A rectangle has an area of 8z² - 12z. If its width is 4z, what is its length? What is the perimeter of the rectangle?

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Answer 1 · 2018-05-10T00:37:54

$l = 2 z - 3$ $l=2z-3$

$P = 12 z - 6$ $P=12z-6$

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$A = 8 z^{2} - 12 z$ $A=8z^2-12z$

Area of a rectangle is:

$A = l \cdot w$ $A=l*w$ where $l$ $l$ is the length and $w$ $w$ is the width.

$8 z^{2} - 12 z = l \cdot w$ $8z^2-12z=l*w$

$w = 4 z$ $w=4z$

$8 z^{2} - 12 z = 4 z \cdot l$ $8z^2-12z=4z*l$

$4 z (2 z - 3) = 4 z \cdot l$ $4z(2z-3)=4z*l$

$l = \frac{4 z (2 z - 3)}{4 z} = 2 z - 3$ $l=(4z(2z-3))/(4z)=2z-3$

Perimeter of a rectangle is:

$P = 2 l + 2 w$ $P=2l+2w$

$P = 2 (2 z - 3) + 2 (4 z) = 4 z - 6 + 8 z = 12 z - 6$ $P=2(2z-3)+2(4z)=4z-6+8z=12z-6$