What is other ways to 2 + 3

3600 dollars is placed in an account with an annual interest rate of 9%. How much will be in the account after 25 years, to the

Mnenie [13.5K]

Assuming the interest is simple interest, as opposed to compound interest,

$\bf ~~~~~~ \textit{Simple Interest Earned Amount}\\\\
A=P(1+rt)\qquad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to& \$3600\\
r=rate\to 9\%\to \frac{9}{100}\to &0.09\\
t=years\to &25
\end{cases}
\\\\\\
A=3600(1+0.09\cdot 25)\implies 3600(3.25)$

3 0

3 years ago

What is the solution set of x for the given equation? x^2/3=x^1/3+4=6 A. -2, -1 B. 2, -1 C. -8, -1 D. 8, -1 E. 2, 8

Valentin [98]

Answer:

Interpreting as: x^2/3=x^1/3+4=6 A

Input:

x^2/3 = x^(1/3) + 4 = 6 A

3 0

3 years ago

Problem 10: A tank initially contains a solution of 10 pounds of salt in 60 gallons of water. Water with 1/2 pound of salt per g

AysviL [449]

Answer:

The quantity of salt at time t is $m_{salt} = (60)\cdot (30 - 29.833\cdot e^{-\frac{t}{10} })$ , where t is measured in minutes.

Step-by-step explanation:

The law of mass conservation for control volume indicates that:

$\dot m_{in} - \dot m_{out} = \left(\frac{dm}{dt} \right)_{CV}$

Where mass flow is the product of salt concentration and water volume flow.

The model of the tank according to the statement is:

$(0.5\,\frac{pd}{gal} )\cdot \left(6\,\frac{gal}{min} \right) - c\cdot \left(6\,\frac{gal}{min} \right) = V\cdot \frac{dc}{dt}$

Where:

$c$ - The salt concentration in the tank, as well at the exit of the tank, measured in $\frac{pd}{gal}$ .

$\frac{dc}{dt}$ - Concentration rate of change in the tank, measured in $\frac{pd}{min}$ .

$V$ - Volume of the tank, measured in gallons.

The following first-order linear non-homogeneous differential equation is found:

$V \cdot \frac{dc}{dt} + 6\cdot c = 3$

$60\cdot \frac{dc}{dt} + 6\cdot c = 3$

$\frac{dc}{dt} + \frac{1}{10}\cdot c = 3$

This equation is solved as follows:

$e^{\frac{t}{10} }\cdot \left(\frac{dc}{dt} +\frac{1}{10} \cdot c \right) = 3 \cdot e^{\frac{t}{10} }$

$\frac{d}{dt}\left(e^{\frac{t}{10}}\cdot c\right) = 3\cdot e^{\frac{t}{10} }$

$e^{\frac{t}{10} }\cdot c = 3 \cdot \int {e^{\frac{t}{10} }} \, dt$

$e^{\frac{t}{10} }\cdot c = 30\cdot e^{\frac{t}{10} } + C$

$c = 30 + C\cdot e^{-\frac{t}{10} }$

The initial concentration in the tank is:

$c_{o} = \frac{10\,pd}{60\,gal}$

$c_{o} = 0.167\,\frac{pd}{gal}$

Now, the integration constant is:

$0.167 = 30 + C$

$C = -29.833$

The solution of the differential equation is:

$c(t) = 30 - 29.833\cdot e^{-\frac{t}{10} }$

Now, the quantity of salt at time t is:

$m_{salt} = V_{tank}\cdot c(t)$

$m_{salt} = (60)\cdot (30 - 29.833\cdot e^{-\frac{t}{10} })$

Where t is measured in minutes.

7 0

3 years ago

Which number line best shows the position of 113?

sveta [45]

This question is extremely hard for me to understand I don’t know why there are so many + and -. Have a good day.

8 0

3 years ago

Read 2 more answers

A company makes flower pots. The flower pots have an expense equations of E=2.75q+$24,582. What is the fixed cost in the expense

pantera1 [17]

Answer: $24,582 I believe

5 0

3 years ago