Two positive integers between 51 and 999 are both rounded to the nearest hundred. If these two numbers are multiplied together, which of the following must be true? F
The product is less than 1000.

G
The product has a zero only in the ones and tens places.

H
The product has a digit greater than zero in the thousands place.

J
The product has a zero in the ones, tens, and hundreds places.

K
The product has a digit greater than zero in the ten-thousands place.

the answer is K . the product of 51 and 999 is 50949 and 5 is greater than 0.

is should be y = 5x

step-by-step explanation:

correct me if im wrong

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)