PLEASE HELP Evaluate 3x^3y-2/5x^-2y^3 for x=2 and y=3 32/405 160/81 2/5

Question

asked Dec 14, 2019 80.5k views

1 Answer

Angelo Giuffredi · Answer 1 · 2019-12-21T15:26:17+0000

Let's try to simplify that a bit. Rewriting it in its current form it looks like this:

(3x^3y^(-2))/(5x^(-2)y^3) .

We could split it up and deal with one problem at a time if we do this:

(3)/(5)*(x^3)/(x^(-2))*(y^(-2))/(y^3) .

The rule with like bases and division of the exponents is that we subtract the lower exponent from the upper so that would be

x^(3-(-2))*y^(-2-3)

which gives us new exponents of

x^5*y^(-5) .

We rewrite to make that negative exponent a positive by putting it under a 1, so the whole problem with no negatives looks like this now:

(3x^5)/(5y^5) .

If x = 2 and y = 3, we sub in accordingly:

(3(2)^5)/(5(3)^5) =(3(32))/(5(243))

which simplifies to
(96)/(1215)=(32)/(405) , the first choice above.