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3 years ago

Sharon read a 300-page book. She read at a rate of 15 pages per day in d days. Write an equation to describe this situation.

Mathematics

1 answer:

777dan777 [17]3 years ago

3 0

Answer:

300 ÷ 15 = D

D=20

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F = 1.8C + 32 what is c

zepelin [54]

Step-by-step explanation:

right.

and therefore c is in this case the name of the variable in the equation (or function).

add on.

ah, do you mean (and does this strange problem description mean) that you want the reverse formula : to get C out of F ?

in that case

F = 1.8c + 32 = 9/5 × c + 32

F - 32 = 9/5 × c

5/9 × (F - 32) = c

c = 5/9 × (F - 32)

that's it.

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2 years ago

What is the area of the parallelogram? A parallelogram with a base of 14 centimeters and a height of 5 centimeters. 32 cm2 65 cm

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Answer:

70 cm2

Step-by-step explanation:

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3 years ago

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What is the Least common multiple (LCM) of the denominators in the equation below?

xz_007 [3.2K]

Answer:

Step-by-step explanation:

because

32 : 32 =1

32 : 16 = 2

32 : 8 = 4

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3 years ago

Max is thinking of a double fact. It has a sum greater than 10 but less than 13. What number fact could it be?

igor_vitrenko [27]

The answere would be 11 and 12

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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 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

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

By the midpoint rule

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

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

Sharon read a 300-page book. She read at a rate of 15 pages per day in d days. Write an equation to describe this situation.​

Sharon read a 300-page book. She read at a rate of 15 pages per day in d days. Write an equation to describe this situation.