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Anton [14]
3 years ago
8

X = 53. What is the value of x ?

Mathematics
1 answer:
kompoz [17]3 years ago
7 0
53



20 words do b annoying
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um pai tem hoje 54 anos e seus quatro filhos tem juntos 39 anos. dentro de quantos anos a idade do pai será a soma da idade dos
Elan Coil [88]

a1 + r.N = b1 + R.N  

54 + 1.N = 39 + 4.N  

4.N - 1.N = 54 - 39  

3.N = 15  

N = 15 / 3  

N = 5 anos

8 0
3 years ago
Which number is located between 15 and 20 on a number line?
jeyben [28]

Answer: 17.5  is located between them.

Step-by-step explanation:

To find out which number is located between the numbers, find their mean average which sum of the numbers divide by the number of values.

15 + 20 = 35

35 / 2 = 17.5

8 0
3 years ago
If F, G, and H are the midpoints of the sides
V125BC [204]

Answer:FH=24 JL=76 KJ=60 FJ=30

Step-by-step explanation:

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3 years ago
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

8 0
3 years ago
Does a rectangle have one line of symmetry
Thepotemich [5.8K]

No, it has more than one. A rectangle is symmetric with respect to its diagonals, and the lines connecting the midpoints of two opposite sides.

8 0
3 years ago
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