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Flura [38]
4 years ago
6

Please answer these questions as soon as possible!!! If done by Wed i will award you 25 points! And don't answer something silly

just for the points or you will be reported!! Thanks!!
Look at the picture plz (all of the questions).

Mathematics
1 answer:
vodka [1.7K]4 years ago
7 0
\sqrt{a} =a^{\frac{1}{2}}\\\\ \sqrt[3]{c} =c^{\frac{1}{3}}\\\\ \sqrt[5]{x^4} =x^{\frac{4}{5}}\\\\ \frac{1}{ \sqrt{e} } = e ^{\frac{-1}{2}}\\\\ \frac{1}{ \sqrt[7]{x^2} } =x^{\frac{-2}{7}}\\\\ \sqrt[4]{(16a^4)^5} = 32a^5\\\\ \sqrt[8]{(b^{\frac{3}{5}})^5} = b^{\frac{3}{8}}

\sqrt[3]{ \frac{c^{15}}{c^9} } =c^2\\\\ \sqrt[3]{27d^2 \times 27d^{-4}} = 9d^{\frac{-2}{3}}


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(-3, 1)

The solution is the point at which both lines intersect.

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Step-by-step explanation:

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4 years ago
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lys-0071 [83]

\displaystyle
(x+y)^n=\sum_{k=0}^n\binom{n}{k}x^{n-k}y^k

<em>-------------------------------------------------------------</em>


\displaystyle&#10;(x^2+y)^n=\sum_{k=0}^n\binom{n}{k}x^{2n-2k}y^k\\&#10;n=5\\&#10;k=3\\\\\binom{5}{3}=\dfrac{5!}{3!2!}=\dfrac{4\cdor5}{2}=10

<u>It's 10.</u>

----------------------------------------------------

\displaystyle&#10;(3x^2+y)^n=\sum_{k=0}^n\binom{n}{k}(3x)^{2n-2k}y^k=\sum_{k=0}^n\binom{n}{k}\cdot 3^{2n-2k}\cdot x^{2n-2k}y^k\\\\&#10;n=5\\&#10;k=4\\\\&#10;\binom{5}{3}\cdot3^{2\cdot5-2\cdot4}=10\cdot3^{2}=10\cdot9=90

<u>It's 90</u>

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This is your answer thanks!


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