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3 years ago

Can somebody help me this math problem please correctly

Mathematics

1 answer:

Vinil7 [7]3 years ago

8 0

Answer: the answer is 1.8 x $10^{-3}$ because their is 3 (0) on the left side and left Side is negative

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The approximate proportion of all university students that prefer to eat organic foods is 18 percent.

Step-by-step explanation:

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3 years ago

Answer this pleaseeeee past due

Airida [17]

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3 years ago

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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

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

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3 years ago

The function rule h=18+1.5n

vichka [17]

What inputs do you want?

n | h
----------------------
5 | 25.5
6 | 27
7 | 28.5
8 | 30
9 | 31.5
10 | 33

7 0

4 years ago