Another thing we learned a while back is ellipse and this how you work and after graph it.
The equation: ((x-h)^2/(length of x/2)^2) + ((y-k)^2/(length of y/2)^2)
Center: (h, k)
Major axis has the larger denominator
Vertex is on major axis
focus: smaller # = larger # - focus^2
focus is on major axis, also
How to graph:
1. center
2. major axis - bigger demoninator x or y
3. vertex is plus or minus sqrt of big denominator
4. other intercepts is plus or minus sqrt of small denominator
5. focus
6. length of major axis is 2 times the sqrt of big denominator
7. length of minor axis is 2 times the sqrt of small denominator
8. then graph it
Saturday, January 2, 2010
Reflection #19
Okayyyyyy, this is one of my holliday blogs. I really don't know what to talk about, so i'm just going t go back and review my notes and give some examples. hah.
_________________________________________________________________________
quadrant rules:
I to II---make it negative and add 180 degrees
I to III---add 180 degrees
I to IV---make it negative and add 360 degrees
II to IV---add 180 degrees
^^when you are supposed to find an inverse, you are looking for two angles. If you are given the inverse of cos(1/2), you just plug cos^-1(1/2) into your calculator and you get an angle. Cosine is positive in the first and fourth quadrant. So when you get the first angle when you plug it into your calculator, you make it negative and add 360. Then you have both of your angles. And both of those angles are your answer.
_________________________________________________________________________
I don't really get the amplitude problems. I tried some of them on the exam, but i did not really understand how to graph them.
Amplitude = how high the graph is
period = how long it takes for the graph to repeat
p/4 = what you add to each value to eventually get the period
phase shift = horizontal shift of the whole graph
vertical shift = up and down movement of the whole graph
^^I know how to get this part, but after that, i'm stuck. Can someone please help me do one of these problems? I would greatly appreciate it. Thankkssss :)
_________________________________________________________________________
quadrant rules:
I to II---make it negative and add 180 degrees
I to III---add 180 degrees
I to IV---make it negative and add 360 degrees
II to IV---add 180 degrees
^^when you are supposed to find an inverse, you are looking for two angles. If you are given the inverse of cos(1/2), you just plug cos^-1(1/2) into your calculator and you get an angle. Cosine is positive in the first and fourth quadrant. So when you get the first angle when you plug it into your calculator, you make it negative and add 360. Then you have both of your angles. And both of those angles are your answer.
_________________________________________________________________________
I don't really get the amplitude problems. I tried some of them on the exam, but i did not really understand how to graph them.
Amplitude = how high the graph is
period = how long it takes for the graph to repeat
p/4 = what you add to each value to eventually get the period
phase shift = horizontal shift of the whole graph
vertical shift = up and down movement of the whole graph
^^I know how to get this part, but after that, i'm stuck. Can someone please help me do one of these problems? I would greatly appreciate it. Thankkssss :)
Reflection #20
AHHHHH, i don't want to go back to school. I love sleeping in late, and not having to worry about homework and other school stuff. Now i can't wait until the Mardi Gras Break, lol. Okayyyyy, this is my second holliday blog, hah. I am so glad that we do not have to do comments, and we just have to do the blogs. Everybody is probably waiting until the last day anyway, hah. I am so glad i spaced my blogs out. I really didn't feel like doing them all in one night. Anyways, i don't know what else to talk about, so i am just going to give more examples.
_______________________________________________________________________
Here is something that i did understand on the exam. The functions:
Adding or subtracting functions:
(f+g)(x)
f(x)+g(x)
(f-g)(x)
f(x)-g(x)
example:
f(x)=x+1
g(x)=x-4
(f+g)(x)= x+1+x-4
(f+g)(x)=2x-3
Multiplying functions:
(fxg)(x)
f(x) x g(x)
f(x)=2x+4
g(x)=4x-6
(2x+4)(4x-6)
F.O.I.L.= (8x^2-12x+16x-24)
=8x^2+4x-24
Finding a function of another function:
(fog)(x)=
f(g(x))
f(x)= 2x+4
g(x)= x-4
plug in the g(x) equation into the f(x) equation where all of the x's appear.
2(x-4)+4
distribute.
