Were suppose to multiply. I got 8y - 12 but I don't think it's right. Could someone tell me if it is right or wrong and if it's wrong could you please tell me what I'm not doing right.
2007-03-11 11:49:32 · 8 answers · asked by topcook15 1 in Science & Mathematics ➔ Mathematics
2y(4y-6)
2y*4y = 8y^2
2y*-6 = -12y
2y(4y-6) = 8y^2-12y
2007-03-11 11:56:16 · answer #1 · answered by Anonymous · 1⤊ 0⤋
multiply each term in the brackets (4y and -6) by 2y
to give 2y * 4y = (2 * 4) * (y * y) = 8y^2
and 2y * -6 = (2 * -6) * y = -12y
then add the terms together
8y^2 + -12y
=8y^2 -12y
2007-03-11 19:00:37 · answer #2 · answered by rg 3 · 1⤊ 0⤋
what you got left out a factor of y, therefore:
2y(4y -6) = y(8y -12) = 8y^2 - 12y
2007-03-11 18:59:31 · answer #3 · answered by The Yeti 3 · 1⤊ 0⤋
2y(4y-6) = 8y² - 12y = 8y-12.
y = 12 - 8. y = 4
2007-03-11 19:06:05 · answer #4 · answered by Norrie 7 · 0⤊ 1⤋
8Y^2 - 12Y
Distributive property
2007-03-11 18:55:37 · answer #5 · answered by davidosterberg1 6 · 1⤊ 0⤋
8y(2)-12y
Eight y squared minus twelve y
Or y=1.5
When you multiply any variable times a variable, you end up with both being represented.
Ex. 2x(x-4)
Answer would be 2x(2)-8x (two x squared minus eight x)
2007-03-11 18:59:44 · answer #6 · answered by pinktoenails 2 · 0⤊ 1⤋
8y² - 12y is the answer.
2007-03-11 19:05:10 · answer #7 · answered by Como 7 · 1⤊ 0⤋
you are right
2007-03-11 18:52:48 · answer #8 · answered by pcmpq386 1 · 0⤊ 3⤋