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Llana [10]
2 years ago
15

Please help me please help me please help me please help me

Mathematics
1 answer:
ehidna [41]2 years ago
7 0

Answer:10%= 310

5%= 155

15%=465

Step-by-step explanation: Solution for what is 15% of 3100

3100/x=100/15

(3100/x)*x=(100/15)*x       - we multiply both sides of the equation by x

3100=6.66666666667*x       - we divide both sides of the equation by (6.66666666667) to get x

3100/6.66666666667=x

465=x

x=465

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If your balance after 8 years in the bank at a rate of interest of 11%% is 3783193 how much was your original deposit? 29 Origin
Wittaler [7]

Answer:

The original amount deposit is $1641627.68.

Step-by-step explanation:

Given : If your balance after 8 years in the bank at a rate of interest of 11%% is 3783193.

To find : How much was your original deposit?  

Solution :

Applying interest formula,

A=P(1+r)^t

Where, A is the amount i.e. A=$3783193

P is the principal value i.e. the original deposit

r is the interest rate r=11%=0.11

t is the time t=8 years.

Substitute the value in the formula,

3783193=P(1+0.11)^8

3783193=P(1.11)^8

3783193=P\times 2.30

P=\frac{3783193}{2.30}

P=1641627.68

Therefore, The original amount deposit is $1641627.68.

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
The coordinates of the vertices of a right triangle are (0,0), (5,0) and (0,3) what statement is true
Tom [10]
What grade level is this
7 0
3 years ago
Polyhedron P is a cube with a corner removed and relocated to the top of P. Polyhedron Q is a cube. How do their surface areas c
tatiyna

Answer:

  • C. P’s surface area is greater than Q’s surface area.

Step-by-step explanation:

<em>Refer to attached</em>

With a corner removed, it doesn't change surface area of the cube as same area is exposed equal to the area of 3 small faces.

With the corner relocated to the top of the cube, it covers the area of one small face but adds up small 5 faces (4 sides and one top). In total the P has more surface area than the Q.

Correct answer choice is C.

4 0
3 years ago
733x9,600,121=<br> I need help hurry its due rn<br> Also I will give you 25 pts and brainliest
insens350 [35]

Answer:

7036888693

Step-by-step explanation:

6 0
3 years ago
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