Question

exponent question

Answer 1

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Answer 2

what part of it do you ahve questions about?

Answer 3

with the first one, \[\frac{ 3xy ^{2}(xy)^{1} }{ z ^{4} }\]\[\frac{ 3x ^{2}y ^{3} }{ z ^{4} }\]

Answer 4

just want to make sure im doing these right

Answer 5

its good so far

Answer 6

isnt that as far are you can simplify

Answer 7

yep

Answer 8

\[(3m^2)^3(2mn)^{-1}(8n^3)^\frac{ 2 }{ 3 }\]

Answer 9

so you'd have \[(3m^6)(\frac{ 1 }{ 2mn})\]

Answer 10

but what do you do with the 8n stuff?

Answer 11

you missed something

Answer 12

what? the last group?

Answer 13

a^3 = a*a*a

but a = 3mm

3mm*3mm*3mm = 3^3 m^6 not 3m^6

Answer 14

so 27m^6

Answer 15

yes

Answer 16

\[\sqrt[3]{8n^3}^2\]

Answer 17

would that be the last group?

Answer 18

\[b^{m/n} =(b^m)^{1/n}=\sqrt[n]{~b^m~}\]

Answer 19

\[(3m^6)(\frac{ 1 }{ 2mn })(\sqrt[3]{8n^3})^2\]

Answer 20

cuberoot is good

cbrt(8.8.nnn.nnn) = 2.2.n.n

Answer 21

does tha tmake any sense?

Answer 22

\[(3^3m^6)(\frac{ 1 }{ 2mn })\sqrt[3]{(8n^3)^2~}\]

Answer 23

3.3.3.m.m.m.m.m.m.1.2.2.n.n
--------------------------
2.m.n


3.3.3.m.m.m.m.m.2.n

Answer 24

\[\sqrt[3]{8n^3}^2 \sqrt[3]{64^6} 4n^2\]

Answer 25

so \[(27m^6)(\frac{ 1 }{ 2mn })(4n^2)\]

Answer 26

yep

Answer 27

any way to simplify that or is that it?

Answer 28

well, if we simplify it to the max ....


3.3.3.m.m.m.m.m.m.1.2.2.n.n
--------------------------
2.m.n


3.3.3.m.m.m.m.m.2.n

Answer 29

a 2,m,n cancel with a 2,m,n on top

Answer 30

\[27m^52n\]

Answer 31

almost there, 27*2 is?

Answer 32

54

Answer 33

good

Answer 34

\[54m^5n\]

Answer 35

\[(3m^2)^3(2mn)^{-1}(8n^3)^\frac{ 2 }{ 3 }\]
\[27m^{2*3}~(2^{-1}m^{-1}n^{-1})~ (64n^{2*3})^\frac{ 1 }{ 3 }\]
\[27m^{6}~(2^{-1}m^{-1}n^{-1})~64^{1/3}n^{2*3/3}\]
\[27m^{6}~(2^{-1}m^{-1}n^{-1})~4n^{2}\]
\[27m^{6-1}~\frac42n^{2-1}\]
\[27(2)~m^{5}~n\]
yes

Answer 36

that was a bit of a pain. :P thanks

Answer 37

ye sit was lol