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Kazeer [188]
4 years ago
10

Please help me and show your work!!

Mathematics
1 answer:
11Alexandr11 [23.1K]4 years ago
4 0
\bf \cfrac{\sqrt{250x^{16}}}{\sqrt{2x}}\qquad 
\begin{cases}
250=2\cdot 5\cdot 5\cdot 5\\
\qquad 2\cdot 5\cdot 5^2\\
x^{16}=x^{8\cdot 2}\\
\qquad (x^8)^2
\end{cases}\implies \cfrac{\sqrt{2\cdot 5\cdot 5^2\cdot (x^8)^2}}{\sqrt{2\cdot  x}}

\bf \cfrac{5x^8\sqrt{2\cdot 5}}{\sqrt{2}\cdot \sqrt{x}}\implies \cfrac{5x^8\underline{\sqrt{2}}\cdot \sqrt{5}}{\underline{\sqrt{2}}\cdot \sqrt{x}}\implies \cfrac{5x^8\sqrt{5}}{\sqrt{x}}\impliedby 
\begin{array}{llll}
\textit{and rationalizing the}\\
\textit{denominator}
\end{array}

\bf \cfrac{5x^8\sqrt{5}}{\sqrt{x}}\cdot \cfrac{\sqrt{x}}{\sqrt{x}}\implies \cfrac{5x^8\sqrt{5}\cdot \sqrt{x}}{(\sqrt{x})^2}\implies \cfrac{5x^8\sqrt{5x}}{x}\implies 5x^8x^{-1}\sqrt{5x}
\\\\\\
5x^{8-1}\sqrt{5x}\implies 5x^7\sqrt{5x}
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Hello,

1)
a_{1}=7
a_{2}=7*3
a_{3}=7*3^2

a_{n}=7*3^{n-1}

s=a_{1}+a_{2}+a_{3}+a_{4}+a_{5}=7*(1+3^1+3^2+3^3+3^4)
=7*\frac{3^{5}-1}{3-1}
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2)
b_{8}=6*8^{8-1}=12582912




















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