Question

44

Answer 1

dont get help from this guy hes no help at all

Answer 2

whats the difference between
x^3 + x^3 = ?
or
(x^3)*(x^3) =?

Answer 3

amorfide is no help at all really wait and see

Answer 4

\[a^{m} \times a^{n}=a^{m+n}\]

although he was correct in attempting to simplify he could have changed 3+3+3+3 to be 12
so his simplification is not correct
\[x^{3+3+3+3}=x^{12}\]

Answer 5

ofcourse the addition of powers rule only works if you are multiplying

since the end of the question asks if adding the expressions together is the same answer, that is not true as if you had

2+2+2+2 you get 8
but
\[2^{4}\] is not equal to 8
(I am just using this as an example)

Answer 6

it doesn't say 12 anywhere which is what I am saying
he was half correct in simplifying but it can be simplified further

Answer 7

if this makes sense to you I can continue to question 2

Answer 8

\[x^{-6} = \frac{ 1 }{ x^{6} }\]

so we can rewrite
\[\sqrt[3]{x^{-6}}\]

to be

\[\sqrt[3]{\frac{ 1 }{ x^{6} }}\]
but we know the cuberoot of 1 is 1 so
\[\frac{ 1 }{ \sqrt[3]{x^{6}} }\]

NOTE
this is the bottom of the fraction, we are only working with the denominator at this time

Answer 9

\[\sqrt[3]{x^{6}}= \sqrt[3]{x^3 \times x^{3}}\]
\[\sqrt[3]{x^{3} \times x^{3}}= \sqrt[3]{x^{3}} \times \sqrt[3]{x^{3}}\]

obviously the cube root of x^3 = x

so the bottom of our fraction is now
\[\frac{ 1 }{ x^{2}}\]

Answer 10

and ofcourse you can rewrite this
\[\frac{ 1 }{ x^{2} }\]
to be
\[x^{-2}\]

Answer 11

so since this is the bottom of the fraction, just replace it in the original question to get

\[\frac{ 1 }{ \sqrt[3]{x^{-6}} }=\frac{ 1 }{ x^{-2} }\]

Answer 12

if this makes sense I will do question 3

Answer 13

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

Answer 14

so we subtract the powers on this one

Answer 15

\[\frac{ x^{\frac{ 2 }{ 3}} }{ x^{\frac{ 4 }{ 9} } } = x^{\frac{ 2 }{ 3 } - \frac{ 4 }{ 9 }}\]

Answer 16

\[\frac{ 2 }{ 3 } - \frac{ 4 }{9 }= \frac{ 6 }{ 9 } - \frac{ 4 }{ 9 }\]

Answer 17

so our answer so far is

\[x^{\frac{ 2 }{ 9 }}\]

Answer 18

now we want radical form

Answer 19

\[a^{\frac{ m }{ n }} = \sqrt[n]{a^{m}}\]

Answer 20

so we have

\[x^{\frac{ 2 }{ 9 }}= \sqrt[9]{x^{2}}\]

Answer 21

if this makes sense we can continue

Answer 22

we know that
\[\sqrt[n]{a^{m}} = a^{\frac{ m }{ n }}\]
so we have
\[\sqrt[3]{x^{3}} = (x^{3})^{\frac{ 1 }{ 3 }}\]

we know that
\[(a^{m})^{n} = a^{m \times n}\]

\[(x^{3})^{\frac{ 1 }{ 3 }} = x^{3 \times \frac{ 1 }{ 3 }}\]
which becomes
\[x^{1}=x\]

Answer 23

we know that

\[a^{m} \times a^{n} = a^{m+n}\]

so

\[x^{\frac{ 1 }{ 3 }} \times x^{\frac{ 1 }{ 3 }} \times x^{\frac{ 1 }{ 3 }} = x^{\frac{ 1 }{ 3 } +\frac{ 1 }{ 3 } +\frac{ 1 }{ 3 }}\]

which becomes

\[x^{1}=x\]

Answer 24

we know that
\[a^{ -m}=\frac{ 1 }{ a^{m} }\]

so we have

\[\frac{ 1 }{ x^{-1} } = \frac{ 1 }{ \frac{ 1 }{ x } }\]

this is dividing by a fraction
we know that

\[\frac{ a }{ b } \div \frac{ c }{ d } = \frac{ a }{ b } \times \frac{ d }{ c }\]

so we have

\[1 \div \frac{ 1 }{ x } = 1 \times x = x\]

Answer 25

since we know that
\[\sqrt[n]{a^{m}} = a^{\frac{ m }{ n }}\]
and
\[a^{m} \times a^{n} = a^m+n\]
and
\[(a^{m})^{n}=a^{m \times n}\]

then

\[\sqrt[11]{x^{5} \times x^{4} \times x^{2}} = x^{\frac{ 11 }{ 11 }}=x\]

Answer 26

the coding messed up that should be

\[a^{m} \times a^{n} = a^{m+n}\]

Answer 27

would appreciate a medal for best answer is this helped you :)

Answer 28

if you need anything cleared up let me know

Answer 29

anytime! I try my best to help :)