(nearest hundredth) 0.523 x 8 =

Mathematics

1 answer:

seropon [69]3 years ago

8 0

Answer:

4.18

Step-by-step explanation:

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This is so frustrating i keep submitting only getting this answer partially correct, help pls

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3 years ago

Find the sum \[\sum_{k = 1}^{2004} \frac{1}{1 + \tan^2 (\frac{k \pi}{2 \cdot 2005})}.\]

vampirchik [111]

Answer:1002

Step-by-step explanation:

$\Rightarrow \sum_{k=1}^{2004}\left ( \frac{1}{1+\tan ^2\left ( \frac{k\pi }{2\cdot 2005}\right )}\right )$

and $1+\tan ^2\theta =\sec^2\theta$

and $\cos \theta =\frac{1}{\sec \theta }$

$\Rightarrow \sum_{k=1}^{2004}\left ( \cos^2\frac{k\pi }{2\cdot 2005}\right )$

as $\cos ^2\theta =\cos ^2(\pi -\theta )$

Applying this we get

$\Rightarrow \sum_{1}^{1002}\left [ \cos^2\left ( \frac{k\pi }{2\cdot 2005}\right )+\cos^2\left ( \frac{(2005-k)\pi }{2\cdot 2005}\right )\right ]$

every $\theta$ there exist $\pi -\theta$

such that $\sin^2\theta +\cos^2\theta =1$

therefore

$\Rightarrow \sum_{1}^{1002}\left [ \cos^2\left ( \frac{k\pi }{2\cdot 2005}\right )+\cos^2\left ( \frac{(2005-k)\pi }{2\cdot 2005}\right )\right ]=1002$

6 0

3 years ago

There is a photo attached<br><br>tan 270 is incorrect!

wlad13 [49]

$\dfrac{1-\cos135^\circ}{\sin135^\circ}=\dfrac{1-\left(\cos^2 67.5^\circ-\sin^2 67.5^\circ\right)}{2\sin67.5^\circ\cos67.5^\circ}=\dfrac{1-\left(1-\sin^2 67.5^\circ-\sin^2 67.5^\circ\right)}{2\sin67.5^\circ\cos67.5^\circ}=\medskip\\=\dfrac{2\sin^2 67.5^\circ}{2\sin67.5^\circ\cos67.5^\circ}=\dfrac{\sin 67.5^\circ}{\cos 67.5^\circ}=\tan 67.5^\circ$