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lisabon 2012 [21]

3 years ago

5

Benjamin invests $400 in a savings account that earns 5% interest each year. Which graph shows the balance of Benjamin's savings

y(in dollars) after x years

1 answer:

KengaRu [80]3 years ago

4 0

Answer: it’s D

Step-by-step explanation:

Starts at 400 and goes up 5%

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B
because rise over run

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3 years ago

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6. Mary worked 27 hours last week. If she gets paid $12/hour, what is her<br> gross pay?

kupik [55]

Answer:

324

Step-by-step explanation:

you would multiply 27 × 12 and that would equal 324

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3 years ago

Find the circumference of the circle.<br> 7 m

Nutka1998 [239]

Answer:

43.96m is the answer

Step-by-step explanation:

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2 years ago

6) Find the measure of ∠G.

topjm [15]

Answer:

m<G = 71

Step-by-step explanation:

Find x first

7x + 1 + 6x - 6 + 55 = 180

Combine like terms:

13x + 50 = 180

- 50 - 50

13x = 130

/13 /13

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Then apply to problem

7 (10) + 1

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8 0

2 years ago

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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

4 years ago