carlenstar281
help me please Algebra 1
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ghazi
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2\[(-2)^{-3*24}=(-2)^{72}\]
whpalmer4
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I think that should be
\[(-m)^{-3}n\]
Yes?
ghazi
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what exactly is your question, what are you supposed to find?
whpalmer4
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First thing to remember is that \[x^{-m} = \frac{1}{x^m}\]If we rewrite the expression in that fashion, we get \[\frac{n}{(-m)^3} = \frac{-24}{(-2)^3}\]Can you evaluate that?
whpalmer4
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\[(-2)^3 = (-2)*(-2)*(-2)\]
ghazi
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well no
carlenstar281
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its negative though right?
carlenstar281
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its -3 right?
whpalmer4
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@ghazi what do you mean, "well no"?
carlenstar281
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he means its not -4
whpalmer4
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Did someone say it was?
\[\frac{-24}{(-2)*(-2)*(-2)}\]What does the expression on the bottom equal?
ghazi
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its not -4 , its 4 ,if the explanation of @whpalmer4 is true and second i am not sure if that explanation is true, it would be great if you could take snapshot and post it here :D @carlenstar281
carlenstar281
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I cant but thank you for helping me :)
ghazi
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:) YW
whpalmer4
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Yes, all of my statements are true and correct. The answer is not 4.
carlenstar281
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are you sure i get 4?
whpalmer4
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What is -2 * -2 * -2?
ghazi
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\[\frac{ -24 }{ -2*-2*-2 }=\frac{ -24 }{ -8 }---> -A*-A*-A=-A\]
ghazi
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and \[\frac{ -24 }{ -8 }=\frac{ 24 }{ 8 }=3\] LOL
ghazi
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that was the silliest mistake i've ever done
carlenstar281
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sorry computer froze haha its okay i did that to still thanks for helping!
ghazi
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np :)
carlenstar281
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also thank you to @whpalmer4 as well haha
whpalmer4
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Another way of looking at this is \[\frac{-24}{(-1*2)^3} = \frac{-24}{(-1)^3*2^3}\]\(-1^n = -1, n\) odd so we have
\[\frac{-24}{-1*2^3} = \frac{-24}{-8} = 3\]
whpalmer4
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We can factor out that -1 because \[(ab)^n = a^nb^n\]