It would be the 2nd answer....your initial value (read from the left of the line graph) is 4.....following the line to the first even point you can see it has dropped 3 for the 4 points it has moved to the right....giving -3/4 as the rate of change.

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3 years ago

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The length of a rectangle is 59 inches greater than twice the width. If the diagonal is 2 inches more than the length, find the

Paul [167]

So.. hmm notice the picture here

we know the length is 59 more than twice the width,
twice the width, 2 * w, or 2w
59 more than that
2w + 59

we also know that, whatever that is, the diagonal of it is,
2 more inches than that, or
(2w + 59) + 2

now.. use the pythagorean theorem
$\bf c^2=a^2+b^2\implies \qquad 
\begin{cases}
a=2w+59\\
b=w\\
c=(2w+59)+2\to 2w+61
\end{cases}
\\\\\\
(2w+61)^2=(2w+59)^2+(w)^2
\\ \qquad\qquad \uparrow\qquad \qquad\qquad \qquad \uparrow \\
\textit{expand using binomial theorem, or FOIL}$

you'd end up with a quadratic, after simplifying, solve for "w"