of numbers.
Each sequence is either arithmetic (adding the same number every term) or geometric (multipling by the same number for every term).
For this we have two formulas:
arithmetic formula: t*n = t*1 + (n-1)d
geometric formula: t*n = t*1 X r^n-1
For these formulas d is the number being added or subtracted and r is the number being multiplied by.
Or you may have to find certain terms.
Fo this you get a formula like t*n = #n -# and you make n the term you're looking for.
Some Examples:
#1: Find the first 5 terms of t*n = 2n - 9
t*1 = 2(1) - 9 = 2 - 9 = - 7
t*2 = 2(2) - 9 = 4 - 9 = -5
t*3 = 2(3) - 9 = 6 - 9 = -3
t*4 = 2(4) - 9 = 8 - 9 = -1
t*5 = 2(5) - 9 = 10 - 9 = 1
#2: 2, 4, 6, 8, 10
t*n = t*1 + (n-1)d
t*n = 2 + (n-1)2
t*n = 2 + 2n -2
t*n = 2n
the chapter is about sequences, which are a listing of numbers.
Each sequence is either arithmetic (adding the same number every term) or geometric (multipling by the same number for every term).
For this we have two formulas:
arithmetic formula: t*n = t*1 + (n-1)d
geometric formula: t*n = t*1 X r^n-1
For these formulas d is the number being added or subtracted and r is the number being multiplied by.
Or you may have to find certain terms.
Fo this you get a formula like t*n = #n -# and you make n the term you're looking for.
Some Examples:
#1: Find the first 5 terms of t*n = 2n - 9
t*1 = 2(1) - 9 = 2 - 9 = - 7
t*2 = 2(2) - 9 = 4 - 9 = -5
t*3 = 2(3) - 9 = 6 - 9 = -3
t*4 = 2(4) - 9 = 8 - 9 = -1
t*5 = 2(5) - 9 = 10 - 9 = 1
#2: 2, 4, 6, 8, 10
t*n = t*1 + (n-1)d
t*n = 2 + (n-1)2
t*n = 2 + 2n -2
t*n = 2n
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