<img src="https://tex.z-dn.net/?f=%5Cfrac%7B%281-tan%5E%7B2%7Dx%29tan%282x%29%20%7D%7Btanx%7D" id="TexFormula1" title="\frac{(1-

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Gnesinka [82]

3 years ago

tan^{2}x)tan(2x) }{tanx}" alt="\frac{(1-tan^{2}x)tan(2x) }{tanx}" align="absmiddle" class="latex-formula"> simplify

Mathematics

1 answer:

Ierofanga [76]3 years ago

6 0

Step-by-step explanation:

$\frac{(1-tan^{2}x)tan(2x) }{tanx} \\ but \: \tan(2x) = \frac{2 \tan(x) }{1 - { \tan }^{2}x } \\ therefore : \\ = \frac{(1 - { \tan ^{2}x)(2 \tan(x)) }}{ (\tan(x) )(1 - { \tan ^{2}x })} \\ = \frac{2 \tan(x) }{ \tan(x) } \\ = 2$

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Y is inversely proportional to the square root of x if y=3 when x=25 find y when x is 9

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Answer:

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Step-by-step explanation:

$\textsf{If }y \textsf{ is \underline{inversely proportional} to the square root of }x, \textsf{ then}:$

$y \propto\dfrac{1}{\sqrt{x}} \implies y=\dfrac{k}{\sqrt{x}}\quad \textsf{(where k is some constant)}$

$\textsf{When }x=25, y = 3:$

$\implies 3=\dfrac{k}{\sqrt{25}}$

$\implies 3=\dfrac{k}{5}$

$\implies k=15$

Inputting the <u>found value of k</u> into the equation:

$\implies y=\dfrac{15}{\sqrt{x}}$

To find the value of y when x is 9, <u>substitute</u> x = 9 into the found equation:

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Group like terms:

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Add similar elements:

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