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PLZPLZPLZ HELP ME!!!

The length of a particular rectangle is 2 inches less than twice its width. If the area of the rectangle is 40 square inches, find the length and width.

(A.) length is 6 in, width is 4 in

(B.) length is 2 in, width is 2 in

(C.) length is 5 in, width is 8 in

(D.) length is 8 in, width is 5 in

Respuesta :

LammettHash
If [tex]\ell[/tex] is the length and [tex]w[/tex] the width, then [tex]\ell=2w-2[/tex].

Since the area of the rectangle is

[tex]\ell w=40[/tex]

you have

[tex](2w-2)w=40\implies 2w^2-2w-40=0\implies w^2-w-20=0[/tex]

Factoring the left hand side gives

[tex](w-5)(w+4)=0\implies w=5,w=-4[/tex]

Obviously the width can't be negative, so [tex]w=5[/tex]. This means D is the answer, since it's the only option with this measure.

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