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QveST [7]
3 years ago
11

A motorcycle purchased for 12,500 today will be valued 7% less each year. How much could you expect to receive for the motorcycl

e at the end of 4 years of you wanted to sell it?
Mathematics
1 answer:
Mice21 [21]3 years ago
8 0
Year 1
$12500 x .07 = $875
12500-875=11625
Year 2
11625 x .07 = $813.75
11625-813.75=10811
Year 3
10811 x .07 = $756.79
10811-756.79= 10054.21
Year 4
10054.21 x .07= 703.79
10054.21-703.79= 9350.42
So after 4 years expect to get $9350.42
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How would you round this to the nearest mile 244.1399784294165944?
KonstantinChe [14]
It would be rounded down to 244 as its only .1 so its closer to 244 than it is to 245
4 0
3 years ago
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Urgent please help..
Katyanochek1 [597]

Answer:

18π

Step-by-step explanation:

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Plug in the numbers into the equation.

but leave the pi

6 0
3 years ago
What’s the LMC of 11 and 44
juin [17]

Answer:

The answer is 44.  The least common mutiple (lcm) of 11 and 44 is 44.  Please mark me brainliest! :)

3 0
3 years ago
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Pleaseeee helppp I don’t understand
Rudiy27

Answer:

c

Step-by-step explanation:

5 0
2 years ago
A 500500-gallon tank initially contains 200200 gallons of brine containing 100100 pounds of dissolved salt. brine containing 11
mixer [17]
Let A(t) denote the amount of salt in the tank at time t. We're given that the tank initially holds A(0)=100 lbs of salt.

The rate at which salt flows in and out of the tank is given by the relation

\dfrac{\mathrm dA}{\mathrm dt}=\underbrace{\dfrac{11\text{ lb}}{1\text{ gal}}\times\dfrac{44\text{ gal}}{1\text{ min}}}_{\text{rate in}}-\underbrace{\dfrac{A(t)}{200+(44-11)t}\times\dfrac{11\text{ gal}}{1\text{ min}}}_{\text{rate out}}
\implies A'(t)+\dfrac{11}{200+33t}A(t)=484

Find the integrating factor:

\mu(t)=\exp\left(\displaystyle\int\frac{11}{200+33t}\,\mathrm dt\right)=(200+33t)^{1/3}

Distribute \mu(t) along both sides of the ODE:

(200+33t)^{1/3}A'(t)+11(200+33t)^{-2/3}A(t)=484(200+33t)^{-1/3}
\bigg((200+33t)^{1/3}A(t)\bigg)'=484(200+33t)^{-1/3}
A(t)=484\displaystyle\int(200+33t)^{-1/3}\,\mathrm dt
A(t)=22(200+33t)^{2/3}+C

Since A(0)=100, we get

100=22(200)^{2/3}+C\implies C\approx-652.39

so that the particular solution for A(t) is

A(t)=22(200+33t)^{2/3}-652.39

The tank becomes full when the volume of solution in the tank at time t is the same as the total volume of the tank:

200+(44-11)t=500\implies 33t=300\implies t\approx9.09

at which point the amount of salt in the solution would be

A(9.09)\approx733.47\text{ lb}
4 0
3 years ago
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