this is my question x to the second and y to the negative third when x=3 y=-4 x^2/y^-3

Question

x=3 y=-4 and the problem is x to the second divided by y to the negative third.

Answer 1

If you're asking how to evaluate it, let's substitute the values into your expression: x²/y^-3

(3)²/(-4)^-3

Let's do each part separately, and then I'll talk about how you can evaluate these expressions a bit faster:

(3)² = 3*3 = 9

(-4)^-3 = 1/(-4)³ (notice how the exponent has become a positive 3 now) = 1/[-4 * -4 * -4] = 1/-64 = -1/64

So to evaluate the entire expression, we are calculating: 9 ÷ (-1/64) = 9 × -64 = -576

** Slightly faster method**

As you become more familiar with exponential expressions, you'll find that it's faster to evaluate them (i.e. substitute values in and then calculate) if you've simplified the original algebraic expression first.

So let's start again with our original expression: x²/y^-3

The fact that there is a term with a negative exponent in the denominator means that we can rewrite it with that term with a POSITIVE exponent in the numerator.

So that means that x²/y^-3 = x²y³ (so the product of x^2 and y^3)

Now we can easily substitute both values for x and y at the same time:

x²y³ = (3)²(-4)³ = 9 × -64 = -576