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find the area of a rectangle with a length of (x^2+2x-3) and a width of (x+6)
oobleck
answered2 years ago
Anonymous
answered2 years ago
(x^2+2x-3)(x+6)
which is
= x (x^2+2x-3)
+ 6 (x^2+2x-3)
------------------------
Explain Bot
answered7 months ago
To find the area of a rectangle, you need to multiply its length by its width. In this case, the length is given as (x^2+2x-3) and the width is given as (x+6).
To find the area, you can use the distributive property to expand the expression for the length:
Length = (x^2+2x-3)
Width = (x+6)
Area = Length * Width
= (x^2+2x-3) * (x+6)
Now, you can multiply the expressions using the distributive property and simplify the expression further:
Area = (x^2+2x-3) * (x+6)
= x^2(x+6) + 2x(x+6) - 3(x+6)
= x^3 + 6x^2 + 2x^2 + 12x - 3x - 18
= x^3 + 8x^2 + 9x - 18
Therefore, the area of the rectangle with a length of (x^2+2x-3) and a width of (x+6) is represented by the expression x^3 + 8x^2 + 9x - 18.