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3 years ago

If you have $10,000 in a savings account and you earn 8% simple interest, how much money will be in the account after 3 years?

Mathematics

1 answer:

ludmilkaskok [199]3 years ago

4 0

Answer:

$ 12400

Step-by-step explanation:

P =$10,000

r= 10%

t = 3 year

$I = P*r*t=\dfrac{10000*8*3}{100}$

I = $ 2400

Amount = P + I = 10000 + 2400 = $ 12,400

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If B(x)=-1,then what is x

jekas [21]

Step-by-step explanation:

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3 years ago

What is an example of when you might need to use decimals and other rational numbers in everyday life?

nignag [31]

Well, in the store, when you buy something, there are decimals, also, speed limit for your car, and time, but that is not an exact thing for this question, because 60 can't be a decimal. you cant make it so you can show 60 seconds as a decimal without an irrational number. sometimes, the number just goes on forever.

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4 years ago

(1 point) A very large tank initially contains 100L of pure water. Starting at time t=0 a solution with a salt concentration of

Paraphin [41]

1. dy/dt is the net rate of change of salt in the tank over time. As such, it's equal to the difference in the rates at which salt enters and leaves the tank.

The inflow rate is

(0.4 kg/L) (6 L/min) = 2.4 kg/min

and the outflow rate is

(concentration of salt at time t) (4 L/min)

The concentration of salt is the amount of salt (in kg) per unit volume (in L). At any time t > 0, the volume of solution in the tank is

100 L + (6 L/min - 4 L/min) t = 100 L + (2 L/min) t

That is, the tank starts with 100 L of pure water, and every minute 6 L of solution flows in and 4 L is drained, so there's a net inflow of 2 L of solution per minute. The amount of salt at time t is simply y(t). So, the outflow rate is

(y(t)/(100 + 2t) kg/L) (4 L/min) = 2 y(t) / (50 + t) kg/min

and the differential equation for this situation is

$\dfrac{dy}{dt} = 2.4 \dfrac{\rm kg}{\rm min} - \dfrac{2y}{50+t} \dfrac{\rm kg}{\rm min}$

There's no salt in the tank at the start, so y(0) = 0.

2. Solve the ODE. It's linear, so you can use the integrating factor method.

$\dfrac{dy}{dt} = 2.4 - \dfrac{2y}{50+t}$

$\dfrac{dy}{dt} + \dfrac{2}{50+t} y = 2.4$

The integrating factor is

$\mu = \displaystyle \exp\left(\int \frac{2}{50+t} \, dt\right) = \exp\left(2\ln|50+t|\right) = (50+t)^2$

Multiply both sides of the ODE by µ :

$(50+t)^2 \dfrac{dy}{dt} + 2(50+t) y = 2.4 (50+t)^2$

The left side is the derivative of a product:

$\dfrac{d}{dt}\left[(50+t)^2 y\right] = 2.4 (50+t)^2$

Integrate both sides with respect to t :

$\displaystyle \int \dfrac{d}{dt}\left[(50+t)^2 y\right] \, dt = \int 2.4 (50+t)^2 \, dt$

$\displaystyle (50+t)^2 y = \frac{2.4}3 (50+t)^3 + C$

$\displaystyle y = 0.8 (50+t) + \frac{C}{(50+t)^2}$

Use the initial condition to solve for C :

$y(0) = 0 \implies 0 = 0.8 (50+0) + \dfrac{C}{(50+0)^2} \implies C = -100,000$

Then the amount of salt in the tank at time t is given by the function

$y(t) = 0.8 (50+t) - \dfrac{10^5}{(50+t)^2}$

so that after t = 50 min, the tank contains

$y(50) = 0.8 (50+50) - \dfrac{10^5}{(50+50)^2} = \boxed{70}$

kg of salt.

7 0

2 years ago

Simply -3 3/4 divided by 2 1/2

Dima020 [189]

Answer:

$=-1\frac{1}{2}$

Step-by-step explanation:

Given the following question:

$-3\frac{3}{4} \div2\frac{1}{2}$

In order to find the answer, we first convert the mixed numbers into improper fractions. Then we use KCF (Keep, Change, Flip) and solve.

$-3\frac{3}{4} \div2\frac{1}{2}$
$-3\frac{3}{4} =12+3=-\frac{15}{4}$
$2\frac{1}{2}=2\times2=4+1=\frac{5}{2}$
$-\frac{15}{4} \div\frac{5}{2}$
$-\frac{15}{4} \times\frac{2}{5}$
$15\times2=30$
$4\times5=20$
$=-\frac{30}{20}$
$-\frac{30}{20} \div10=-\frac{3}{2}$
$-\frac{3}{2}=3\div2=-1\frac{1}{2}$
$=-1\frac{1}{2}$

Hope this helps.

4 0

2 years ago

If 2 and 4 are the factors of a number than is 8 always a factor? true or false

kotegsom [21]

Answer:

hmmm I think it's true, if it's not my bad.

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3 years ago