# How do you find the dimensions of a rectangle whose perimeter is 52 m and whose area is 160 m?

The dimension of rectangle is

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Let ( l ) be the length and ( w ) be the width of the rectangle.

Given: Perimeter ( P = 2l + 2w = 52 ) Area ( A = lw = 160 )

From the perimeter equation: [ 2l + 2w = 52 ] [ l + w = 26 ] [ l = 26 - w ]

Substitute ( l ) from the perimeter equation into the area equation: [ (26 - w)w = 160 ] [ 26w - w^2 = 160 ] [ w^2 - 26w + 160 = 0 ]

Using quadratic formula: [ w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ] where ( a = 1 ), ( b = -26 ), and ( c = 160 ) [ w = \frac{-(-26) \pm \sqrt{(-26)^2 - 4(1)(160)}}{2(1)} ] [ w = \frac{26 \pm \sqrt{676 - 640}}{2} ] [ w = \frac{26 \pm \sqrt{36}}{2} ] [ w = \frac{26 \pm 6}{2} ]

The possible values for ( w ) are: [ w_1 = \frac{26 + 6}{2} = 16 ] [ w_2 = \frac{26 - 6}{2} = 10 ]

When ( w = 16 ), then ( l = 26 - 16 = 10 ) When ( w = 10 ), then ( l = 26 - 10 = 16 )

Thus, the dimensions of the rectangle are either ( 16 , \text{m} \times 10 , \text{m} ) or ( 10 , \text{m} \times 16 , \text{m} ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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