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saveliy_v [14]
3 years ago
9

How do you determine the y-intercept and the x-intercept in an algebraic equation? I'm trying to figure out how to tell which is

which. I already know the rule; y=mx+b
Mathematics
1 answer:
natka813 [3]3 years ago
4 0
If you want to find the x-intercept, you replace y in the equation with 0 and solve the equation for x.
If you want to find the y-intercept, you replace x in the equation with 0 and solve the equation for y.

Example:
y=2x+5\\\\
\hbox{x-intercept}\\
0=2x+5\\
2x=-5\\
x=-2.5\\
\hbox{x-intercept is } (-2.5,0)\\\\
\hbox{y-intercept}\\
y=2\cdot0+5\\
y=5\\
\hbox{y-intercept is }(0,5)
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Find the sum \[\sum_{k = 1}^{2004} \frac{1}{1 + \tan^2 (\frac{k \pi}{2 \cdot 2005})}.\]
vampirchik [111]

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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 \thetathere 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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