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fiasKO [112]
4 years ago
12

As mars revolves around the sun, it travels at a rate of approximately 15 miles per second. convert this rate to kilometers per

second. at this rate, how many kilometers will mars travel in 20 seconds? assume that 1 mile is equal to 1.6 kilometers. don't round your answer
Mathematics
1 answer:
mash [69]4 years ago
6 0
First count how many miles for 20 seconds
15*20=300
Now assuming that a mile is 1.6 km than
300*1.6=480km

Methode 2
Fimd the rate km/s
15*1.6=1*15+0.6*15=15+9=24
than multiply by 12
24*12=480
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Fifteen years ago, you deposited $12,500 into an investment fund. Five years ago, you added an
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Answer:

a) 8.16%

b) $65,762.50

c) $39,701.07

Step-by-step explanation:

Given:

15 years ago, Initial investment = $12500

5 years ago, Investment = $20000

Nominal interest = 8% semi annually for first 10 years

Interest2= 6.5% compunded annually for last five years

a) for the effective annual interest rate (EAR) in the first 10 years, let's use the formula:

[1+(nominal interest rate/number of compounding periods)]^ number of compounding periods-1

EAR = [1 + (\frac{0.08}{2})]^2 - 1

= (1 + 0.04)^2 - 1

= (1.04)^2 - 1

= 1.0816 - 1

= 0.0816

= 8.16%

The effective annual interesting rate, EAR = 8.16 %

b) To find the amount in my account today.

Let's first find the amount for $12500 for 10 years compounded semi annually

= 12,500 +( 12,500 * 8.160% * 10)

= $ 22,700

Let's also find the amount for $32,500($12,500+$20,000) for 5 Years compoundeed annually

$32,500 + ($32,500 * 6.5% *5)

= $ 43,062.50

Money in account today will be:

$22,700 + $43,062.50

= $65,762.50

c) Let's the amount I should have invested to be X

For first 10 years at 8.160%, we have:

Interest Amount = ( X * 8.160% * 10 ) = 0.8160 X

For next 5 years at 6.5%, we have:

Interest Amount = (X * 6.5% * 5) = 0.325 X

Therefore the total money at the end of 15 Years = 85000

0.8160X + 0.3250X + X = $85,000

= 2.141X = $85,000

X = 85,000/2.141

X = 39,701.074 ≈ $39,700

If I wish to have $85,000 now, I should have invested $39,700 15 years ago

7 0
3 years ago
What is the Laplace Transform of 7t^3 using the definition (and not the shortcut method)
Leokris [45]

Answer:

Step-by-step explanation:

By definition of Laplace transform we have

L{f(t)} = L{{f(t)}}=\int_{0}^{\infty }e^{-st}f(t)dt\\\\Given\\f(t)=7t^{3}\\\\\therefore L[7t^{3}]=\int_{0}^{\infty }e^{-st}7t^{3}dt\\\\

Now to solve the integral on the right hand side we shall use Integration by parts Taking 7t^{3} as first function thus we have

\int_{0}^{\infty }e^{-st}7t^{3}dt=7\int_{0}^{\infty }e^{-st}t^{3}dt\\\\= [t^3\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(3t^2)\int e^{-st}dt]dt\\\\=0-\int_{0}^{\infty }\frac{3t^{2}}{-s}e^{-st}dt\\\\=\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt\\\\

Again repeating the same procedure we get

=0-\int_{0}^{\infty }\frac{3t^{2}}{-s}e^{-st}dt\\\\=\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt\\\\\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt= \frac{3}{s}[t^2\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(t^2)\int e^{-st}dt]dt\\\\=\frac{3}{s}[0-\int_{0}^{\infty }\frac{2t^{1}}{-s}e^{-st}dt]\\\\=\frac{3\times 2}{s^{2}}[\int_{0}^{\infty }te^{-st}dt]\\\\

Again repeating the same procedure we get

\frac{3\times 2}{s^2}[\int_{0}^{\infty }te^{-st}dt]= \frac{3\times 2}{s^{2}}[t\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(t)\int e^{-st}dt]dt\\\\=\frac{3\times 2}{s^2}[0-\int_{0}^{\infty }\frac{1}{-s}e^{-st}dt]\\\\=\frac{3\times 2}{s^{3}}[\int_{0}^{\infty }e^{-st}dt]\\\\

Now solving this integral we have

\int_{0}^{\infty }e^{-st}dt=\frac{1}{-s}[\frac{1}{e^\infty }-\frac{1}{1}]\\\\\int_{0}^{\infty }e^{-st}dt=\frac{1}{s}

Thus we have

L[7t^{3}]=\frac{7\times 3\times 2}{s^4}

where s is any complex parameter

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svp [43]

Answer:

Xs=30cm and As=900 cm^{2}

Step-by-step explanation:

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