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July 22, 2014

July 22, 2014

Posted by **Ann** on Saturday, September 10, 2011 at 12:33am.

Evaluate the following:

x^3+y^-2 for x=-3 and y=4

I don't understand how to solve this equation. I solved it and got -11 as my answer but the books answer is -26 15/16

Thanks For Your Help

- Algebra 2 -
**Jai**, Saturday, September 10, 2011 at 1:27amx^3 + y^(-2)

to evaluate means you substitute the value of x and y which are given in the problem.

first, recall some laws of exponents. if a term is raised by a negative number, it is also equal to its reciprocal raised to the that number but positive in sign. for example,

5^(-3) = (1/5)^3

thus we can actually rewrite this equation as

x^3 + (1/y)^2

substituting x = -3 and y = 4,

(-3)^3 + (1/4)^2

(-3)*(-3)*(-3) + (1/4)*(1/4)

-27 + 1/16

-26 15/16

hope this helps~ :)

- Algebra 2 -
**Ann**, Saturday, September 10, 2011 at 2:26amthanks so much :)) That was really helpful. Just one quick question. How did you get 15/16?

- Algebra 2 -
**Jai**, Saturday, September 10, 2011 at 3:14amyou can rewrite -27 as -26 16/16 (since 16/16 is just equal to 1). thus,

-26 16/16 + 1/16

-26 + (16-1)/16

-26 15/16

- Algebra 2 -
**Ann**, Saturday, September 10, 2011 at 3:03pmohh okk thanks so much :)

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