Help with this question please!

Step-by-step explanation:

6cos^2(x) + 7cos(x) +2 = 0

Call cos(x) = y to change this to something a little bit more familiar.

6y^2 + 7y + 2 =0

This equation is a quadratic. Solve it first. Then we'll go back to solving for cos(x)

This equation factors into 2 possible answers

(3y + 2)(2y + 1) = 0

3y + 2  = 0

3y = - 2

y = - 2/3

2y + 1 = 0

2y = - 1

y = - 1/2

Now let's start with

cos(x) = - 1/2

That means that x is in quadrant two or three.

cos-1(1/2) = 60 degrees.

Quad 2: x = 180 - 60 =  120

Quad 3: x = 180 + 60 =  240

Now cos(x) = - 2/3

cos-1(2/3) = 48.19

Quad 2: x = 180 - 48.19 = 131.81

Quad 3: x  = 180 + 48.19 = 228.19

f(2x−x2)=−3x2+6x−1

step-by-step explanation:

set up the composite function and evaluate.

f(2x−x2)=−3x2+6x−1

answer: 270

step-by-step explanation:

you have to times 200 by 1.35 to get your answer