how can i simplify n^-2 / n^3?

Answer 1

When multiplying numbers with like bases add the exponents, and when dividing, subtract.
So...
-2-3=-5
n^-5 or 1/n^5

You can think about it many ways if it is easyer...
The negative expontent flips it position in the fraction so it could also be written as:
(n^-2)(n^-3)
1/(n^2)(n^3)

Answer 2

Having a negative exponent moves the value to the other side of the fraction, so

n^-2/n^3
=> 1/(n^2 * n^3)
=> 1/n^(2+3)
=> 1/n^5

1/n^5 or n^-5 is the answer

Answer 3

n^(- 2 - 3) = n^(-5) = 1 / n^5

Answer 4

n^ -2 = 1/n^2

therefore,

n^-2/n^2
= 1/(n^2*n^3)
=1/n^(2+3)
= 1/n^5

Answer 5

with negative exponents, you take the base (in this case "n") and take it to the other side (down in this case) of the fraction bar.

if you take n^2 to the bottom, you end up with 1/(n^3*n^2) which results in 1/n^5

Answer 6

1/n^5 b/c you can't have negative exponents in a mathematical sentence, and negatives change into positve when you switch them (up/down). You have to simplify.

Answer 7

the least complicated situation for convergence is that if all words are +ve and the Lim n--->oo u(n)=non-0 then it particularly is divergent. indoors the given case Lim n--->oou(n)=one million/2 so the sequence is divergent. the area on convergence Lim n---_oou(n)=0 is mandatory yet not sufficient situation.There are a sequence of attempt neede to show convergence at the same time with De'Alembert,Couchy attempt to so forth. IVAN IVAN

Answer 8

n^-2 / n^3
= 1 / n^2 n^3
= 1 / n^5
or
n^(-5).

Answer 9

n^-2 / n^3
= n^(-2-3)
= n^-5

I hope this helps!