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what is the area of a rectangle with a length of (x^2-2) and a width of(2x-x+2)

Sagot :

a386
The area of a rectangle can be found through the formula [tex]A=lw[/tex], where l is the length and w is the height.  In this situation, the answer can be found by multiplying the length and width provided:

[tex](x^2-2)*(2x-x+2)[/tex]

The distributive property must be applied here:

[tex](x^2-2)*(2x-x+2) \\ 2x^3-x^3+2x^2-4x+2x-4[/tex]

Now simplify by combining like terms:

The area is
[tex] x^3+2x^2-2x-4[/tex]
Panoyin
A = lw
A = (x² - 2)(2x - x + 2)
A = (x² - 2)(x + 2)
A = x²(x + 2) - 2(x + 2)
A = x²(x) + x²(2) - 2(x) - 2(2)
A = x³ + 2x² - 2x - 4
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