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3 years ago

Help meeeeeeeeeeessss

Mathematics

1 answer:

saveliy_v [14]3 years ago

6 0

Answer:

I think its c but just so you know I'm not really smart so I might be wrong

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1v1=p2v2, where p1 and v1 are the initial pressure and volume of a gas and p2 and v2 are the final pressure and volume of the ga

Dima020 [189]

Step-by-step explanation:

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.... wrong answer

7 0

3 years ago

someone please help me with this if one person already answers then dont answer i only have a few points left and need it for sc

Harrizon [31]

Hi !! the correct answer would be the last option, due to the fact that -2 and 2 both have the same absolute value, hope it helps !! :]

3 0

3 years ago

Read 2 more answers

Please help me answer ASAP

bagirrra123 [75]

Answer: the answer is approximately 41.65428, not rounded (maybe re do all the steps just in case)

Step-by-step explanation:

1. label the space between X and W the unknown.

2. You will be using SOH-CAH-TOA

3. Label the X as the opposite because it is the opposite side of where your degree is located.

4. Label five as adjacent because it’s next to the degree.

5. You will use TOA since you have adjacent and opposite values.

6. Write the equation tan58=x/5, the unknown goes on the top because O is first, the five goes on the bottom because A is last on TOA.

7. After you have the equation time 5 on both sides leaving you out with (5)tan58=x

8. On your calculator time (5)tan58 which will give you x.

6 0

2 years ago

HELP!!!

Hoochie [10]

Answer:

AT = 8 $\sqrt{2}$

Step-by-step explanation:

Using the sine ratio in the right triangle

and sin45° = $\frac{1}{\sqrt{2} }$

sin45° = $\frac{opposite}{hypotenuse}$ = $\frac{AB}{AT}$ = $\frac{8}{AT}$

Multiply both sides by AT

AT × sin45° = 8, that is

AT × $\frac{1}{\sqrt{2} }$ = 8

Multiply both sides by $\sqrt{2}$

AT = 8 $\sqrt{2}$

3 0

3 years ago

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

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Help meeeeeeeeeeessss