the length of a rectangle is 4 feet longer than its width. its area is 32 square feet. what are its dimensions?

Length=8, Width=4

Step-by-step explanation: I did this by trial and error, but you can also use an equation. The length is x+4, the width is x. Your equation would be x(x+4)=32. That is equal to x^2+4x=32. Solve this by completing the square. X^2+4x+ (4/2)^2=32+ (4/2)^2

X^2+4x+4=36. (X+2)(x+2)=36. X+2=6, x=4.

hello from mrbilldoesmath!

answer:   (7x + 6y) ( 49x^2 - 42xy + 36y^2)

discussion:

in general, the factorization of a^3 + b^3 is

a^3 + b^ 3 = (a + b) ( a^2 - ab + b^2)

in our case note that 343 = 7^3 and 216 = 6^3 so the equation becomes:

343x^3 + 216y^3 = (7x)^3 + (6y)^3

if you think of a (above) as 7x and b as 6y the factorization becomes

343x^3 + 216y^3 =

((7x + 6y) (   (7x)^2 - 7x*6y + (6y)^2)   =

(7x + 6y) ( 49x^2 - 42xy + 36y^2)

you,

mrb

(a + b) (a^2 - a b + b^2)

multiply length by height. multiply the length by the height, and then double the product. work out 10 x 10 results = 100 square inches, then work out 100 x 2 = 200. the answer is 200 square inches.

step-by-step explanation: