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Snezhnost [94]
3 years ago
13

Y = 40x + 600 in point-slope form

Mathematics
2 answers:
bezimeni [28]3 years ago
6 0

Answer:

y - 600 = 40(x - 0)

Step-by-step explanation:

y = 40x + 600

Slope: 40

Find any point on the line:

Let x = 0

y = 600

(0,600)

Point slope form:

y - y1 = m(x - x1)

y - 600 = 40(x - 0)

qaws [65]3 years ago
5 0
Answer: y-600=40(x-0)
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Replace x with π/2 - x to get the equivalent integral

\displaystyle \int_{-\frac\pi2}^{\frac\pi2} \cos(\cot(x) - \tan(x)) \, dx

but the integrand is even, so this is really just

\displaystyle 2 \int_0^{\frac\pi2} \cos(\cot(x) - \tan(x)) \, dx

Substitute x = 1/2 arccot(u/2), which transforms the integral to

\displaystyle 2 \int_{-\infty}^\infty \frac{\cos(u)}{u^2+4} \, du

There are lots of ways to compute this. What I did was to consider the complex contour integral

\displaystyle \int_\gamma \frac{e^{iz}}{z^2+4} \, dz

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\displaystyle \left|\int_{z=Re^{i0}}^{z=Re^{i\pi}} f(z) \, dz\right| \le \frac{\pi R}{|R^2-4|}

which vanishes as R goes to ∞. Then by the residue theorem, we have in the limit

\displaystyle \int_{-\infty}^\infty \frac{\cos(x)}{x^2+4} \, dx = 2\pi i {} \mathrm{Res}\left(\frac{e^{iz}}{z^2+4},z=2i\right) = \frac\pi{2e^2}

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