Introduction to proofs (4 points)
prove the claim, "x^2 is odd < --> x + 1 is even", where x e z by
proving the following two conditional statements:
(1) prove x^2 is odd -> x + 1 is even
(hint use proof by contrapositive)
(a) (0.5 points) what are you assuming is true:
(b) (0.5 points) what are you proving is true:
(c) (1 point) complete the proof:

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)

answer: 486in^2

step-by-step explanation:

im guessing (23,14). hope this : )

32.5 is the answer i got thats the answer