CPhill

8.8

sin (7pi /12)   =

sin (pi/4  + pi/3) =

sin (pi/4) cos (pi/3)  + sin(pi/3)cos(pi/4)  =

√2/2 * 1/2  +  √3/2 * √2/2  =

√2/4 + √6/4 =

[ √2 + √6 ] / 4

f(x)  =  x^2

Proof

f(g)  =

f([2+x^9]/[2-x^9] )  =   

([2+x^9]/[2-x^9] )^2  =  h(x)

BTW...the answer may not  be unique.........

A ( x^2 + x + 1) + (Bx + C) x   = 4x^2 + 3

Ax^2 + Ax + A  + Bx^2 + Cx  = 4x^2 + 3

(A + B)x^2 + ( A + C)x + A  =  4x^2 + 0x +  3       equate coefficients

A + B = 4

A + C  = 0

A  = 3     →    C  = - 3  →  B  = 1

Proof 

3 / x   + [1x -  3] / [ x ^2 + x + 1] =

[3 (x^2 + x + 1)  + x (1x - 3) ] / [ x ( x^2 + x + 1 ]  =

[ 3x^2 + 3x + 3 + 1x^2 - 3x ]  / [ x^3 + x^2 + x]  =

[4x^2 + 3] / [x^3 + x^2 + x]

L = 2W - 3    so

464  = L * W

464 = [2W - 3] * W

464  = 2W^2 - 3W        rearrange  as

2W^2 - 3W - 464 =  0    factor

(2W + 29) (W - 16)  = 0     set the second factor to 0 and W  = 16 m

And L = 2W  - 3  =  2(16) - 3  =  29 m

g(x) just shifts the point down 1 unit →  (-14, 1) 

5x^2+13x-6  =

Factors of 5  = 1 and 5

Factors of - 6  we need are -3 and 2

And    5(-3) + 1(2)  = -13  ....so we have....

(5x + 2) (1x - 3)

We can "solve" this

[-4√6 ] / [8√3]  =

[-4 / 8]  *  √6 / √3  =

[ -1 / 2] * √[6 /3]  =

[ -1 / 2] * √[2]  =

 -  √2 / 2

f(x)  = x + 1

Proof

A → B  increases the "A" size by 4

B → C decreases the "B" size by 3

Net change in size from A to C  = 1

Symmetric with the y  axis  only