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svet-max [94.6K]
3 years ago
10

Prime time factorization multiples of 63.

Mathematics
1 answer:
11111nata11111 [884]3 years ago
3 0
63 = 3,3,7 = 3squared,7
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Romashka-Z-Leto [24]
Subtract 3 from the right and do the same to the left which leaves you to x = -1
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Cathy lives in a state where speeders fined $ 10 for each mile per hour over the s speed limit cathy was given a fine for $80for
LuckyWell [14K]

Answer:

58 miles per hour

Step-by-step explanation:

First you need to divide 80 by 10 to see how many miles she was over the speed limit.

80/10 = 8 miles over the speed limit.

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So Cathy was driving at 58 miles per hour

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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

\bf d=10^{-4}

\bf \lambda = 632.8*10^{-9}

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}

So, the intensity is given by the function

\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 <em>total light intensity</em> 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 <em>even function</em>

\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}

<em>By the midpoint rule</em>

\bf \int_{0}^{10^{-6}}I(\theta)d\theta\approx\frac{10^{-6}}{5}[I(m_1)+I(m_2)+...+I(m_5)]

computing the values of I:

\bf I(m_1)=I(10^{-5})=\frac{10^8sin^2(\frac{\pi 10^9sin(10^{-5})}{632.8})}{(\frac{\pi 10^9sin(10^{-5})}{632.8})^2}=13681.31478

\bf I(m_2)=I(3*10^{-5})=\frac{10^8sin^2(\frac{\pi 10^9sin(3*10^{-5})}{632.8})}{(\frac{\pi 10^9sin(3*10^{-5})}{632.8})^2}=4144.509447

Similarly with the help of a calculator or spreadsheet we find

\bf I(m_3)=3.09562973\\I(m_4)=716.7480066\\I(m_5)=211.3187228

and we have

\bf \int_{0}^{10^{-6}}I(\theta)d\theta\approx\frac{10^{-6}}{5}[I(m_1)+I(m_2)+...+I(m_5)]=\frac{10^{-6}}{5}(18756.98654)=0.003751395

Finally the the total light intensity

would be 2*0.003751395 = 0.007502795

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3 years ago
PLEASE HELP URGENT DUE IN 2 HOURS
Kaylis [27]

Answer:

look it up on slader.com thats the sams problem that i have had before and it worked on slader.

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Determine the number of cement blocks needed to put up a wall measuring 20 meters in length and 1.5 meters in height. Each block
Montano1993 [528]

calculate area of wall

20 x 1.5 = 30 square meters

 now calculate area of 1 block

12 x 18 =216 square centimeters

 convert cm to meters  1 cm^2 = 0.0001m^2

 so 216 x 0.0001 = 0.0216 sq meter for 1 block

 divide area of wall by area of block

30/0.0216 = 1388.88 so 1389 blocks are required.

8 0
3 years ago
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