Question

Please Help Me

Answer 1

Which expression is equivalent to

Answer 2

A. ab

Answer 3

we can write this step:
\[\huge {\left( {{a^6}{b^{ - 3}}} \right)^{1/3}} = {a^{6/3}}{b^{ - 3/3}} = ...?\]

Answer 4

just a sec

Answer 5

-3/3?

Answer 6

\(-3/3=-1\)

Answer 7

oh ok so C

Answer 8

are you sure?

Answer 9

hint: \(6/3=2\)

Answer 10

I thought me solved that into -1

Answer 11

please we have:
\(6/3=2\) and \(-3/3=-1\)

Answer 12

I see what I did i said 6/3 and -3/1 came to -/3 because 6 +-3 =-3 and I kept the bottom

Answer 13

ok I get it sorry

Answer 14

here is the next step:
\[\Large {\left( {{a^6}{b^{ - 3}}} \right)^{1/3}} = {a^{6/3}}{b^{ - 3/3}} = {a^2}{b^{ - 1}} = ...?\]

Answer 15

furthermore, we can write this:
\[\huge {b^{ - 1}} = \frac{1}{b}\]

Answer 16

2/a and 1/b = 3 because 2 +1 =3

Answer 17

another step:
\[\Large {\left( {{a^6}{b^{ - 3}}} \right)^{1/3}} = {a^{6/3}}{b^{ - 3/3}} = {a^2}{b^{ - 1}} = \frac{{{a^2}}}{b}\]

Answer 18

Ok so d if you don't add 2+1

Answer 19

no, because I can add only exponents of powers with the same basis

Answer 20

So a and b could never be added because their different varibles

Answer 21

correct! \(a\) and \(b\) are factors

Answer 22

D is the answer (a 2/b)

Answer 23

yes!

Answer 24

Thank you sorry I took long

Answer 25

no worries!! :)