Anthonyk2013
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8-1/4t^3=0
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solving for t
- Thread starter Anthonyk2013
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D
Deleted member 4993
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8-1/4t^3=0
What are your thoughts?
Please share your work with us ...even if you know it is wrong
If you are stuck at the beginning tell us and we'll start with the definitions. Do you know the definition of functor? Where is it used?
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Anthonyk2013
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I know this incorrect but not sure what to do.
-1/4t3 =8
-1/4=8t3
-1/32=t3
-1/4t3 =8
-1/4=8t3
-1/32=t3
D
Deleted member 4993
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I know this incorrect but not sure what to do.
-1/4t3 =8
-1/4=8t3
-1/32=t3
So your expression is:
\(\displaystyle - \dfrac{1}{4t^3} \ = 8\)
\(\displaystyle t^3 = - \dfrac{1}{32} = - 2^{-5}\)
\(\displaystyle \displaystyle{t = -2^{\left [-\frac{5}{3}\right ]}}\)
Anthonyk2013
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So your expression is:
\(\displaystyle - \dfrac{1}{4t^3} \ = 8\)
\(\displaystyle t^3 = - \dfrac{1}{32} = - 2^{-5}\)
\(\displaystyle \displaystyle{t = -2^{\left [-\frac{5}{3}\right ]}}\)
Yes up until \(\displaystyle = - 2^{-5}\) Don't understand this part where -5 came from
\(\displaystyle = - 2^{-5}\)
\(\displaystyle \displaystyle{t = -2^{\left [-\frac{5}{3}\right ]}}\)
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Deleted member 4993
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Yes up until \(\displaystyle = - 2^{-5}\) Don't understand this part where -5 came from
\(\displaystyle = - 2^{-5}\)
\(\displaystyle \displaystyle{t = -2^{\left [-\frac{5}{3}\right ]}}\)
\(\displaystyle 2^5 \ = \ 32\) → \(\displaystyle \dfrac{1}{32} \ = \ 2^{-5}\)
Anthonyk2013
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8 - 1/(4t^3) = 0 is what I should have wrote.
Anthonyk2013
Junior Member
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Could you explain the procedure of what is happening a bit more, is there algebra, indices rules going on there?
I'm assuming you're still confused about where the 2^(-5) came from. Think about what negative exponents mean. n^(-1) can be rewritten as 1/n, so what does n^(-2) mean? Apply that same logic to 2^(-5). Once you have your new way of writing it, do you see that term in your original equation? Then when you get to t^3 = 2^(-5), you can simplify by taking the cube root of both sides. Like so:
\(\displaystyle t=\sqrt[3]{2^{-5}}\)
And since n^(1/3) means the same thing as the cube root of n, do you see why you can rewrite that as t = 2^(-5/3)?
\(\displaystyle t=\sqrt[3]{2^{-5}}\)
And since n^(1/3) means the same thing as the cube root of n, do you see why you can rewrite that as t = 2^(-5/3)?
Anthonyk2013
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I'm assuming you're still confused about where the 2^(-5) came from. Think about what negative exponents mean. n^(-1) can be rewritten as 1/n, so what does n^(-2) mean? Apply that same logic to 2^(-5). Once you have your new way of writing it, do you see that term in your original equation? Then when you get to t^3 = 2^(-5), you can simplify by taking the cube root of both sides. Like so:
\(\displaystyle t=\sqrt[3]{2^{-5}}\)
And since n^(1/3) means the same thing as the cube root of n, do you see why you can rewrite that as t = 2^(-5/3)?
\(\displaystyle t=\sqrt[3]{2^{-5}}\)
Ya that make sense now, I just couldn't picture what was happening.
Thanks you