Question

Please help my brain needs assistance asap!!

Answer 1

If x and y are both positive then \[\sqrt{72x ^{3}y ^{16}}\]

Answer 2

\[\sqrt{72x ^{3}y ^{16}} = (72x ^{3}y ^{16})^{\frac{ 1 }{ 2 }}\]

Answer 3

\[72^{\frac{ 1 }{ 2 }}x ^{3 (\frac{ 1 }{ 2 })}y ^{16 (\frac{ 1 }{ 2 })}\]

Answer 4

@yomamabf

Answer 5

\[6xy ^{8}\sqrt{2x}\]

Answer 6

thats the answer but i don't know how to get it

Answer 7

@freckles @Directrix

Answer 8

2 goes in to 16 8 times
that explains the \(y^8\) out front

Answer 9

2 goes in to 3 1 time with a remainder of 1
that explains the \(x^1\) on the outside and the \(x^1\) on the inside

Answer 10

i got lost with the x

Answer 11

and finally \(72=2\times 36\) so
\[\sqrt{72}=\sqrt{36}\sqrt2=6\sqrt2\] so a 6 comes out and a 2 stays in

Answer 12

yea i get all that but it's just the x i don't understand why theres 1 outside and one inside

Answer 13

suppose you had \(\sqrt8\) could you do that one?

Answer 14

yes its \[2\sqrt{2}\]

Answer 15

replace \(2\) by \(x\) and do exactly the same thing with \(\sqrt{x^3}\) instead of \(\sqrt{2^3}\)

Answer 16

actually the x and the y .... both i don't understand

Answer 17

if you can do it with \(2^3\) you can do it with \(x^3\) it is identical

Answer 18

replace it? what will that do?

Answer 19

how did you turn
\[\sqrt8\] in to
\[2\sqrt2\]?

Answer 20

i split the 8 to 4*2 then split the 2 in 4 and put it outside

Answer 21

do the same thing with the x
split it in to \(x^2\times x\)

Answer 22

wait why would you do that? its not a perfect square

Answer 23

so
\[\sqrt{x^3}=\sqrt{x^2\times x}=\sqrt{x^2}\sqrt{x}=x\sqrt{x}\]

Answer 24

\(x^2\) is not a perfect square ?!

Answer 25

hmmmm.... I didn't know you can make a x cube into that there....

Answer 26

of course you do
what is \(x^2\times x\)

Answer 27

ohhhhh okay gotcha

Answer 28

whew