Reflection15

uhh we didnt have school this week so i guess the things we learn this week i understood. But i have a question about our test .

Do we have to know the Trig Chart again or be able to use it when needed???

reflection14

To find a line you use a formula
Formula for a line:
1) m=tan (alpha)
m=slope
alpha=angle of inclination

Ex>
1. Find the angle of inclination of the line that passes through (4,8) and (7,1).

First find the slope by plugging in the two points into the slope formula, y-y/x-x.
(8-1) divided by (4-7) = -7/3
Plug into formula.
-7/3 = tan (alpha)
alpha = tan^-1(-7/3)

the next thing you do is find quadrants: II & IV
start by plugging inverse into calculator.
66.801 degrees
Make negative and add 180.
113.199 degrees
Make negative and add 360.
293.199 degrees
Angle of Inclination = 113.199 degrees, 293.199 degrees

reflection 15

soz, the holidays were.......BLAH...........VERY TIRING. woohoo for sleepin late. and the football team lost :( and turkey makes u very tired!
anyways, we didnt rly have school this week, so thers nothing i didnt or did understand this week :P

Reflection 14

Ok. Now we have trig involving lines and conices. The basic formula for a line in this section is m=tan(fish)
m=slope
(fish)= angle of inclination.

And don't forget that when you are finding an invers that you are finding two angles. And don't forget that you must also check the quadrants. The positive quadrants for tangent are one and three. Making two and four the negative quadrants for tangent.

I'll use the first example she gave us in class: 2x+5y=15 find the angle of iniclination (fish). Find your slope. Slope=-A/B so that is -2/5 referring to the standard formula Ax+By=C. now (-2/5)=tan(fish) since we dont divide by trig functions you take an inverse of tangent to find your angle. With your calc the inverse tan of -2/5 is -21.801. Ignore the negative so that you have 21.801. We need this answer in the quadrants where tan is negative. Because our slop was negative. To move 21.801 to quadrant two you make it negative then add 180. To move it to the fourth quadrant you make it negative then add 360. Those are your two angles of inclination and equal fish.

The basic formula for a conic is tan(2x(fish))=B/(A-C)

To make life easier you should remember that if A=C then (fish)=Pi/4

The basic formula for a conic is Ax^2 + Bxy +Cy^2 +Dx + Ey + F=0
Learn that. Its vital.

reflection 14

wow, this week was very.......bad
drama! was the main focus of the week, but meh, got through it, not much really happened tho, except we on holiday break now

anyway, here's the math part:

one thing i did understand this week was the formula for a conic: tan 2(alpha)=B/A-C

but one thing i didnt understand was the whole, graph thing and where u have to move it and stuff, any help?

Reflection #14

Well, to begin, I understand the formulas we learned:

Formula for a line:
1) m=tan alpha
m=slope
alpha=angle of inclination

Examples:
1. Find the angle of inclination of the line that passes through (4,8) and (7,1).
First find the slope by plugging in the two points into the slope formula, y-y/x-x.
8-1/4-7 = -7/3
Plug into formula.
-7/3 = tan alpha
alpha = tan^-1(-7/3)
Find quadrants: II & IV
Plug inverse into calculator.
66.801 degrees
Make negative and add 180.
113.199 degrees
Make negative and add 360.
293.199 degrees
Angle of Inclination = 113.199 degrees, 293.199 degrees

Now, what I don't understand are the graphs. I'm having some trouble with that. I think I'm okay with looking at a graph and forming an equation. But I'm a little iffy on drawing my own graph. Any help? An example would be great, though I know it's hard to show graphs on here.

REFLECTION #14

Another week has gone by. Thank goodness!!! I've been ready for the holidays to get another break from math. hah. Well, not so much, because of course we have our bridges that we have to build. And I dont know about you guys, but I'm not much of a construction person. Anyway this week didnt go by fast for me, but it was alright I guess. I don't completely understand the new things we learned this week. Hopefully we review them a good bit when we get back from the holidays, because I know I'll completely forget all the stuff by the end of this week. And I'm also glad to say that I didn't do half bad on my Chapter 9 test that we took last week. To be honest, I was expecting a much worse grade than I got. P.S. Thank you Edee!! (He taught me how to work all the word problems. and I actually understood most of them. ha) Oh and just throwing this out there, but I left my advanced math binder at school, so I'm just going straight off the top of my head. (which is losing information verrrryyy quickly because of holiday things and whatnot. ha) So first of all, I believe this week we started Chapter 8. I remember reviewing some things from Chapter 7 also. Like dealing with trig inverses and things like that. Those come pretty easy to me now. Now for what we started learning for chapter 8. We learned a couple formulas that you use in different situations. (I havent really figured out when to use those without looking at my notes, sooooo thats something to work on) Then we learned how to find the angle of inclination, while we also learned about amplitude, vertical shifts, and period.

So here is an example:
(*By the way, if i did it wrong, dont blame me. remember I don't really have a clue what i'm doing, i left my binder at school, and I just made up this problem) So I appologize if this isn't correct. but i tried :)

Ex. 1.) Find the angle of inclination for the following line equation: 10x + 5y = 15

*Okay if i remember correctly, the first thing you want to do is find your slope. which is -2 when you solve the linear equation.
then you use your formula >> m = tan alpha
so you get>> tan alpha = -2
then the inverse of tan (2) (since you dont enter a negative in your calc) is 63.435 degrees
then you check off the quadrants in which tangent is negative (which would be the 2nd and 4th)
then you take 63.435 make it negative and add 180 and get >> 116.565 degrees
then you take 63.435 make it negative and add 360 and you get >> 296.565 degrees

so tan alpha is approximately 116.565 degrees and 296.565 degrees
(that is if i did this right) hope so. ha

anyway, for what i dont understand is pretty much everything else that we learned. considering i dont remember any of it. so if anyone would like to jog my memory and kinda explain to me what else we learned this past week that would be great :)

reflection 14

this week went by extremely fast for some reason. and now its thanksgiving break and im sure that everyone is excited about that. but dont forget about building your bridges..
anyway this week we learned more formulas about trig and other stuff.
heres some formulas.....

for lines:
m=tan where m=slope, alpha=angle of inclination


for conics:
tan 2(alpha)=B/A-C

also for conics, if A=C then alpha=pi/4


example:
find the angle of inclination...

2x+5y=15
m=-2/5
tan=-2/5
alpha=tan^-1(-2/5)

alpha=158.199degrees, 338.199degrees

Reflection

This past week felt soo long and tiring. Of course because it was the week before the holidays. But anyways, we started learning new stuff from chapter 8. We learned how to find the angle of inclination, which i found was really easy compared to some stuff we learn. We also learned about amplitudes, periods, vertical shifts, etc. There are some formulas we had to learn to be able to work these problems:

1.) For a line
m=tan alpha where m=slope and alpha=angle of inclination
2.) For a conic
tan 2 alpha=B/A-C
3.) For a conic if A=C then
a=pi/4

EXAMPLE:

Find the angle of inclination.
2x+5y=15
m=-2/5 tan alpha=-2/5 Checks are in the II and IV area and 21.801 degrees in I
alpha=tan^-1(-2/5)
180-21.801 alpha ~ 158.199 degrees, 338.199 degrees
158.199+180