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Damm [24]
4 years ago
5

Help solve this math problem please

Mathematics
1 answer:
quester [9]4 years ago
7 0

Answer:

=536870912

Step-by-step explanation:

\frac{\left(2^3\right)^{10}\cdot \:2^{-8}}{2^{-7}}\\\mathrm{Cancel\:}\frac{\left(2^3\right)^{10}\cdot \:2^{-8}}{2^{-7}}:\quad \frac{\left(2^3\right)^{10}}{2}\\\frac{\left(2^3\right)^{10}\cdot \:2^{-8}}{2^{-7}}\\\mathrm{Apply\:exponent\:rule}:\quad \frac{x^a}{x^b}=\frac{1}{x^{b-a}}\\\frac{2^{-8}}{2^{-7}}=\frac{1}{2^{-7-\left(-8\right)}}\\=\frac{\left(2^3\right)^{10}}{2^{-7-\left(-8\right)}}\\\mathrm{Subtract\:the\:numbers:}\:-7-\left(-8\right)=\\=\frac{\left(2^3\right)^{10}}{2}

\mathrm{Simplify\:}\left(2^3\right)^{10}:\quad 2^{30}\\\left(2^3\right)^{10}\\\mathrm{Apply\:exponent\:rule}:\quad \left(a^b\right)^c=a^{bc}\\=2^{3\cdot \:10}\\\mathrm{Multiply\:the\:numbers:}\:3\cdot \:10=30\\=2^{30}\\=\frac{2^{30}}{2}\\\mathrm{Cancel\:the\:common\:factor:}\:2\\=2^{29}\\2^{29}=536870912\\=536870912

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Your constant goes up if the wheel diameter goes down. Think about this. Do you ride a bicycle? I do. It makes perfect sense to me that if the wheel is small, it will have to turn more often to go a mile. No matter where that 0.00125 comes from or how it was derived, the constant will have to go up if the wheel gets smaller.


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Step-by-step explanation:

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