SOLUTION: Let width = w Area = {{{(length)*(width)}}} So {{{3x^2 + 5x = x*w}}} or {{{w = (3x^2 + 5x)/x = 3x + 5}}} So required width is (3x + 5).
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Polynomials-and-rational-expressions
-> SOLUTION: Let width = w Area = {{{(length)*(width)}}} So {{{3x^2 + 5x = x*w}}} or {{{w = (3x^2 + 5x)/x = 3x + 5}}} So required width is (3x + 5).
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Question 40846
:
Let width = w
Area =
So
or
So required width is (3x + 5).
Answer by
psbhowmick(878)
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