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2 years ago

A family buys a studio apartment for 120,000. They pay down payment of 18,000.

Mathematics

2 answers:

Gelneren [198K]2 years ago

7 0

Answer:

A) 15% B) 40%

Step-by-step explanation:

Schach [20]2 years ago

4 0

Answer:

Step-by-step explanation:

A. 15 Percent

B. 40 Percent

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What is 8 divided by 2.5 (In fraction form)

Alika [10]

8/2.5

= 80/25
= 16/5
= 3 1/5

4 0

3 years ago

Naviah buys 8 but no more than 12 gallons of gas each week. The price of gas ranged from $2.25 to $2.50 each week. Write a compo

aleksklad [387]

Answer:

Step-by-step explanation:

8 0

3 years ago

38600 mg = n dag what does n equal???

Vlad [161]

Answer:
n = 3.68

Explanation:
We know that:
1 mg is equivalent to 0.0001 dag
To convert 38600 mg to dag, we will simply use cross multiplication as follows:
1 mg ..............> 0.0001 dag
38600 mg ......> n

n = (38600*0.0001) / 1
n = 3.68

Hope this helps :)

3 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

2 years ago

(x - 3) (3x + 7) = 0 Find the values of x.

expeople1 [14]

(x-3)(3x+7)=0
x-3=0 or 3x+7=0

1. x-3=0
x₁=3

2. 3x+7=0
3x=-7
x₂=-7/3 = -2(1/3)

8 0

3 years ago

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