# Homework help?? 3 challenging problems?

Question:Suppose that a rectangle has an area of 77 square meters. Express the perimeter P as a function of the length X of one of the sides.
P(x)=???

Find the slope of the line passing through the points
(a, -3a+2) and (a+h, -3(a-4h)+2))
M=??

An open rectangular box (no top) with volume 9 cubic meters has a square base. Express the surface area S of the box as a function of the length x of one of the sides.
S(x)=??

No. 1 has been done.

2] Let the other side of the rectangle be y, so the area is:

xy, which is 77 so,

xy = 77, so

y = 77/x

The perimeter, P = 2x + 2y, so substitute y = 77/x, and you get

P = 2x + 2*77/x, so,

P = 2x + 154/x

3] I`m assuming that x is the side of the base. Let y be the

height of the box.

V = 9 = x^2y, so,

y = 9/x^2

Total surface area of the box = area of base add area of the

four sides, so,

S = x^2 + 4xy, so substitute y = 9/x^2, and you get:

S = x^2 + 4x*9/x^2,so,

S = x^2 + 36/x

Hope this helps, Twiggy.
idk about the 1st or the 3rd one...but the 2nd one is 6/1...

if you need to know how i got that...

M=(y1)-(y2) / (x1)-(x2)
M=(-3a+2) - (-3(a-4h)+2) / (a) - (a+h)
M=(-3a+2 + 3a+4h+2) / (a) - (a+h)
M=(2 + 4h+2) / (h)
M=(4h+2) / (h)
M=(4+2) / 1
M=6/1

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