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abruzzese [7]
3 years ago
9

The length of a rectangle is five less than its width. The area of the rectangle is 84 square feet. Write a quadratic equation i

n standard form, ax^2+bx+c=0
Mathematics
1 answer:
salantis [7]3 years ago
4 0

<u>Given</u>:

The length of a rectangle is 5 less than its width.

The area of the rectangle is 84 square feet.

We need to determine the quadratic equation in standard form that represents the area of the rectangle.

<u>Dimensions of the rectangle:</u>

Let l denote the length of the rectangle.

Let w denote the width of the rectangle.

Since, it is given that the length is 5 less than its width, it can be written as,

l=5-w and w=w

<u>Area of the rectangle:</u>

The area of the rectangle can be determined using the formula,

A=length \times width

Substituting A = 84, l=5-w and w=w, we get

84=(5-w)\times w

84=5w-w^2

Adding both sides of the equation by w², we have;

w^2+84=5w

Subtracting by 5w on both sides, we get;

w^2-5w+84=0

Thus, the quadratic equation in standard form for the area of the rectangle is w^2-5w+84=0

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Find the measurements (the length L and the width W) of an inscribed rectangle under the line with the 1st quadrant of the x &am
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The question is incomplete. Here is the complete question.

Find the measurements (the lenght L and the width W) of an inscribed rectangle under the line y = -\frac{3}{4}x + 3 with the 1st quadrant of the x & y coordinate system such that the area is maximum. Also, find that maximum area. To get full credit, you must draw the picture of the problem and label the length and the width in terms of x and y.

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Step-by-step explanation: The rectangle is under a straight line. Area of a rectangle is given by A = L*W. To determine the maximum area:

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To maximize, we have to differentiate the equation:

\frac{dA}{dx} = \frac{d}{dx}(-\frac{3}{4}.x^{2}  + 3x)

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