Question

(2x -1/2)^2, I got x^2-2xy+y^2, it was counted wrong,, help pleaasee .!

Answer 1

there is no y?

Answer 2

i don't know where i got that.... haha x^2-2x?

Answer 3

\[(2x -\frac{1}{2}y)^2=(2x -\frac{1}{2}y)(2x -\frac{1}{2}y)=(2x)(2x -\frac{1}{2}y)-(\frac{1}{2}y)(2x -\frac{1}{2}y)\]

Answer 4

theres no y in the problem... i imagined it or something, (2x-1/2)^2 sorry!!

Answer 5

\[(2x-\frac{1}{2})^2=(2x-\frac{1}{2})(2x-\frac{1}{2})=(2x)(2x-\frac{1}{2})-\frac{1}{2}(2x-\frac{1}{2})=??\]

Answer 6

on a side note
(x-y)^2= x^2-2xy+y^2

Answer 7

simplifies to 4x^2-2x-1?

Answer 8

Nevermind, I got an A on the assignment, thanks! :)

Answer 9

1/2 *1/2 = 1/4

Answer 10

-1/2*-1/2= +1/4

Answer 11

Thanks!

Answer 12

4x^2-2x+1/4