Question

14ab^3/7a^-2b^-1

Answer 1

\[\frac{14ab ^{3}}{7a ^{-2}b ^{-1}}\]

Answer 2

yes

Answer 3

Hang on a sec

Answer 4

k

Answer 5

\[2a ^{3}b ^{4}\]

Answer 6

thx

Answer 7

Do you understand?

Answer 8

can you explain?

Answer 9

The factors that are in the denominator have negative exponents (except for the 7) When you move them to the numerator, you change the exponents to positive.

Answer 10

\[\frac{1}{a ^{-n}}=a ^{n}\]

Answer 11

\[\frac{a}{a ^{-2}}=a(a ^{2})=a ^{3}\]

Answer 12

a^3b^3c/1^-3b^-3c^-1

Answer 13

would it be a^6b^6c^2?

Answer 14

And\[\frac{b ^{3}}{b ^{-1}}=b ^{3}b ^{1}=b ^{4}\]

Answer 15

\[\frac{a ^{3}b ^{3}c}{1^{-3}b ^{-3}c ^{-1}}\]

Answer 16

Is that the problem?

Answer 17

or is there a typo?

Answer 18

ya

Answer 19

thats the problem

Answer 20

So the denominator contains 1^-3?

Answer 21

and my answer is A^6b^6c^2

Answer 22

ya

Answer 23

yes. If that is a^-3 in the denominator instead of 1^-3

Answer 24

thx :) i get it now

Answer 25

yw