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2 years ago

9

Sally borrowed $9,000 from the bank to buy a new car. She has to pay back the money in 5 years. After 5 years, she will owe $1,8

00 in interest. What the simple interest rate on her loan? *

1 answer:

Aneli [31]2 years ago

6 0

7.42 would be her interest rate.

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Solve this 4xX3X5x=541

Allisa [31]

Answer:

Can you re-type the question. This doesn't make any sense.

Step-by-step explanation:

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3 years ago

A 75-gallon tank is filled with brine (water nearly saturated with salt; used as a preservative) holding 11 pounds of salt in so

Debora [2.8K]

Let $A(t)$ = amount of salt (in pounds) in the tank at time $t$ (in minutes). Then $A(0) = 11$ .

Salt flows in at a rate

$\left(0.6\dfrac{\rm lb}{\rm gal}\right) \left(3\dfrac{\rm gal}{\rm min}\right) = \dfrac95 \dfrac{\rm lb}{\rm min}$

and flows out at a rate

$\left(\dfrac{A(t)\,\rm lb}{75\,\rm gal + \left(3\frac{\rm gal}{\rm min} - 3.25\frac{\rm gal}{\rm min}\right)t}\right) \left(3.25\dfrac{\rm gal}{\rm min}\right) = \dfrac{13A(t)}{300-t} \dfrac{\rm lb}{\rm min}$

where 4 quarts = 1 gallon so 13 quarts = 3.25 gallon.

Then the net rate of salt flow is given by the differential equation

$\dfrac{dA}{dt} = \dfrac95 - \dfrac{13A}{300-t}$

which I'll solve with the integrating factor method.

$\dfrac{dA}{dt} + \dfrac{13}{300-t} A = \dfrac95$

$-\dfrac1{(300-t)^{13}} \dfrac{dA}{dt} - \dfrac{13}{(300-t)^{14}} A = -\dfrac9{5(300-t)^{13}}$

$\dfrac d{dt} \left(-\dfrac1{(300-t)^{13}} A\right) = -\dfrac9{5(300-t)^{13}}$

Integrate both sides. By the fundamental theorem of calculus,

$\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac1{(300-t)^{13}} A\bigg|_{t=0} - \frac95 \int_0^t \frac{du}{(300-u)^{13}}$

$\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac{11}{300^{13}} - \frac95 \times \dfrac1{12} \left(\frac1{(300-t)^{12}} - \frac1{300^{12}}\right)$

$\displaystyle -\dfrac1{(300-t)^{13}} A = \dfrac{34}{300^{13}} - \frac3{20}\frac1{(300-t)^{12}}$

$\displaystyle A = \frac3{20} (300-t) - \dfrac{34}{300^{13}}(300-t)^{13}$

$\displaystyle A = 45 \left(1 - \frac t{300}\right) - 34 \left(1 - \frac t{300}\right)^{13}$

After 1 hour = 60 minutes, the tank will contain

$A(60) = 45 \left(1 - \dfrac {60}{300}\right) - 34 \left(1 - \dfrac {60}{300}\right)^{13} = 45\left(\dfrac45\right) - 34 \left(\dfrac45\right)^{13} \approx 34.131$

pounds of salt.

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2 years ago

The period of a pendulum is given by the equation

sattari [20]

Answer: g > 0

Step-by-step explanation:

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3 years ago

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The table below shows data for a class's mid-term and final exams:

lina2011 [118]

Mid term :
Q1 = (88 + 85)/2 = 86.5
Q2 = (92 + 95)/2 = 93.5
Q3 = 100
IQR = Q3 - Q1 = 100 - 86.5 = 13.5

final exams :
Q1 = (65 + 78)/2 = 71.5
Q2 = (88 + 82)/2 = 85
Q3 = (95 + 93)/2 = 94
IQR = Q3 - Q1 = 94 - 71.5 = 22.5

so the final exams has the largest IQR

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3 years ago

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Round to the nearest tenth if neccessary<br> 69.904 divided by 34

Cloud [144]

Answer:

2.056

rounded to the nearest tenth: 2.1

Plz mark brainliest!! :DD

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3 years ago

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