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dev problem # 1     https://prezi.com/i7crpme7xrwm/1999-radian/





dev problem # 2    https://prezi.com/zpwxj4rasp8y/dev/

D.E.V Project

I decided to choose these problems because I wanted to review them before the final exam. I also chose them because I believe I did fairly well when learning how to solve them in class. I now believe I am more prepared for the exam as well as precalc classes in the future. I think precalc trig has helped me a lot. I learned a lot of new concepts that I didn't know about before this class. 

D.E.V. Project

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DEV

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D.E.V- Austyn Killebrew & Allena Mulville

Math Puns are the first SINE of madness! :) Hope you enjoy www.faqmath.weebly.com

D.E.V

Rationals problem

7 5

4x+3 - x+4   

----------------------  

4x+3  + x+4

2 4x+3

1) The first thing you want to do is get a common denominator for the top set of fractions.

2) The common denominators for this first set are (4x+3) and (x+4).

3) Now what you’ll have to do is multiply the numerators by the same factor. So, it would be

7 * x+4, which equals, 7x + 28 because you have to distribute thoroughly. Now you’ll do the same on the other side. So you’ll take the -5 * 4x+3 = -20x -15.

4) You now can combine the numerators all over the same denominator. As shown below:

7x + 28 - 20x -15       

   (4x+3)(x+4) Remember to simplify as much as possible!

5) -13x + 13  

      (4x+3)(x+4)   Is now what you are left with on the top! We are halfway done.

6) We can start by repeating step one. Try and get a common denominator for the fractions.

7) Let’s take 4x+3 and multiply by 4x+3

1. 4x*4x = 16x^2
2. 4x*3 = 12x
3. 4x*3 = 12x
4. 3*3 = 9 So, we end up with, 16x^2 + 24x + 9

8) The next thing you’ll want to do is multiply x+4 by 2.

     x+4 * 2 = 2x + 8.

9) Combine the 16x^2+24x+9 and 2x+8 over the common denominator.

     16x^2+26x+17

            2(4x+3)

10) You now can combine it into one whole fraction! --> 

   -13x +  13

      (4x+3)(x+4)

--------------------------

    16x^2+26x+17

         2(4x+3)

11) To undo division, you can now multiply by the reciprocal.

12)  -13x+13                2(4x+3)

   (4x+3)(x+4)     *  (16x^2+26x+17)    

13) Notice that we have a common factor of 4x+3 in both the numerator and denominator allowing us to get rid of it because they cancel each other out.

14) -13x+13              2

        (x+4)     *   (16x^2+26x+17)  

15) We have to make sure we factor everything out completely.

we can factor out a -13 from -13x+13 leaving us with -13(x-1)

16) -13(x-1)              2

        (x+4)   *  (16x^2+26x+17)

17) All that is left to do now is multiply straight across leaving you with the final answer of:

-26(x-1)

(x+4)(16x^2+26x+17).

An Extreme solving problem

12 - 27sin(3Θ+5π/4) = 21 - 9sin(3Θ+5π/4)

1) We can start solving by first subtracting 21 from both sides.

-9 - 27sin(3Θ+5π/4) = -9sin(3Θ+5π/4)

2) The next step is to add 27sin(3Θ+5π/4 ) to -9sin(3Θ+5π/4).

-9 = 18sin(3Θ+5π/4)

3) Now that we almost have sin alone, our next step is to divide the -9 by 18, which will leave us with → -1/2 = sin(3Θ+5π/4)

4) Now you find all values that make this statement true. Where on the unit circle is sin -1/2? Well it happens to be at 7π/6 and 11π/6.

5) What you want to do next is take the radian measures you found and add or subtract 2πn and then set it equal to our theta expression in our parenthesis.

1. 11π/6 +/- 2πn= 3Θ+5π/4

6) We are trying to get theta by itself so the first thing you do is subtract 5π/4 from 11π/6 by getting common denominators.

1.     2*  11π/6   -    5π/4  *3  

2.      22π/12 -   15π/12 =  7π/12

7) So now you can take 7π/12 +/- 2πn= 3Θ and divide it by 3 to get theta alone.

8) After dividing by 3 you should end up with, 7π/36 +/- 2/3πn.

9) One possible solution is 7π36. To find more solutions (this can be done in a calculator or just on paper) you can take (7/36) and add (1/2) which equals (25/36). 2π is equivalent to 72π/36 so you find as many solutions as possible without going over 2π.

1. (25/36) + (1/2) = (43/36)
2. (43/36) + (1/2) = (61/36)

10) Our last solution for this particular radian would be 61/36 because if you add another 1/2 you exceed 72π/36.

11) Now we can start the other radian measure.

1. 7π/6 +/- 2πn = 3Θ+5π/4

12) We are aiming at getting the theta value by itself so we’re going to subtract 5π/4 from 7π/6 by getting common denominators.

1.       2* 7π/6   -   5π/4 *3

2.      14π/12 -  15π/12 = -π/12

13) Now you can take your radian measure of -π/12 +/- 2πn = 3Θ and divide it by 3 to get theta alone.

14) After you have divided by 3 you should end up with, -π/36 +/- 2/3πn.

15) You already have one of your solutions, that being -π/36, to find more solutions all you have to do is add 1/2 to your previous solution but keep in mind that 2π is equivalent to 72π/36.

1. (1/36) + (1/2) = (19/36)
2. (19/36) + (1/2) = (37/36)
3. (37/36) + (1/2) = (55/36)

16) Our last solution is 55π/36 because if we were to add another 1/2 we would exceed our limit of 2π.

17) So all of the possible solutions for this particular problem are:

7π/36, 25π/36, 43π/36, 61π/36, -π/36,19π/36, 37π/36, and 55π/36.

I chose to do these problems because I had a hard time understanding them when we first went over them. There are so many steps in figuring out a rationals problem. You have to be careful and make sure you completely simplify before you take further steps. I would often forget that to undo multiplication you would have to multiply by the reciprocal of the fraction and it would get me every time. I finally got it though! I also chose to do an extreme solving problem because I remembered how annoying and long and hard they were. It's extremely rewarding going through and being able to create and solve a problem you once thought was nearly impossible. Anyways, I'd also like to thank you for always being there for me when I needed it! See ya! :)