The length of a rectangle is four times its width. If the perimeter is at most 106 centimeters, what is the greatest possible value for the width?
Which inequality models this problem?
2w + 2 • (4w) < 106
2w+ 2. (4w) 106
2w + 2. (4w) < 106
2w+2. (4w) > 106

answer: f = 32.2m

step-by-step explanation: that is the answer

not sure but it may be b, as i said i'm not sure.

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)