If two distinct planes intersect then they intersect in exactly one line

Find m<br> (3x-70)<br><br> Help me please

soldi70 [24.7K]

Answer:

cannot see your diagram

5 0

3 years ago

Mary Catherine makes 2 gallons of punch how many cups of punch did she make

Tresset [83]

There are 16 cups in the gallon

6 0

3 years ago

It is now 3:15 pm. Is it possible to drive 135 miles and arrive before 5:00 pm if ypu drove 55 mph? Explain your answer

Troyanec [42]

Well, how long will it take you to drive 135 miles at 55mph?

at 55 mph, you're doing 55 miles every hour, so we can simply get the quotient of 135/55 and that's how many hours it'll take you to drive 135 miles at that speed.

$\bf \cfrac{135}{55}\implies \cfrac{27}{11}\implies 2\frac{5}{11}~hours\implies \textit{about 2hrs, 27 minutes}$

so, it takes you that long, however, from 3:15 to 5:00pm there are only 45mins + 60mins or 1hr and 45 minutes, namely 1¾ hr.

so 2hrs and 27 minutes is much later than 1¾ hr, so, no dice, you can't arrive at 5pm, actually you'll arrive around 5:42pm.

6 0

3 years ago

Read 2 more answers

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