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What is the aera of the rectangle whose length is(× +5)and width (× - 5)?​

Sagot :

ANSWER :

A = 0

STEP BY STEP EXPLANATION

A = lenght × width

length = x+5

width = x-5

SUBSTITUTE

A = lenght × width

A= (x+5)(x-5)

A= x²-5x +5x -25

COMBINE THE LIKE TERMS (which is -5x and 5x)

A= x²-25

TRANSPOSE THE (-25)

25 = x²

USE SQUARE ROOT BOTH SIDE

[tex] \sqrt{25 } = {x}^{2} [/tex]

5 = x

SUBSTITUTE IT

A= length × width

A = (5+5)(5-5)

A = (10)(0)

A = 0