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BlackZzzverrR [31]
3 years ago
10

M = {2, 3, 5, 7, 11, 13}​

Mathematics
1 answer:
qwelly [4]3 years ago
3 0

Answer:

what to do buddy

just write a valid question

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At the fair, the ratio of males to females was 4:5. If there are 180 people at the fair, how mamy males are there?​
Setler79 [48]

Answer:

80 males

Step-by-step explanation:

180 / 9 = 20

20 x 4 = 80

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Reese is selling lemonade at the parade. He gets to keep 50% of the money he collects. A large lemonade is $6.00 and a small lem
Ivenika [448]

Answer:

3l + 1s

Step-by-step explanation:

Given that:

The expression to represent 50% of the money collected : 0.50(6l + 2s)

Expanding the expression :

0.50(6l + 2s)

(0.50 * 6l) + (0.50 * 2s)

3l + 1s

Hence, the simplified expression is 3l + 1s

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
4 years ago
Solution to the system of equations
lara31 [8.8K]

Answer:

huh

Step-by-step explanation:

7 0
4 years ago
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