Exponential Notation

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

x^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~ :)

thanks so much :)) That was really helpful. Just one quick question. How did you get 15/16?

you 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

ohh okk thanks so much :)

To evaluate the expression x^3 + y^-2 for x = -3 and y = 4, you need to substitute the given values into the expression and perform the necessary calculations.

Let's break it down step by step:

1. Substitute the values of x and y into the expression:
(-3)^3 + 4^(-2)

2. Simplify the exponents:
(-3)^3 = -3 * -3 * -3 = -27
4^(-2) = 1 / (4 * 4) = 1/16

Substituting these values back into the expression, we have:
-27 + 1/16

3. To add these two terms, we need to find a common denominator:
Multiplying -27 by 16/16, we get -432/16.

Now, we can combine the fractions:
-432/16 + 1/16 = -431/16

Therefore, the answer is -431/16.

The book's answer of -26 15/16 is incorrect.

So, the correct answer is -431/16.