2x-8+4
(fog)(x)=2x-12
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I understood this right away, when we first learned this. It wasn't that difficult. I still need help with the domain and range problems if anyone wants to help me. I would greatly appreciate it. I really need to learn how the domain and range problems work, hah. I don't know why i can not grasp the concept of domain and range. But if you want to show me a problem, go ahead. THANKSSSSS :)
_______________________________________________________________________
Here is something that i did understand on the exam. The functions:
Adding or subtracting functions:
(f+g)(x)
f(x)+g(x)
(f-g)(x)
f(x)-g(x)
example:
f(x)=x+1
g(x)=x-4
(f+g)(x)= x+1+x-4
(f+g)(x)=2x-3
Multiplying functions:
(fxg)(x)
f(x) x g(x)
f(x)=2x+4
g(x)=4x-6
(2x+4)(4x-6)
F.O.I.L.= (8x^2-12x+16x-24)
=8x^2+4x-24
Finding a function of another function:
(fog)(x)=
f(g(x))
f(x)= 2x+4
g(x)= x-4
plug in the g(x) equation into the f(x) equation where all of the x's appear.
2(x-4)+4
distribute.
2x-8+4
(fog)(x)=2x-12
__________________________________________________________________
I understood this right away, when we first learned this. It wasn't that difficult. I still need help with the domain and range problems if anyone wants to help me. I would greatly appreciate it. I really need to learn how the domain and range problems work, hah. I don't know why i can not grasp the concept of domain and range. But if you want to show me a problem, go ahead. THANKSSSSS :)
Friday, January 1, 2010
Reflection 18 I Think
For my first holiday blog I am going all the way back to the begging weeks and show how the i chart works. It is very simple if anybody has forgotten.
This is the i chart:
.25 = i
.5 = -1
.75 = -i
1or any whole number = 1
Example: i^498+i^185+i^2003+i^16+i^89=
So first step you divide each exponent by 4: i^498=124.5, i^185=46.25, i^2003=500.75, i^16=4, i^89= 22.25
Next step look at the decimals or whole numbers you got and plug them into the chart and next into the equation : (-1) + i + (-i) + 1 + i =
This is the i chart:
.25 = i
.5 = -1
.75 = -i
1or any whole number = 1
Example: i^498+i^185+i^2003+i^16+i^89=
So first step you divide each exponent by 4: i^498=124.5, i^185=46.25, i^2003=500.75, i^16=4, i^89= 22.25
Next step look at the decimals or whole numbers you got and plug them into the chart and next into the equation : (-1) + i + (-i) + 1 + i =
Monday, December 28, 2009
Reflection #19
This week, I'm going to review the areas of different triangles:
Area of a right triangle: A=(1/2)(base)(height)
Area of a non-right triangle: A=(1/2)(leg)(other leg)sin(angle in between the two legs)
Examples:
visualize a right triangle: DEF (left to right)
D=90 degrees, d=8, e=6...Find the area.
So, you know your base is e, which is 6.
Now, to find your height, or f, use: a^2 + b^2 = c^2
c is always to hypotenuse, which is d (8).
6^2 + b^2 = 8^2
36 + b^2 = 64
b^2 = 28
b = 5.292
Your height is 5.292
Now, plug into the formula.
A = (1/2)(6)(5.292)
A = 15.875
Now, visualize a non-right triangle: HIJ (left to right)
H = 65 degrees, j = 2, i = 6
Plug into the formula.
A = (1/2)(2)(6)sin(65)
A = 5.438
Area of a right triangle: A=(1/2)(base)(height)
Area of a non-right triangle: A=(1/2)(leg)(other leg)sin(angle in between the two legs)
Examples:
visualize a right triangle: DEF (left to right)
D=90 degrees, d=8, e=6...Find the area.
So, you know your base is e, which is 6.
Now, to find your height, or f, use: a^2 + b^2 = c^2
c is always to hypotenuse, which is d (8).
6^2 + b^2 = 8^2
36 + b^2 = 64
b^2 = 28
b = 5.292
Your height is 5.292
Now, plug into the formula.
A = (1/2)(6)(5.292)
A = 15.875
Now, visualize a non-right triangle: HIJ (left to right)
H = 65 degrees, j = 2, i = 6
Plug into the formula.
A = (1/2)(2)(6)sin(65)
A = 5.438
Sunday, December 27, 2009
Holiday Reflection
So heres a refelection on the law of sines and cosines and the formula for the area of an isosceles triangle.
