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2 years ago

14

michael borrowed $4,500 from the bank. the back charged a simple interest rate of 10.5% for the 3 years he had the back loan. wh

at was the total cost of the loan ? HURRY .

1 answer:

elixir [45]2 years ago

4 0

Answer:

$1417.50+$4500= 5917.50

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The .............. from the regular selling price to the discounted sale price is 20%

11Alexandr11 [23.1K]

Answer:

80%

Step-by-step explanation:

regular price is 100%

so, after the discounted price is 20%

the discount is 80%

4 0

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.

7 0

1 year ago

Ray BD bisects ∠ABC so that m∠DBC = (x + 7) and m∠ABD = (2x − 11). What is x? A. 6 B. 18 C. 29 D. 54

nata0808 [166]

Answer:

B

Step-by-step explanation:

Here, we want to the find the value of the x.

By bisecting the angle, the ray in question divides it into two other angles which are equal.

We proceed as follows;

x + 7 = 2x -11

Collect like

terms

2x -x = 7 + 11

x = 18

6 0

3 years ago

Cyclic quadrilateral.

ad-work [718]

What do you mean, you need more information in your queston p,ease try to ask it again, it makes no Sence

5 0

3 years ago

Read 2 more answers

What is the x-intercept of the line with this equation 1/3x+y= -15?<br><br> (_,0)

Sidana [21]

( - 45, 0)

to find the x-intercept let y = 0 in the equation

x = 0 → $\frac{1}{3}$ x = - 15

multiply both sides by 3

x = 3 × - 15 = - 45

x-intercept = ( - 45, 0)

4 0

3 years ago