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zalisa [80]
3 years ago
9

Plz help someone products of exponents

Mathematics
2 answers:
algol [13]3 years ago
7 0
2. would be 7^11 because when you’re multiplying with exponents on the same base number you add them together.
3. would be 54^3 because they have the same exponent, so you just multiply the base numbers
4. is true because when you multiply exponents on the same base number you add them, and 5+4 and 4+5 are both 9 so the order doesn’t matter
Airida [17]3 years ago
6 0
1. 7^30 because 5*6 equals 30 and u keep the base
2. 54^3 you multiply the base and keep the exponent
3. Yes because the associative property of multiplication still applies to exponents
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3 years ago
Find the area of the shaded region ​
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so hmmm let's get the area of the whole hexagon, and then get the area of the circle inside it, then <u>subtract the area of the circle from that of the hexagon's</u>, what's leftover is what we didn't subtract, namely the shaded part.

\textit{area of a regular polygon}\\\\ A=\cfrac{1}{4}ns^2\cot\stackrel{\stackrel{degrees}{\downarrow }}{\left( \frac{180}{n} \right)}~ \begin{cases} n=\textit{number of sides}\\ s=\textit{length of a side}\\[-0.5em] \hrulefill\\ n=\stackrel{hexagon}{6}\\ s=\frac{9}{2} \end{cases}\implies A=\cfrac{1}{4}(6)\left( \cfrac{9}{2} \right)^2 \cot\left( \cfrac{180}{6} \right)

A=\cfrac{1}{4}(6)\cfrac{9^2}{2^2} \cot(30^o)\implies A=\cfrac{243}{8}\cot(30^o)\implies A=\cfrac{243\sqrt{3}}{8} \\\\[-0.35em] ~\dotfill\\\\ \textit{area of circle}\\\\ A=\pi r^2~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=\frac{4}{5} \end{cases}\implies A=\pi \left( \cfrac{4}{5} \right)^2\implies A=\cfrac{16\pi }{25} \\\\[-0.35em] ~\dotfill

\stackrel{\textit{area of the hexagon}}{\cfrac{243\sqrt{3}}{8}}~~ - ~~\stackrel{\textit{area of the circle}}{\cfrac{16\pi }{25}}\implies \cfrac{6075\sqrt{3}-128\pi }{200}

5 0
2 years ago
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you move 3 down and 1 to the right. So basically -3/1 which can be simplified to -3!!!!

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