how would you solve a^-2*a*a^0?

Question

the carrotops (^) representing exponents of course.

i remember doing it in algebra last year but i'm not sure how it's done. like doesn't a^0 equal 1?

if someone can tell me how to solve it (and not just give me the answer) it'll be greatly appreciated and i'll choose the first person to do that as the best answer rather than someone that just gives me the answer.

so, please help! thanks.

Answer 1

Yes, you are correct. Anything, no matter what the base is raised to the power zero is one.
A negative exponent is the same as the reciprocal of the base with the absolute value of the exponent. In easier terms that would mean a^-b= 1/a^b
Do you guys still do "Please excuse my dear Aunt Sally" for order of operations? Anyway, just an easy sentence to help you remember.
We have no parentheses here, so move on to exponents:
a^-2*a*a^0
(1/a^2)(a)(1) now we do multiplication and division
(1/a^2)(a)= a/a^2
When you divide exponents with the same base, you subtract the exponents. Remember to always subtract the top from the bottom, so a is the same as a^1 so 1-2= -1
You can express your answer as a^-1 or 1/a; same thing.

Answer 2

to multiply terms with the same base, add the exponents.
-2+1+0 = -1
anything (except 0) to the power of 0 = 1
a^-1

Answer 3

(1 / a ²) (a) (1) = 1 / a

Answer 4

a^0=1
sp we have 1/a^2*a*1
a's cancel out.
so 1/a.