Simplify the quantity of a times b times c to the fourth power divided by the quantity 6 times a to the negative second power times b to the third power all to the negative second power. HELP Answers: six times b to the fifth power divided by a to the sixth power times c to the eighth power thirty six times b to the fourth power divided by a squared times c to the eighth power six times b to the fourth power divided by a to the sixth power times c to the eighth power thirty six times b to the fourth power divided by a to the six power times c to the eighth power
Could you write out the actual numbers for the question? @chocodropa7
Ill try
(abc^4 over 6a^-2b^-3) to the -2 power
^ represent power
\[(abc ^{4}/6*a ^{-2}b ^{3})^{-2}\]
Is it basically like what I posted?
yes
the three is negative though
o wait no it's not your right
Okay
So first we have to multiply a, b, & c by the -2
\[a ^{-2}b ^{-2}c ^{-8}/36a ^{4}b ^{-6}\]
Do you understand that so far?
yes
Now, we must remember that NEGATIVE exponents are forced to the denominator... and turn positive. Any/Everything else stays in the numerator.
But before we do that we must simplify the 'a' and the 'b'
\[a ^{-2}-a ^{4} = a ^{-6}\] \[b ^{-2}-b ^{-6}=b ^{4}\]
The 'c' remains since it can not be simplified anymore than that.
So our a and c are both negative which means they must be shifted to the denominator.
\[\frac{ 36b ^{4} }{ a ^{6}c ^{8} }\]
Understand?
So your last choice... is your answer. :) @chocodropa7
Yes, so the answer would be d thank you!
You are welcome. :)
Could you also help me with how to write cube root of fifteen in exponential form
Honestly, I do not know how to do that.
\[\sqrt[3]{15}\]
That equals.... 2.466212074
oh ok Thank You anyways
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