More problems ugh! (Just 3)

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More problems ugh! (Just 3)

Mathematics
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What is the value of the expression \[\frac{ 3 }{ 3^{-2} }\]
\[\huge\rm \frac{ x^m }{ x^{-n} }=x^{m+n}\] exponent rule

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Other answers:

????
\[\frac{1}{b^{-n}}=b^n\] therefore \[\huge\frac{3}{3^{-2}}=3\times 3^2=3\times 3\times 3\]
so b.27
if there are same bases at the numerator and 't the denominator you should move the denominator exponent to the top |dw:1439563849463:dw|
yes it is
what @Nnesha said
yay ty
what you are supposed to be learning is that \[\huge \frac{b^m}{b^n}=b^{m-n}\]
What value of k solves the equation? \[k ^{-5}=\frac{ 1 }{ 32 }\]
so \[\frac{3^1}{3^{-2}}=3^{1-(-2)}=3^3\]
i dont really think she understands the equations
i get it now
i am sure of it, but we can soldier on
\[k^{-5}=\frac{1}{k^5}=\frac{1}{32}\iff k^5=32\]
think of a number that when raised to the power of 5 gives 32
to much to think, you should write that too!
6.4?
no
she is too focused on the answer
\[6.4^5\]is way too big pick a smaller number hint: it is a whole number
oh i get it hmm
2
bingo
yay!
could u help me with 1 more
why not?
nvm i got it ill still post to see if im right. What value of m solves the equation?\[2^{m} = \frac{ 1 }{ 8 }\] 3
close
\[2^3=8\] not \(\frac{1}{8}\)
make the exponent negative
no thats the problem my answer is 3
yes i know, but 3 is not the right answer
you don't want 8 you want 1/8, i.e. the "reciprocal" to get that, make the exponent negative
-3
yes
say hi to christina perry for me
huh
"perri"
i got an 80 i passed
shoot for a 100 next time
tysm
@satellite73 make me an admin lol

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