Mathematics
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OpenStudy (anonymous):
simplify 6n^6/3n^4 help please !!!!!
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jimthompson5910 (jim_thompson5910):
the coefficients are 6 and 3
you divide them to get 6/3 = 2
jimthompson5910 (jim_thompson5910):
the variable expressions are: n^6 and n^4
you divide them by subtracting the exponents: 6 - 4 = 2
the final exponent is 2
jimthompson5910 (jim_thompson5910):
which means that
6n^6/3n^4 = 2n^2
OpenStudy (anonymous):
thank you so so much you are very nice C:
jimthompson5910 (jim_thompson5910):
you're welcome
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OpenStudy (anonymous):
hey can you help me with another one ?
@jim_thompson5910
jimthompson5910 (jim_thompson5910):
sure one more
OpenStudy (anonymous):
\[[6x ^{2}y ^{4}/3x ^{4}y ^{3} ]^{3}\]
OpenStudy (anonymous):
help please c; @jim_thompson5910
jimthompson5910 (jim_thompson5910):
first i would simplify the stuff on the inside of the brackets
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jimthompson5910 (jim_thompson5910):
so using what I showed you above, what do you get?
OpenStudy (anonymous):
2x^-2
jimthompson5910 (jim_thompson5910):
close, but you're forgetting about the y term
jimthompson5910 (jim_thompson5910):
y^4/y^3 = ???
OpenStudy (anonymous):
idk the answer to that
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jimthompson5910 (jim_thompson5910):
subtract the exponents
OpenStudy (anonymous):
1.3?
jimthompson5910 (jim_thompson5910):
the exponents are 4 and 3, you subtract them to get 4-3 = 1
jimthompson5910 (jim_thompson5910):
so that means y^4/y^3 = y^1 = y
make sense?
OpenStudy (anonymous):
oh ok i got it
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jimthompson5910 (jim_thompson5910):
so this means that
\[\large \left[\frac{6x ^{2}y ^{4}}{3x ^{4}y ^{3}} \right]^{3}\]
turns into
\[\large \left[2x^{-2}y \right]^{3}\]
jimthompson5910 (jim_thompson5910):
what's next?
OpenStudy (anonymous):
multiply them all by 3?
jimthompson5910 (jim_thompson5910):
close, very close
jimthompson5910 (jim_thompson5910):
you raise each term to the 3rd power
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jimthompson5910 (jim_thompson5910):
you do that to the following terms: \(\large 2\), \(\large x^{-2}\), and \(\large y\)
OpenStudy (anonymous):
so what would be the answer ?
OpenStudy (skullpatrol):
$$2^3*(x^{-2})^3*y^3$$
jimthompson5910 (jim_thompson5910):
what do you get when you raise 2 to the 3rd power?
OpenStudy (anonymous):
8
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OpenStudy (skullpatrol):
Yes.
jimthompson5910 (jim_thompson5910):
what happens when you raise \(\large x^{-2}\) to the 3rd power
OpenStudy (anonymous):
would it be 2n^2
OpenStudy (skullpatrol):
$$b^n=b*b*b*...*b$$n factors
the base is b and the exponent is n
OpenStudy (skullpatrol):
$$(x^{-2})^3=x^{-2}*x^{-2}*x^{-2}$$
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OpenStudy (anonymous):
would it be 2n^2
OpenStudy (skullpatrol):
No.
OpenStudy (anonymous):
2n^10 ?
OpenStudy (skullpatrol):
No, try and use the definition I posed above.
OpenStudy (anonymous):
n^2/2
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OpenStudy (skullpatrol):
No,recall when you multiply two powers having the same base you add the exponents.
OpenStudy (anonymous):
n^2
jimthompson5910 (jim_thompson5910):
or you could multiply exponents
jimthompson5910 (jim_thompson5910):
(n^-2)^3 = n^(-2*3) = ??
OpenStudy (anonymous):
so it would be n^2
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OpenStudy (skullpatrol):
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