Arectangle is inscribed under an arc of the graph of y = cos x.
your goal is to find the largest possible area for this rectangle.
a. give the coordinates (x, y) of the vertices of the rectangle
that touch the arc in terms of x.
b. in terms of x, what is the length of the rectangle?
c. in terms of x, what is the width of the rectangle?
d. find an expression f(x) for the area of the rectangle in
terms of x.
e. what interval of x-values should be considered when
finding the area?
f. find f(0.2), f(1), and xplain what each means in the
context of the problem.
g. graph fover the domain you found in part e, and find the
maximum value of the rectangle.

Step-by-step explanation:

a. The coordinates of the vertices are (x, y) and (-x, y).  Since y = cos(x), we can write the coordinates in terms of x as (x, cos(x)) and (-x, cos(x)).

b. The length of the rectangle is the horizontal distance between the vertices, or 2x.

c. The width of the rectangle is the vertical distance between the vertices and the x-axis, or cos(x).

d. The area of the rectangle is width times length:

f(x) = 2x cos(x)

e. The width of the rectangle must be greater than 0, so:

cos(x) > 0

-π/2 < x < π/2

f. f(0.2) = 2(0.2) cos(0.2) = 0.392

f(1) = 2(1) cos(1) = 1.081

f(π/2) = 2(π/2) cos(π/2) = 0

f(x) is the area for each rectangle where x is half the rectangle's length.

g. Graph: desmos.com/calculator/f7xhwfy2dj

f(x) reaches a maximum value of 1.122.

ddd

step-by-step explanation:

answer: (c) m∠qpo + (2x + 16)° = 180°

step-by-step explanation:

the opposite angles of a quadrilateral are supplementary.

so, ∠o + ∠q = 180°   and   ∠p + ∠r = 180°

since ∠r = 2x + 16°, we can use substitution as follows:

∠p + ∠r = 180°

∠p + (2x + 16)° = 180°