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3 years ago

Today, a boat is valued at $45,000. The

Mathematics

1 answer:

aleksley [76]3 years ago

5 0

Question:

Today, a boat is valued at $45,000. The value is expected to decrease at a rate of 8.5% each year. What is the value of the boat expected to be 3 years from now?

Answer:

C.) 34,472.74

Hope this helps!

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Find the value of z

Stells [14]

The value of z is 17

w=24 , 24-7 is 17

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3 years ago

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Given:3x<9 choose the solution set

8_murik_8 [283]

You would divide both sides by 3 and would be left with x<3

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4 years ago

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PLEASE HELP GUYS I’m being timed

fomenos

The question is the base of the triangle, so we use the formula : b²=c²–a²

——————————————————————

 ${b}^{2} = \sqrt{c^{2} - {a}^{2} }$ 

 ${b}^{2} = \sqrt{ {16}^{2} - {8}^{2} }$ 

 ${b}^{2} = \sqrt{256 - 64} = \sqrt{192}$ 

 $\sqrt{192} =8 \sqrt{3}$ 

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3 years ago

John invests $500 in a savings account with a nominal annual interest rate of 10% compounded quarterly. How many years will it t

Illusion [34]

It will take 10 years for the investment to reach $1,000.

Sure hope this helps you

8 0

2 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