Asked by Ann
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
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
Answers
Answered by
Jai
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~ :)
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~ :)
Answered by
Ann
thanks so much :)) That was really helpful. Just one quick question. How did you get 15/16?
Answered by
Jai
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
-26 16/16 + 1/16
-26 + (16-1)/16
-26 15/16
Answered by
Ann
ohh okk thanks so much :)
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