7.75 ➗ 125 = 0.062
Simplify ➡ 31 / 4
31 over 4 ➗ 125
Then, divide ➡31 over 1 by 125
and it should leave you with the answer 0.062
x-coordinates for the maximum points in any function f(x) by f'(x) =0 would be x = π/2 and x= 3π/2.
<h3>How to obtain the maximum value of a function?</h3>
To find the maximum of a continuous and twice differentiable function f(x), we can firstly differentiate it with respect to x and equating it to 0 will give us critical points.
we want to find x-coordinates for the maximum points in any function f(x) by f'(x) =0
Given f(x)= 4cos(2x -π)

In general 
from x = 0 to x = 2π :
when k =0 then x = π/2
when k =1 then x= π
when k =2 then x= 3π/2
when k =3 then x=2π
Thus, X-coordinates of maximum points are x = π/2 and x= 3π/2
Learn more about maximum of a function here:
brainly.com/question/13333267
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Answer:
DF is your answer. Hope this helps!
Step-by-step explanation: