Write as a sum, difference or multiple of logarithms. In(8x) = [ ? ]8+ [ ]x

Mathematics

1 answer:

deff fn [24]3 years ago

8 0

Answer:

In(8x)=[In]8 + [In]x

Step-by-step explanation:

By product rule:

$In(8x)=[In]8 + [In]x$

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Answer:

750 minutes

Step-by-step explanation:

8 pages in 20 minutes

300 / 8 = 37.5

37.5 * 20 = 750

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Write the given expression as an algebraic expression in x. tan(2 cos^-1(x))

miv72 [106K]

$\bf tan\left[ 2cos^{-1}(x) \right]\implies tan\left[ 2cos^{-1}\left( \frac{x}{1} \right) \right]
\\\\\\
\textit{if we say }cos^{-1}\left( \frac{x}{1} \right)=\theta\textit{ that means }tan\left[ 2cos^{-1}\left( \frac{x}{1} \right) \right]\iff tan(2\theta)\\\\
-----------------------------\\\\
cos^{-1}\left( \frac{x}{1} \right)=\theta\implies cos(\theta)=\cfrac{x}{1}\cfrac{\leftarrow adjacent=a}{\leftarrow hypotenuse=c}\\\\
-----------------------------\\\\
$

$\bf \textit{again, using the pythagorean theorem to get the opposite side}
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c^2=a^2+b^2\implies \pm\sqrt{c^2-a^2}=b\implies \pm\sqrt{1^2-x^2}=b
\\\\\\
\pm\sqrt{1-x^2}=b\\\\
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tan(2\theta)=\cfrac{2tan(\theta)}{1-tan^2(\theta)}\implies tan(2\theta)=\cfrac{2\cdot \frac{\pm\sqrt{1-x^2}}{x}}{1-\left( \frac{\pm\sqrt{1-x^2}}{x} \right)^2}
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$\bf tan(2\theta)=\cfrac{\frac{\pm2\sqrt{1-x^2}}{x}}{1-\frac{1-x}{x}}\implies 
tan(2\theta)=\cfrac{\frac{\pm2\sqrt{1-x^2}}{x}}{\frac{x-1+x}{x}}
\\\\\\
tan(2\theta)=\cfrac{\pm2\sqrt{1-x^2}}{x}\cdot \cfrac{x}{2x-1}
\implies 
tan(2\theta)=\cfrac{\pm 2\sqrt{1-x^2}}{2x-1}$

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Step-by-step explanation:

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Question is attached

RoseWind [281]

Answer: A, 63 minutes

Step-by-step explanation:

We can start by making this into two ratios. Isa finished 30% in 27 minutes, so 30/27 is our first ratio. We want to know how long it will take for the rest of the homework to be completed. 30% is finished already, which leaves 70% (100-30=70). The time left is x. Our second ratio is 70/x. Now, we can set the two ratios equal to each other to solve for x.

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