Law of Sines: sin(opp. angle)/Leg=sin(Opp. angle)/Leg. Set up a proportion.
Example. triangle ABC where A=36 degrees a=3 and B=56 degrees. find b
sin36/3=sin56/x 3sin56/sin36= x Plug into your calculator and done.
Law of Cosines:
(opposite leg)^2=(adjacent leg)^2+(other opposite leg)^2-2(leg)(leg)cos(angle in between)
example:
for a triangle with C=36 degrees a=5 b=6
c^2=5^2+6^2-2(5)(6)cos36
c=Square root of(25+36-2(5)(6)cos36)
c= 3.53
Are of non-right triangle:
1/2(leg)(leg)sin(angle between)
Example: Triangle ABC has sides a=5 b=3 and C=40 degrees
therefore your formula will be 1/2)(5)(3)(sin(40)) type that in your calculator and you will get the answer.
Law of Sines: sin(opp. angle)/Leg=sin(Opp. angle)/Leg. Set up a proportion.
Example. triangle ABC where A=36 degrees a=3 and B=56 degrees. find b
sin36/3=sin56/x 3sin56/sin36= x Plug into your calculator and done.
Law of Cosines:
(opposite leg)^2=(adjacent leg)^2+(other opposite leg)^2-2(leg)(leg)cos(angle in between)
example:
for a triangle with C=36 degrees a=5 b=6
c^2=5^2+6^2-2(5)(6)cos36
c=Square root of(25+36-2(5)(6)cos36)
c= 3.53
Are of non-right triangle:
1/2(leg)(leg)sin(angle between)
Example: Triangle ABC has sides a=5 b=3 and C=40 degrees
therefore your formula will be 1/2)(5)(3)(sin(40)) type that in your calculator and you will get the answer.
REFLECTION #19
Time for another blog! yay. Hope everyone is enjoying their holidays because I'm not (: Anyway, I think I'll just review the first things that pop into my head. This should be quite interesting...
Well, let's see I think way back in like Chapter 3 or something we learned inequalities so I'll start off with something easy like that. hah. To start off, there are two types of inequalities, regular inequalities and absolute value inequaties. In absolute value inequalities there are two different types determined by the signs if it is not an equal sign. They can either be "and" or "or" inequalities. "And" inequalities always have a less than symbol (<) and "or" inequalities always have a greater than symbol (>). Also, (something important to remember) in absolute value inequalities you always get two answers. I'm sure everyone remembers how to work these problems but here's a few examples:
Ex. 1.) 4x + 1 > 13
*you would just solve this like a regular equation even though it does not have an equal sign. *and for this problem you would only get one answer because it does not have an absolute value symbol.
*so you subtract 1 from 13 and you get 4x > 12
*divide by 4 and you get that x > 3
Ex. 2.) 2y - 4 = 12 (*assume there is an absolute value thing around 2y-4)
*the first thing you want to do with this problem is set up two equations. since in absolute value you know you will get two answers
*your first equation will be the original equation but without the absolute value symbol like this:
2y - 4 = 12
*and your second equation will be the same equation except you change the 12 to -12 like this:
2y - 4 = -12
*then you solve both equations for y and you get that y = -4 and 8
Ex. 3.) 3x - 4 + 5 < 27 (*assume there is an absolute value thing around only 3x - 4)
*the first thing you have to do is isolate the absolute value and to do that you have to get rid of the 5. so you subtract the 5 over from the 27 and your new equation is:
3x - 4 < 22
*the next thing you notice is that this equation has a < sign and that means it is an "and" equation so you have to set it up a certain way like this:
-22 < 3x - 4 < 22
*then you solve the equation first by adding 4 over to all sides then you get this:
-18 < 3x < 26
*then you divide by 3 on all sides
your final answer is: -6 < x < 26/3
Next I think I will go over slope. First of all, there are 3 different equations of slope. There's slope intercept form, point slope form, and standard form. Slope intercept is y = mx + b. Point slope form is y - y1 = m (x - x1). And standard form is Ax + By = C
Here are a few examples of problems:
Ex. 4.) Find the slope of the two points (4,1) (3,0)
*to solve this you use the formula for slope which is: m = y2 - y1/x2 - x1
*so you get...0 - 1/3 - 4 and that gives you -1/-1 which equals 1
Ex. 5.) Find the equation of the line 3x + 4y = 12 that is perpendicular to the point (3,2).
