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3 years ago

the length of a rectangle is five feet less than its width. If the area of the rectangle is 84 square feet, find its dimensions

Mathematics

1 answer:

KIM [24]3 years ago

6 0

Let
x---------> the length of the rectangle
y--------> the width of the rectangle

we know that
A=84 ft²
[area of rectangle]=x*y-----> 84=x*y-----> equation 1
x=y-5------> equation 2
substitute 2 in 1
84=[y-5]*y-----> 84=y²-5y---------> y²-5y-84=0

using a graph tool-----> to resolve the second order equation
see the attached figure

the solution is
y=12
x=y-5-----> x=12-5-----> x=7

the answer is
the length of the rectangle is 7 ft
the width of the rectangle is 12 ft

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$\large\underline{\sf{Solution-}}$

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So, on substituting the values evaluated above, we get

$\rm \: = \: \displaystyle\lim_{n \to \infty} \frac{\dfrac{ {n}^{n} - 1}{1 - \dfrac{1}{n} }}{{n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}$

$\rm \: = \: \displaystyle\lim_{n \to \infty} \frac{ {n}^{n} - 1}{{n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}$

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