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3 years ago

PLZ HELP ASAP!!!!

1 answer:

8 0

2,356 students would say corn is their favorite because if 62 out of 100 loved corn and there are 3,800 students you can multiply 62 by 38 and get your answer.

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If there are 36 boys and 60 girls in a room, fill out all of the possible ratios of boys to girls that could be made.

marishachu [46]

Answer:

You have the two numbers, so 36/60 is the ratio of boys to girls.

Step-by-step explanation:

You will want to simplify those numbers (reduce to lowest terms).

8 0

3 years ago

Read 2 more answers

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

7.) Find the Unit Rate: 32 flowers per 2 vases (flowers per vase)

IgorLugansk [536]

Answer:

Step-by-step explanation:

I JUST TOOK THE TEST YW

4 0

2 years ago

Lindsay is 5 years younger than Mark. Seven years ago, the sum of their ages was 31.

Rashid [163]

Answer:

The present age of Mark is 25 years.

And present age of Lindsay is 20 years.

Step-by-step explanation:

Let us assume:

The present age of Lindsay = l

The present age of Mark = m

⇒ Age of Lindsay seven years ago = (l -7) years

and Age of Mark seven years ago = (m -7) years

According to question:

Lindsay is 5 years younger than Mark ⇒ l = m - 5

Seven years ago, the sum of their ages was 31 ⇒ (l-7) + (m -7) = 31

So, the given set of equation are:

l = m - 5 ......... (1)

(l-7) + (m -7) = 31

⇒ ((m - 5) -7) + (m -7) = 32 (from (1)

or, m -12 + m - 7 = 31

or, 2m = 31 + 19 = 50

⇒ m = 50 / 2 = 25, or m = 25

Hence,the present age of Mark = 25 years

So,the present age of Lindsay = 25 - 5 = 20 years

7 0

3 years ago

Help Asap

prisoha [69]

Answer:

Step-by-step explanation:

The number of tubes can be found from an inequality that relates the number of square inches that can be painted to the number of square inches that must be painted.

Let t represent the number of tubes.

1400t ≥ 3672 . . . . . the number of tubes that will paint at least 3672 in²

t > 2.623 . . . . . divide by 1400

If only whole numbers of tubes are available, then ...

3 tubes are needed to paint the ramp.

8 0

2 years ago

Read 2 more answers