Wednesday, April 20, 2011

Multiplying Polynomials

Hello class unfortunately I forgot to this blog the night i was assigned to but here is what we learned in section 2 of chapter 11.
We first started out by classifying polynomials by the number or terms
  • Monomial- polynomial with one term (ex. 10 or 14x)
  • Binomial- polynomial with two terms (ex. x+2 or 4x^4-7x)
  • Trinomial- polynomial with three terms (ex. x+2+3x^2)
After getting a little off task with the usual 10-15 minute discussion we began examples on the back page of the notes.
(-y+6)(y^4+2y^2-3) first you must foil it by hand which we learned in some older chapters. when you foil that you get -y^6-2y^4+3y^2+6y^4+12y^2-18... then after you reduce that all down you end with -y^6+4y^4+15y^2-18.
But if you would rather not go through all that fun you can simply type it in to your calculator and get the answer fully reduced and without straining yourself in the slightest.
Hello class, yesterday we learned about factoring. There are 3 special cases of factoring
  • Greatest common factor
  • Difference of Perfect Squares
  • Perfect Square Trinomials
Heres an example of a problem in which we need to solve using the Greatest Common Factor


First we need to find the GCF. All of the numbers are multiples of 5 so 5 would be a GCF so it would come outside the parentheses.
We can also see that every one of them has at least one w in it so that would also come outside the parentheses.
We can also see that there is a z in every number so that too would come out of the parentheses and now you divide everything inside the parentheses by 5wz and your product should look like this

Now heres an example problem with Difference of Powers Squared
the form for solving this kind of problem is a^2-b^2=(a+b)(a-b)
so we need to think. what times what equals x^2? well that would just be x
Now what times what equals 100? well that would just be 10.
and do not worry about the 10x and -10x because they cancel each other out.

And here is an example of a problem with perfect Square Trinomials. They are called that because the first and last numbers are perfect squares.
Now what times what equals 9x^2? 3x so,
Now what times what equals 1? 1 so,
and that is ok because when you distribute it 3x plus 3x equals 6x
Now we all know that something times itself is just that something squared so the final answer would look like this,
I hope this blog is useful in understanding this concept. Thanks.
-Tim Kirby

Thursday, April 14, 2011

10.4 (Day 2)

We learned about more Unit Cricle and Radians on 4/8/11

Radian is..
the angle created by a radius along the circumference of a circle.


To convert from radians to degrees: multiply by 180/ㅠ

To convert from degrees to radians: multiply by ㅠ/180

For example,

11ㅠ/ 7

It's radian, so you multiply 180/ㅠ = 282.9 degrees

If you convert degrees to radians, then you leave ㅠ(pie) in answer

I can't put any images on here, because it doesn't work for me.

But, if you want to see completed Unit circle, go to

Based on 30-60-80 traingle and 45-45-90 tirangle,

you can find out the cosine and the sine of angle of unit circle.

For example,

30' has cosine of square root 3 over 2

You can get this by knowing 30-60-90 triangle

cosine is adj/hyp.

SO! it is squre root 3 over 2.

sine of 30' is 1/2.

sine is opp/hyp

So! it is 1/2

It's same way with 45-45-90 triangle.


In Unit circle,

1 & 2 quadrants is sine POSITIVE.

2& 3 quadrants is cosine NEGATIVE.

you have to becareful of this

-Michelle S.

Wednesday, April 6, 2011


1o.4 ( the unit circle definition of Cosine and Sine)

Today we learned about the unit circle. (shown above)

We continued to explore Trig and the uses of Sin and Cos.


X coordinate- Cos

Y coordinate- Sin

the positive degrees on the unit circle goes counter clockwise starting in the right

(x on the picture above)

they go 0-90-180-270-360

opposite if you are moving in the negative direction

check out the notes for a better explination..

Remember the homework was changed to just the top problems on the assignment sheet ( 1,4-7,13,15-17, 21) because we did not finish the lesson and there is a quiz tomorrow :)


April 5, 2011 Period 2 Adv. Algebra, Lesson 10.2

Hello period 2 Advanced Algebra, I'm here to talk about lesson 10.2 that went down in class on April 5. 10.2 (More Right Triangle Trigonometry) was a lesson composed of Inverse Trig Functions, and how to find the measure of a given angle using the inverse of a trig function.

