CPhill

44.

We have that    y  = Acos (Bθ)

If the max is 4  and the minimun  is -4, the amplitude   =  4  =  A

To find  B, we can solve this

2pi / (2pi / 3)  = B

3 * 2pi  / 2pi  = B

3  = B

So....the function is

y   = 4 cos 3θ

45.  tan  3pi/4   =   - tan pi/4  =    -1

42.

s  = r  * (theta in radians)   ...so we have

s   = 3  *  (5pi/3)    =  5pi ≈  15.7 in

43.   Here's the graph....

https://www.desmos.com/calculator/9btjsjrpqx

40.

Amplitude  =  l  -5  l    =  5

Period....we can solve this......   2pi  / 6  = period   = (1/3) pi    

Last answer

41.  cos 315°  is a 45° reference angle in the 4th quadrant....and the cosine is positive in this quadrant.. so..

  cos 45°   =   √2 / 2

0⁰  is an indeterminant form...here's why ...

Note...if we let n be any real number...we have

0⁰  =  0 ^n-n  =  0ⁿ  / 0ⁿ   =    0  / 0

But also note that    0  / 0   =  pi        or  0 / 0    =  11   or   0 / 0  = e  ......  etc....

Thus....we  can't actually determine  the value  of  0^n-n  = 0⁰  

2x^2 + 15x  + 25

Here's a method that often works for problems like these...it's a little guesswork....but sometimes it pays off

I'm guessing that we can   write  15x   in this  manner......  10x  +  5x

2x^2 + 10x  + 5x  +  25

I see that  the first constant multiplied by 5 will give me the second constant.....and the third constant  multiplied by 5  will give me the last constant...so......let's factor. the first two  terms and the last two  terms....   [factoring  by "grouping " ]

2x ( x + 5 )  + 5 ( x + 5).....  note that  ( x + 5)  is common

So......the factorization  is

( x + 5 )  ( 2x + 5)

We can use the Law of Sines to answer this

sin R  / 30   = sin 96° / 54        multiply both sides by  30

sin R  = 30 * sin 96° / 54

To find  "R"....we  need to use the  sine inverse   ( arcsin )

arcsin [ 30 * sin (96°) / 54  ]  =  "R"   ≈ 33.54°

48.....We can use the tangent inverse to find this

atan (5.67)  =  79.997°  ≈   80°

49.

sin^2  θ                 1   - cos^2  θ            ( 1 - cos  θ) ( 1 + cos  θ)                   1  +  cos  θ

_________   =      __________    =      ____________________   =  

1  - cos  θ               1  - cos  θ                (1  -  cos  θ)

Cubing the  2  and putting it back under the root   would result  in  2^3  = 8

Cubing  the a^3  and putting it back under the root would result  in  (a^3)^3  =  a^9

Cubing  the b^2  and putting it back under the root  would result in  (b^2)^3  = b^6

So....we  must have had

³√  [ 8 * 12 * a^9 * b^6 * b  ] = 

³√ [ 96 * a^9  * b^7 ] 

Does that make some sense, LF  ????

4.

Remember that 

Area  = Width  * Length   ......so we have this.....

[8x^3 + 10x^2 - x  - 2]   =  [2x  + 1]   *   Length     ....so...dividing both sides by   2x  + 1  , we have

[ 8x^3  + 10x^2  - x  - 2   ] / [2x + 1  ]  =  Length 

We can find the length with some  polynomial long  division  (yuck !!!)

               4 x^2  + 3x   - 2

2x + 1  [  8x^3  + 10x^2   -  x   - 2  ] 

               8x^3  +  4x^2

            _______________________

                           6x^2  -    x 

                           6x^2  + 3x

                          ______________

                                      -4x   -  2

                                      -4x   -  2

                                     _________

                                                  0

The Length is the polynomial shown in red   ⇒  "C"