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ser-zykov [4K]
3 years ago
10

Please help me with this :D

Mathematics
1 answer:
Soloha48 [4]3 years ago
5 0
The answer is c perpendicular bisector:)
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What is the tenth term of the geometric sequence that has a common ratio of `1/3` and 36 as its fifth term?
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\bf \begin{array}{llll} term&value\\ \cline{1-2} a_5&36\\ a_6&36\left( \frac{1}{3} \right)\\ a_7&36\left( \frac{1}{3} \right)\left( \frac{1}{3} \right)\\ a_8&36\left( \frac{1}{3} \right)\left( \frac{1}{3} \right)\left( \frac{1}{3} \right)\\ a_9&36\left( \frac{1}{3} \right)\left( \frac{1}{3} \right)\left( \frac{1}{3} \right)\left( \frac{1}{3} \right)\\ a_{10}&36\left( \frac{1}{3} \right)\left( \frac{1}{3} \right)\left( \frac{1}{3} \right)\left( \frac{1}{3} \right)\left( \frac{1}{3} \right) \end{array}

\bf a_{10}=36\left( \frac{1}{3} \right)^5\implies a_{10}=36\cdot \cfrac{1^5}{3^5}\implies a_{10}=\cfrac{36}{243}\implies a_{10}=\cfrac{4}{27}

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3 years ago
Sqrt(2−7x) -sqrt(x 3) =3
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16x^{2} +31x-2=0 \\(x-2)(16x-1)=0 \\ x+2=0 \\ x=-2 \\ 16x-1=0 \\ x= \frac{1}{16}

\sqrt{2-7  \cdot\frac{1}{16} } - \sqrt{\frac{1}{16} +3}  \neq 3 \\ x=-2


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