Algebra 2 help?

Question

1. x^3 times x^-7

2. (-1/5)^0

3. (15x^-1y^2/5xy^-4)

Answer 1

1. When you are multiplying two terms of the same variable, you add their exponents:

x³ * x^-7 = x^{3 + -7} = x^{3 - 7} = x^-4

2. Anything to the 0^th power is 1. The only exception is 0⁰ which is undefined or 1 depending on who you ask. So the answer is 1.

3. We can separate this into the product of three fractions: (15/5) * (x^-1/x) * (y²/y^-4). 15/5 is obviously 3. When dividing two terms of the same variable, you subtract the bottom variable's exponent from the top's. So x^-1/x = x^-1/x¹ = x^{-1 - 1} = x^-2 and y²/y^-4 = y^{2 - (-4)} = y²⁺⁴ = y⁶. Altogether, we have 3 * x^-2 * y⁶ = 3x^-2y⁶

Answer 2

1.  x³(x^-7)

When multiplying the same based (x in this problem) with exponents, you add the exponents together

x^{3 - 7}

x^-4

A negative exponent means that the base should be moved to the other part of the fraction.  In this problem, the x (with the exponent 4) needs to be moved to the denominator

1/x⁴

2.  Anything to the 0 power equals 1

(-1/5)⁰ = 1

3.  15x^-1y²/(5xy^-4)

Negative exponents mean that the base needs to be moved to the other part of the fraction.  In this problem, the x^-1 should become x in the denominator and y^-4 should become y⁴ in the numerator.

15y²*y⁴/(5x*x)

15y⁶/(5x²)

3y⁶/x²