(*by the way, I'm not sure if I worded that problem right but hopefully you should know what I'm talking about right?)
*so the first thing you have to do is find the slope of the equation you are given. 3x+4y=12
*first you subtract 3x over and you get 4y = -3x + 12
*then you divide by 4. and you get that y = -3/4x + 3
*So your perpendicular slope of that equation is 4/3. (because you take the negative reciprocal of the orginal slope of the equation)
*Then to put it in an equation including the point you are given, you use the point slope formula.
*So your final answer is y - 2 = 4/3 (x - 3)
Umm..the last thing I can think of at the moment is reference angles and exact answers. *Remember that reference angles can only be between 0 and 90 degrees. Here are some examples:
Ex. 6.) Find the reference angle of sin 236. (*assuming that there's a degree sign after 236)
*the first thing you have to do is figure out what quadrant 236 is in. it's in the 3rd quadrant.
*sin relates to the y axis and in the third quadrant, the y axis is negative so for the reference angle, sin has to be negative
*and to find the reference angle of 236 you have to subtract 180. and you get 56 degrees.
*so your reference angle is equal to -sin 56. and since this is not on the trig chart, to find the exact answer you would have to type it into your calculator.
Okay well I think I reviewed enough for this week. ha
Well, let's see I think way back in like Chapter 3 or something we learned inequalities so I'll start off with something easy like that. hah. To start off, there are two types of inequalities, regular inequalities and absolute value inequaties. In absolute value inequalities there are two different types determined by the signs if it is not an equal sign. They can either be "and" or "or" inequalities. "And" inequalities always have a less than symbol (<) and "or" inequalities always have a greater than symbol (>). Also, (something important to remember) in absolute value inequalities you always get two answers. I'm sure everyone remembers how to work these problems but here's a few examples:
Ex. 1.) 4x + 1 > 13
*you would just solve this like a regular equation even though it does not have an equal sign. *and for this problem you would only get one answer because it does not have an absolute value symbol.
*so you subtract 1 from 13 and you get 4x > 12
*divide by 4 and you get that x > 3
Ex. 2.) 2y - 4 = 12 (*assume there is an absolute value thing around 2y-4)
*the first thing you want to do with this problem is set up two equations. since in absolute value you know you will get two answers
*your first equation will be the original equation but without the absolute value symbol like this:
2y - 4 = 12
*and your second equation will be the same equation except you change the 12 to -12 like this:
2y - 4 = -12
*then you solve both equations for y and you get that y = -4 and 8
Ex. 3.) 3x - 4 + 5 < 27 (*assume there is an absolute value thing around only 3x - 4)
*the first thing you have to do is isolate the absolute value and to do that you have to get rid of the 5. so you subtract the 5 over from the 27 and your new equation is:
3x - 4 < 22
*the next thing you notice is that this equation has a < sign and that means it is an "and" equation so you have to set it up a certain way like this:
-22 < 3x - 4 < 22
*then you solve the equation first by adding 4 over to all sides then you get this:
-18 < 3x < 26
*then you divide by 3 on all sides
your final answer is: -6 < x < 26/3
Next I think I will go over slope. First of all, there are 3 different equations of slope. There's slope intercept form, point slope form, and standard form. Slope intercept is y = mx + b. Point slope form is y - y1 = m (x - x1). And standard form is Ax + By = C
Here are a few examples of problems:
Ex. 4.) Find the slope of the two points (4,1) (3,0)
*to solve this you use the formula for slope which is: m = y2 - y1/x2 - x1
*so you get...0 - 1/3 - 4 and that gives you -1/-1 which equals 1
Ex. 5.) Find the equation of the line 3x + 4y = 12 that is perpendicular to the point (3,2).
(*by the way, I'm not sure if I worded that problem right but hopefully you should know what I'm talking about right?)
*so the first thing you have to do is find the slope of the equation you are given. 3x+4y=12
*first you subtract 3x over and you get 4y = -3x + 12
*then you divide by 4. and you get that y = -3/4x + 3
*So your perpendicular slope of that equation is 4/3. (because you take the negative reciprocal of the orginal slope of the equation)
*Then to put it in an equation including the point you are given, you use the point slope formula.
