Show a full solution to the inequality [(x+2)/(x-4)]

Mathematics

1 answer:

Papessa [141]4 years ago

3 0

Your sign chart is a bit off. The first two rows are correct. However keep in mind that you factored out a negative (when you factored -2x+14) so you need to negate the signs of the f(x) row.

Another thing to keep in mind: x = 4 is not allowed as its the vertical asymptote. However, x = 7 is valid. So you'll have an open circle at x = 4 and a close circle at x = 7

The answer in interval notation is (-infinity, 4) U [7, infinity)
which translates to this disjoint set of inequalities: x < 4 or x >= 7

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The intensity of light with wavelength λ traveling through a diffraction grating with N slits at an angle θ is given by I(θ) = N

Ymorist [56]

Answer:

0.007502795

Step-by-step explanation:

We have

N = 10,000

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Replacing these values in the expression for k:

$\bf k=\frac{\pi Ndsin\theta}{\lambda}=\frac{\pi10^4*10^{-4}sin\theta}{632.8*10^{-9}}=\frac{\pi 10^9sin\theta}{632.8}$

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$\bf I(\theta)=\frac{N^2sin^2(k)}{k^2}=\frac{10^8sin^2(\frac{\pi 10^9sin\theta}{632.8})}{(\frac{\pi 10^9sin\theta}{632.8})^2}$

The total light intensity is then

$\bf \int_{-10^{-6}}^{10^{-6}} I(\theta)d\theta=\int_{-10^{-6}}^{10{-6}}\frac{10^8sin^2(\frac{\pi 10^9sin\theta}{632.8})}{(\frac{\pi 10^9sin\theta}{632.8})^2}d\theta$

Since $\bf I(\theta)$ is an even function

$\bf \int_{-10^{-6}}^{10^{-6}} I(\theta)d\theta=2\int_{0}^{10^{-6}}I(\theta)d\theta$

and we only have to divide the interval $\bf [0,10^{-6}]$ in five equal sub-intervals $\bf I_1,I_2,I_3,I_4,I_5$ with midpoints $\bf m_1,m_2,m_3,m_4,m_5$

The sub-intervals and their midpoints are

$\bf I_1=[0,\frac{10^{-6}}{5}]\;,m_1=10^{-5}\\I_2=[\frac{10^{-6}}{5},2\frac{10^{-6}}{5}]\;,m_2=3*10^{-5}\\I_3=[2\frac{10^{-6}}{5},3\frac{10^{-6}}{5}]\;,m_3=5*10^{-5}\\I_4=[3\frac{10^{-6}}{5},4\frac{10^{-6}}{5}]\;,m_4=7*10^{-5}\\I_5=[4\frac{10^{-6}}{5},10^{-6}]\;,m_5=9*10^{-5}$