Trig Functions

Inverse Trig Functions

We use the inverse trig functions to solve for an angle.

I.E. if two sides to a right triangle are 9,14. Then you can find the measure of the other two angles by using the inverse of cosine for one angle and the inverse of sine for the other angle.
For angle A, you would plug into the calculator, Cos^1(9/14) and the outcome would be 50. That is your angle measure. You would continue this with the inverse of sine for angle B.

If you are given a right triangle and two sides you can find the rest of the angles using inverse trig functions.

For more information from class just go to moodle and go to Ch. 10 notes and click on 10.2 and you're there.
-Mike s.

Monday, April 4, 2011

bonjour, maintenent, je n'ai pas les chats dans mes pantalons

Howdy-Doo, all. It’s time for some refreshingly correctly punctuated and spelled math commentary, thanks in part to my lovely pal spellcheck. (I spelled the words punctuated, commentary and refreshingly all wrongly just now.)

Anyways…the lesson of 10.1 was presumably an almost a review-type lesson, it being the first of the unit and all. What is this unit that we have begun, one might inquire…. Well Trigonometry of course!

Trigonometry being of course the study of triangles and such, a review of what little trig we have as of yet learned in our geometry course was in order.

Alright, so right triangles are really kind of cool little things in that they have many inherent similarities simply do to the fact that they are constricted in a way so that all of their ratios end up having to be constant despite a change in the size and/or shape of the triangle.

The three special ratios that we have learned to utilize are as such:


The sine ratio describes ratio between the opposite side to the specific angle and the Hypotenuse of the triangle


The Cosine ratio describes the ratio ‘tween the adjacent side to the specific angle and the Hypotenuse of the triangle


The tangential ratio naturally occurring in any given right triangle describes the ratio betwixt the opposite side to the specific angle and the adjacent side to the specific angle

So I am just going to assume that y’all are familiar with the parts of a right triangle, hopefully we all remembered at least this from Gemetry.

Now there are a few more really cool interesting things about right triangles that we learned about today:

For instance, the SIN of any angle will have the equivalent value to the COSIN of said angle’s compliment

Now the next thing that we learned, and this was probably the meat of the lesson, was that you can use these ratios to find missing side lengths in right triangles.

Let me walk you through this example to find the length of the hypotenuse and the last side:

Okay so first we need to decide what ratio we can use to easily find the length of CB

I chose TAN from the 30 angle simply because I like my variable on top of a fraction

So: TAN(30)=

Then multiply by 15

So: 15TAN(30)=x

Then use the calculator to simplfiy and you get: 2.66=x

Thar you go, you all can now solve right triangles using their ratios, Yay!

Have a nice night everybody, I hope nobody is up too late.

Sunday, April 3, 2011

Chapter Eight Review!

On March twenty-third, we reviewed for the upcoming chapter eight test. The previous night's homework was 8.8, so first in class we reviewed that. 8.8 wasn't too hard, but I had some questions about number twelve and number fourteen was on a homework quiz. Next we worked on a worksheet to help us review. Some of the main concepts were simplifying fractions such as number six on the worksheet (for some reason my computer isn't letting me copy/paste from the symbol website >:( )How to do this is to take the complex conjugate of the denominator and multiply both parts of the fraction by it, foil, and solve from there. Another key idea is simplifying multiple-variable nth roots. (such as numbers 15-20 on the worksheet) This is done using the "Jailbreak" method (reference section 8.5). Another concept is solving for all real solutions (numbers 7-11) this isn't too hard, we've been solving for a variable since pre-algebra, but note that for example if you had the fourth root of x that equals ten, you would multiply each to the fourth power and solve from there. Also watch negatives, they might lead to undefined results. A rule to know is that a negative number will never have a real solution if it is to an even root. If it does, it will be undefined. The homework for the day was to do section 8R3, numbers 1-12 and 17-21. Happy spring break!