Question

(-x^2)^3/4z ÷ 3x/8z^2 =

Answer 1

that was wrong i forgot the square

Answer 2

ooh

Answer 3

6x/8y^3 ÷ (2x^2)^3/9x^4 =

Answer 4

start with
\[\frac{-x^6\times 8z^2}{4z\times 3x}\] that should work

Answer 5

cancel a 4, and x and a z to get
\[-\frac{2x^5z}{3}\]

Answer 6

start with
\[\frac{6x\times 9x^4}{8y^3\times 8x^6}\]

Answer 7

sorry. i wrote the problem wrong. it is 9y^4

Answer 8

and how did you get 8x^6 out of (2x^2)^3

Answer 9

then start with
\[\frac{6x\times 9y^4}{8y^3\times 8x^6}\]

Answer 10

\[(2x^2)^3=2^3\times (x^2)^3=8\times x^{2\times 3}=8x^6\]

Answer 11

(2x^2)^3 = 2x^2 * 2x^2 * 2x^2. and 2 * 2 * 2 = 8. oh i see. someone else told me differently.

Answer 12

same thing right?

Answer 13

yup

Answer 14

still not done
\[\frac{6x\times 9y^4}{8y^3\times 8x^6}\]
\[\frac{27y}{32x^5}\] by canceling

Answer 15

okay.

Answer 16

and (a^2)^2/3b / 4a^5/(b^3)^2=

Answer 17

(a^2)^2/3b / 4a^5/(b^3)^2=