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satela [25.4K]
3 years ago
15

A sporting goods company can ship

Mathematics
1 answer:
cricket20 [7]3 years ago
7 0

Answer:

12.5

Step-by-step explanation:

75 ÷ 6

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A 1000-liter (L) tank contains 500 L of water with a salt concentration of 10 g/L. Water with a salt concentration of 50 g/L flo
djverab [1.8K]

Answer:

a) y(t)=50000-49990e^{\frac{-2t}{25}}

b) 31690.7 g/L

Step-by-step explanation:

By definition, we have that the change rate of salt in the tank is \frac{dy}{dt}=R_{i}-R_{o}, where R_{i} is the rate of salt entering and R_{o} is the rate of salt going outside.

Then we have, R_{i}=80\frac{L}{min}*50\frac{g}{L}=4000\frac{g}{min}, and

R_{o}=40\frac{L}{min}*\frac{y}{500} \frac{g}{L}=\frac{2y}{25}\frac{g}{min}

So we obtain.  \frac{dy}{dt}=4000-\frac{2y}{25}, then

\frac{dy}{dt}+\frac{2y}{25}=4000, and using the integrating factor e^{\int {\frac{2}{25}} \, dt=e^{\frac{2t}{25}, therefore  (\frac{dy }{dt}+\frac{2y}{25}}=4000)e^{\frac{2t}{25}, we get   \frac{d}{dt}(y*e^{\frac{2t}{25}})= 4000 e^{\frac{2t}{25}, after integrating both sides y*e^{\frac{2t}{25}}= 50000 e^{\frac{2t}{25}}+C, therefore y(t)= 50000 +Ce^{\frac{-2t}{25}}, to find C we know that the tank initially contains a salt concentration of 10 g/L, that means the initial conditions y(0)=10, so 10= 50000+Ce^{\frac{-0*2}{25}}

10=50000+C\\C=10-50000=-49990

Finally we can write an expression for the amount of salt in the tank at any time t, it is y(t)=50000-49990e^{\frac{-2t}{25}}

b) The tank will overflow due Rin>Rout, at a rate of 80 L/min-40L/min=40L/min, due we have 500 L to overflow \frac{500L}{40L/min} =\frac{25}{2} min=t, so we can evualuate the expression of a) y(25/2)=50000-49990e^{\frac{-2}{25}\frac{25}{2}}=50000-49990e^{-1}=31690.7, is the salt concentration when the tank overflows

4 0
4 years ago
Evaluate 4x^3 +2 for x= -1
padilas [110]

4x³+2

4(-1)³+2

(4 * -1) +2

-4+2 = -2


7 0
3 years ago
Tristan records the number of customers who visit the store each hour on a Saturday. His data representing the first seven hours
Alona [7]

Answer:

Option C is correct.

Step-by-step explanation:

We have been given total 8 data

Data is : 12,15,18,20,23,23,28

Here, n=7

Median is \frac{n+1}{2}=\frac{7+1}{2}=4th term

Median is 20.

When one data is added n=8

The median of  even number of data is  

median=\frac{\frac{n}{2}\text{th and}\frac{n}{2}+1\text{th}}{2}

Median is \frac{8}{2}\text{th and}\frac{8}{2}+1th

Median is 4\text{th and}5th

4th terms is 20

5th terms is 23

Since, median should not be change hence,

Median will be 20.

Therefore, option C is correct.

6 0
3 years ago
Read 2 more answers
Write each fraction as a decimal<br>1. 5/11 <br>2. 1/8 <br>3. 2 1/3​
Kobotan [32]

Answer:

5/11=0.45 (45 repeating)

1/8=0.125

2 1/3=2.3 (3 repeating)

8 0
3 years ago
Find the exact value of the expression.
Tresset [83]

\sin(a-b)=\sin a \cos b-\cos a \sin b. If we let a=\cos^{-1} \left(\frac{5}{6} \right) and b=\tan^{-1} \left(\frac{1}{2} \right), then the given expression is equal to:

\sin \left(\cos^{-1} \left(\frac{5}{6}} \right) \right) \cos \left(\tan^{-1} \left(\frac{1}{2} \right) \right)-\cos\left(\cos^{-1} \left(\frac{5}{6} \right) \right) \sin \left( \tan^{-1} \left(\frac{1}{2} \right) \right)

Using the Pythagorean identities \sin^{2} x+\cos^{2} x=1 and \tan^{2} x+1=\sec^{2} x,

1) \sin^{2} \left(\cos^{-1} \left(\frac{5}{6} \right) \right)+\cos^{2}  \left(\cos^{-1} \left(\frac{5}{6} \right) \right)=1\\\sin^{2} \left(\cos^{-1} \left(\frac{5}{6} \right) \right)+\frac{25}{36}=1\\\sin^{2} \left(\cos^{-1} \left(\frac{5}{6} \right) \right)=\frac{11}{36}\sin \left(\cos^{-1} \left(\frac{5}{6} \right) \right)=\frac{\sqrt{11}}{6}

2) \tan^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)+1=\sec^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)\\\frac{1}{4}+1=\sec^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)\\\frac{5}{4}=\sec^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)\\\sec \left(\tan^{-1} \left(\frac{1}{2} \right) \right)=\frac{\sqrt{5}}{2}\\\implies \cos \left(\tan^{-1} \left(\frac{1}{2} \right) \right)=\frac{2}{\sqrt{5}}=\frac{2\sqrt{5}}{5}

\cos^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)+\sin^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)=1\\\frac{4}{5}+\sin^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)=1\\\sin^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)=\frac{1}{5}\\\left(\tan^{-1} \left(\frac{1}{2} \right) \right)=\frac{1}{\sqrt{5}}=\frac{\sqrt{5}}{5}

This means we can write the original expression as:

\left(\frac{\sqrt{11}}{6} \right) \left(\frac{2\sqrt{5}}{5} \right)-\left(\frac{5}{6} \right) \left(\frac{\sqrt{5}}{5} \right)\\=\frac{2\sqrt{11}\sqrt{5}}{30}-\frac{5\sqrt{5}}{30}\\=\boxed{\frac{\sqrt{5}(2\sqrt{11}-5)}{30}}

6 0
2 years ago
Read 2 more answers
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