The function g(x) varies directly with x^2, and g(x) 144 when x=6 which quadratic variation equation models situation

Mathematics

1 answer:

TEA [102]3 years ago

7 0

Hello,
y=k*x² and y=144 if x=6
==>144=k*36==>k=4

g(x)=4*x²

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Csc (2sin^-1(-12/13))<br> that is 2 times inverse sin of -12/13

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Answer: $-\frac{169}{120}$

I'm assuming you mean this:

$cosec(2sin^{-1}(\frac{-12}{13}))$

Recall what the cosecant function represents. It is the reciprocal of a sine function, and is often written in this form:

$\frac{1}{sinx}$

So, using the sine relationship:

$cosec(2sin^{-1}(\frac{-12}{13})) = \frac{1}{sin(2sin^{-1}(\frac{-12}{13}))}$

Let $u = sin^{-1}(\frac{-12}{13})$
$cosec(2sin^{-1}(\frac{-12}{13})) = \frac{1}{sin(2u)}$
$= \frac{1}{2sinu \cdot cosu}$
$= \frac{1}{2sin(sin^{-1}(\frac{-12}{13})) \cdot cos(sin^{-1}(\frac{-12}{13}))}$
$= \frac{1}{\frac{-24}{13} \cdot \sqrt{1 - (\frac{-12}{13})^{2}}}$
$= \frac{1}{\frac{-24}{13}} \cdot \frac{1}{\sqrt{1 - \frac{144}{169}}}$
$= -\frac{13}{24} \cdot \frac{1}{\sqrt{\frac{25}{169}}}$
$= -\frac{13}{24} \cdot \frac{1}{\frac{5}{13}}$
$= -\frac{13}{24} \cdot \frac{13}{5}$
$= -\frac{169}{120}$