*So your final answer is y - 2 = 4/3 (x - 3)
Umm..the last thing I can think of at the moment is reference angles and exact answers. *Remember that reference angles can only be between 0 and 90 degrees. Here are some examples:
Ex. 6.) Find the reference angle of sin 236. (*assuming that there's a degree sign after 236)
*the first thing you have to do is figure out what quadrant 236 is in. it's in the 3rd quadrant.
*sin relates to the y axis and in the third quadrant, the y axis is negative so for the reference angle, sin has to be negative
*and to find the reference angle of 236 you have to subtract 180. and you get 56 degrees.
*so your reference angle is equal to -sin 56. and since this is not on the trig chart, to find the exact answer you would have to type it into your calculator.
Okay well I think I reviewed enough for this week. ha
reflection 19
This week over the holidays im going to review chapter 9, which is right triangles. There are 3 things that you have to remember: hypotenuse opposite, A=1/2bh, and SOHCAHTOA, which is:
sin=opposite leg/hypotenuse
cos=adjacent leg/hypotenuse
tan=opposite leg/adjacent leg
Here are some examples on using SOHCAHTOA....
solve for b & c.
1) You have an ABC triangle with angle A 28 degrees, a=40, and angle C is a right triangle.
tan28=40/b
btan28=40
b=40/tan28
which equals about 75.229
sin28=40/c
c=40/sin28
which equals about 85.202
sin=opposite leg/hypotenuse
cos=adjacent leg/hypotenuse
tan=opposite leg/adjacent leg
Here are some examples on using SOHCAHTOA....
solve for b & c.
1) You have an ABC triangle with angle A 28 degrees, a=40, and angle C is a right triangle.
tan28=40/b
btan28=40
b=40/tan28
which equals about 75.229
sin28=40/c
c=40/sin28
which equals about 85.202
Reflection 19
Today over the holidays i'm going to review the trig chart and the unit circle, i bet remembering a few of those could of helped me get a few more points on my exam.
Wow its amazing how much or little i forgot of that stuff.
Now the steps in solving an inverse is used to solve for an angle, you cant divdie by a trig function. An inverse will find 2 angles.
The steps determine the quad of your two angles, use the trig chart to find a reference angle, if not on chart or unit circle use the calcualter.
S=rtheta, remember that. SOH CAH TOA is a very helpful tool in solving right triangles. Law of sines is sina/a=sinb/b=sinc/c. Keep enjoying your holidays!!!!!!!!!!!!!!!!!!!!!
Wow its amazing how much or little i forgot of that stuff.
Now the steps in solving an inverse is used to solve for an angle, you cant divdie by a trig function. An inverse will find 2 angles.
The steps determine the quad of your two angles, use the trig chart to find a reference angle, if not on chart or unit circle use the calcualter.
S=rtheta, remember that. SOH CAH TOA is a very helpful tool in solving right triangles. Law of sines is sina/a=sinb/b=sinc/c. Keep enjoying your holidays!!!!!!!!!!!!!!!!!!!!!
Reflection 18
Yea school is over and we get our long two week break. I am looking forward for this break cause i need to have a break. I plan to do a lot of hunting and fishing, and chilling with my friends. Ive been working on a bonfire as well. Christmas night went well to, i enjoyed myself alot and my break has been doing a lot of good for me. But to keep my brain fresh in mind i will review a few math notes that we took over the first semester. To find the intersection you must put them equal to. Example:y=2x+5 y=8-x^2 2x+5=8-x^2 x^2=2x-3=0 (x+3)(x-1)=0 so x=1,-3.
To sketch polynomial function you must first factor completely, set up a # line lable zeros, plug in on either side of your root, posisitve intergers above x-axis, negative below x-axis, check in calculater, max in min calc only you find the vertex which is -b/2a. I also though the exam was kinda haed but then it was easy i just 4got how to do a bunch a simple problems that i'm kicking myself in the ass for.
To sketch polynomial function you must first factor completely, set up a # line lable zeros, plug in on either side of your root, posisitve intergers above x-axis, negative below x-axis, check in calculater, max in min calc only you find the vertex which is -b/2a. I also though the exam was kinda haed but then it was easy i just 4got how to do a bunch a simple problems that i'm kicking myself in the ass for.
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