Question

1

Answer 1

\[{9x^4\over4y^3}\cdot{3x^2\over8y^5}\]?

Answer 2

yes, sorry didn't know how to put it in equation form!

Answer 3

multiply straight across, and remember that\[x^a\cdot x^b=x^{a+b}\]

Answer 4

27x^6/32y^8 is what I got originally but that's not an option for one of the answers.

Answer 5

well then something is wrong

Answer 6

The options are 6x^2y^2, x^2y^2/6, 6/x^2y^2, and 1/6x^2y^2? O:

Answer 7

is the problem maybe\[{9x^4\over4y^3}\cdot{8y^5\over3x^2}\]?

Answer 8

Oh, wait. Dang it, the multiplication sign should have been a division sign, whoops.

Answer 9

the rule is\[{a\over b}\div{c\over d}={a\over b}\cdot{d\over c}\]so in this case we have\[{9x^4\over4y^3}\div{3x^2\over8y^5}={9x^4\over4y^3}\cdot{8y^5\over3x^2}\]

Answer 10

now can you solve it?

Answer 11

Oh, so it's flipped? Yeah, I can solve it now. Thanks. (:

Answer 12

more properly stated:
"dividing by a fraction is the same as multiplying by its reciprocal"
-yep, that's the rule
you're welcome :D