Matt6288
  • Matt6288
14ab^3/7a^-2b^-1
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Mertsj
  • Mertsj
\[\frac{14ab ^{3}}{7a ^{-2}b ^{-1}}\]
Matt6288
  • Matt6288
yes
Mertsj
  • Mertsj
Hang on a sec

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More answers

Matt6288
  • Matt6288
k
Mertsj
  • Mertsj
\[2a ^{3}b ^{4}\]
Matt6288
  • Matt6288
thx
Mertsj
  • Mertsj
Do you understand?
Matt6288
  • Matt6288
can you explain?
Mertsj
  • Mertsj
The factors that are in the denominator have negative exponents (except for the 7) When you move them to the numerator, you change the exponents to positive.
Mertsj
  • Mertsj
\[\frac{1}{a ^{-n}}=a ^{n}\]
Mertsj
  • Mertsj
\[\frac{a}{a ^{-2}}=a(a ^{2})=a ^{3}\]
Matt6288
  • Matt6288
a^3b^3c/1^-3b^-3c^-1
Matt6288
  • Matt6288
would it be a^6b^6c^2?
Mertsj
  • Mertsj
And\[\frac{b ^{3}}{b ^{-1}}=b ^{3}b ^{1}=b ^{4}\]
Mertsj
  • Mertsj
\[\frac{a ^{3}b ^{3}c}{1^{-3}b ^{-3}c ^{-1}}\]
Mertsj
  • Mertsj
Is that the problem?
Mertsj
  • Mertsj
or is there a typo?
Matt6288
  • Matt6288
ya
Matt6288
  • Matt6288
thats the problem
Mertsj
  • Mertsj
So the denominator contains 1^-3?
Matt6288
  • Matt6288
and my answer is A^6b^6c^2
Matt6288
  • Matt6288
ya
Mertsj
  • Mertsj
yes. If that is a^-3 in the denominator instead of 1^-3
Matt6288
  • Matt6288
thx :) i get it now
Mertsj
  • Mertsj
yw

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