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3 years ago

8

you put $3000 into an account earning interest compounded annually. after 6 years you have $4049.58. what percent interest did y

ou earn

1 answer:

kicyunya [14]3 years ago

7 0

Per year? i think you would do 4049.58 - 3000 and then divide by 6, right?

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4(x + 3) = 6-x what is the answer I need help please

Amanda [17]

4(x + 3) = 6 - x

First, expand to remove parentheses.
$4x + 12 = 6 - x$
Second, subtract '6' from both sides.
$4x + 12 - 6 = -x$
Third, subtract '12 - 6' to get 6.
$4x + 6 = -x$
Fourth, subtract '4x' from both sides.
$6 = -x - 4x$
Fifth, since 'x' can be referred to as '1', add it to '4x' to get '-5x'.
$6 = -5x$
Sixth, divide both sides by '-5'.
$\frac{6}{-5} =x$
Seventh, change the whole fraction into a negative.
$-\frac{6}{5} =x$
Eighth, switch your sides.
$x = -\frac{6}{5}$

Answer as fraction: $-\frac{6}{5}$
Answer as decimal: -1.2

4 0

3 years ago

A tank contains 30 lb of salt dissolved in 500 gallons of water. A brine solution is pumped into the tank at a rate of 5 gal/min

Dmitry [639]

At any time $t$ (min), the volume of solution in the tank is

$500\,\mathrm{gal}+\left(5\dfrac{\rm gal}{\rm min}-5\dfrac{\rm gal}{\rm min}\right)t=500\,\mathrm{gal}$

If $A(t)$ is the amount of salt in the tank at any time $t$ , then the solution has a concentration of $\dfrac{A(t)}{500}\dfrac{\rm lb}{\rm gal}$ .

The net rate of change of the amount of salt in the solution, $A'(t)$ , is the difference between the amount flowing in and the amount getting pumped out:

$A'(t)=\left(5\dfrac{\rm gal}{\rm min}\right)\left(\left(2+\sin\dfrac t4\right)\dfrac{\rm lb}{\rm gal}\right)-\left(5\dfrac{\rm gal}{\rm min}\right)\left(\dfrac{A(t)}{50}\dfrac{\rm lb}{\rm gal}\right)$

Dropping the units and simplifying, we get the linear ODE

$A'=10+5\sin\dfrac t4-\dfrac A{10}$

$10A'+A=100+50\sin\dfrac t4$

Multiplying both sides by $e^{10t}$ allows us to identify the left side as a derivative of a product:

$10e^{10t}A'+e^{10t}A=\left(100+50\sin\dfrac t4\right)e^{10t}$

$\left(e^{10}tA\right)'=\left(100+50\sin\dfrac t4\right)e^{10t}$

$e^{10t}A=\displaystyle\int\left(100+50\sin\dfrac t4\right)e^{10t}\,\mathrm dt$

Integrate and divide both sides by $e^{10t}$ to get

$A(t)=10-\dfrac{200}{1601}\cos\dfrac t4+\dfrac{8000}{1601}\sin\dfrac t4+Ce^{-10t}$

The tanks starts off with 30 lb of salt, so $A(0)=30$ and we can solve for $C$ to get a particular solution of

$A(t)=10-\dfrac{200}{1601}\cos\dfrac t4+\dfrac{8000}{1601}\sin\dfrac t4+\dfrac{32,220}{1601}e^{-10t}$

6 0

3 years ago

Ian is going to water his plants is it reasonable to say he will fill the watering can with 3 liters of water explain

Y_Kistochka [10]

This is possible for a smaller watering can. 3 liters is about 3/4 of a gallon (imagine a jug of milk).

It is equivalent to 3000 ml, which would also be 100 ounces of water.(5 -20 oz bottles of water).

4 0

4 years ago

The population of a small town is decreasing exponentially at a rate of 14.3% each year. The current population is 9,400 people.

KonstantinChe [14]

Using an exponential function, the inequality is given as follows:

$9400(0.857)^t < 6000$

The solution is t > 2.9, hence the tax status will change within the next 3 years.

<h3>What is an exponential function?</h3>

A decaying exponential function is modeled by:

$A(t) = A(0)(1 - r)^t$

In which:

A(0) is the initial value.

r is the decay rate, as a decimal.

For this problem, the parameters are given as follows:

A(0) = 9400, r = 0.143.

The population after t years is modeled by:

$A(t) = A(0)(1 - r)^t$

$A(t) = 9400(1 - 0.143)^t$

$A(t) = 9400(0.857)^t$

The tax status will change when:

$A(t) < 6000$

Hence the inequality is:

$9400(0.857)^t < 6000$

Then:

$(0.857)^t < \frac{6000}{9400}$

$\log{(0.857)^t} < \log{\left(\frac{6000}{9400}\right)}$

$t\log{0.857} < \log{\left(\frac{6000}{9400}\right)}$

Since both logs are negative:

$t > \frac{\log{\left(\frac{6000}{9400}\right)}}{\log{0.857}}$

t > 2.9.

The solution is t > 2.9, hence the tax status will change within the next 3 years.

More can be learned about exponential functions at brainly.com/question/25537936

#SPJ1

7 0

2 years ago

An employee compiled sales data for a company once each month. The scatter plot below shows the sales (in multiples of $1000) fo

stepan [7]

Sales aer in multipules of 1000
normally the output is y and input is x

input would be number of months
output would be the sales

so

subsitute 40 for x

y=0.94*(40)+12.5
y=37.6+12.5
y=50.1
this is in thousands
times 1000

$50,100 is the sales after 40 months

4 0

3 years ago

Read 2 more answers