Question

The length of a rectangle is 2 feet longer than the width. The area of the rectangle is 8 square feet. Draw the graph of the function A(w) that represents this problem. Determine the maxima or minima.

Answer 1

will.h

Answer 2

no need, am here

Answer 3

i fanned you BTW

Answer 4

thanks

Answer 5

This question looks sick.

Answer 6

i know its tireing

Answer 7

tiring

Answer 8

what course is this. and what's the name of the lesson

Answer 9

appliations of quadratics

Answer 10

since length is 2ft longer than width then we know that
L= w+2
since the are= 8
then we know that
Length * width = 8
therefore
W+2*W = 8
\[w^2 + 2w=8\]
\[w^2+2w - 8 = 0\]
Now we can factor it like this
\[(w+4) (w-2) =0\]
Now either \[W+4 = 0\] or \[W-2 = 0\]
solve for the 1st one
\[w+4 = 0\]
\[w=-4\]
which don't make sense because dimensions cannot be negative.
solve the other one
\[w-2 = 0\]
\[w=2\]
So the only reasonable width is 2
now substitute the width in the following equation to find the length.
L=w+2
since we know that width = 2 then
L= 2+2
L=4
Therefore the length is 4 and the width is 2.

Answer 11

thanks I would ask u for more help but u said u were tired

Answer 12

i can help with some more. maximum 2 more questions.

Answer 13

k