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Lubov Fominskaja [6]
3 years ago
10

Pls help me with this

Mathematics
2 answers:
yuradex [85]3 years ago
5 0
I think it is 1/18 :)
Akimi4 [234]3 years ago
4 0
I think it’s C. 2 bins

hope this helps! sorry if i’m wrong:)
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Jeremy buys 2 sweaters for $18 each. He can use only one of the following coupons:
katen-ka-za [31]

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3 years ago
A 200-gal tank contains 100 gal of pure water. At time t = 0, a salt-water solution containing 0.5 lb/gal of salt enters the tan
Artyom0805 [142]

Answer:

1) \frac{dy}{dt}=2.5-\frac{3y}{2t+100}

2) y(t)=(50+t)- \frac{12500\sqrt{2} }{(50+t)^{\frac{3}{2} }}

3) 98.23lbs

4) The salt concentration will increase without bound.

Step-by-step explanation:

1) Let y represent the amount of salt in the tank at time t, where t is given in minutes.

Recall that: \frac{dy}{dt}=rate\:in-rate\:out

The amount coming in is 0.5\frac{lb}{gal}\times 5\frac{gal}{min}=2.5\frac{lb}{min}

The rate going out depends on the concentration of salt in the tank at time t.

If there is y(t) pounds of  salt and there are 100+2t gallons at time t, then the concentration is: \frac{y(t)}{2t+100}

The rate of liquid leaving is is 3gal\min, so rate out is =\frac{3y(t)}{2t+100}

The required differential equation becomes:

\frac{dy}{dt}=2.5-\frac{3y}{2t+100}

2) We rewrite to obtain:

\frac{dy}{dt}+\frac{3}{2t+100}y=2.5

We multiply through by the integrating factor: e^{\int \frac{3}{2t+100}dt }=e^{\frac{3}{2} \int \frac{1}{t+50}dt }=(50+t)^{\frac{3}{2} }

to get:

(50+t)^{\frac{3}{2} }\frac{dy}{dt}+(50+t)^{\frac{3}{2} }\cdot \frac{3}{2t+100}y=2.5(50+t)^{\frac{3}{2} }

This gives us:

((50+t)^{\frac{3}{2} }y)'=2.5(50+t)^{\frac{3}{2} }

We integrate both sides with respect to t to get:

(50+t)^{\frac{3}{2} }y=(50+t)^{\frac{5}{2} }+ C

Multiply through by: (50+t)^{-\frac{3}{2}} to get:

y=(50+t)^{\frac{5}{2} }(50+t)^{-\frac{3}{2} }+ C(50+t)^{-\frac{3}{2} }

y(t)=(50+t)+ \frac{C}{(50+t)^{\frac{3}{2} }}

We apply the initial condition: y(0)=0

0=(50+0)+ \frac{C}{(50+0)^{\frac{3}{2} }}

C=-12500\sqrt{2}

The amount of salt in the tank at time t is:

y(t)=(50+t)- \frac{12500\sqrt{2} }{(50+t)^{\frac{3}{2} }}

3) The tank will be full after 50 mins.

We put t=50 to find how pounds of salt it will contain:

y(50)=(50+50)- \frac{12500\sqrt{2} }{(50+50)^{\frac{3}{2} }}

y(50)=98.23

There will be 98.23 pounds of salt.

4) The limiting concentration of salt is given by:

\lim_{t \to \infty}y(t)={ \lim_{t \to \infty} ( (50+t)- \frac{12500\sqrt{2} }{(50+t)^{\frac{3}{2} }})

As t\to \infty, 50+t\to \infty and \frac{12500\sqrt{2} }{(50+t)^{\frac{3}{2} }}\to 0

This implies that:

\lim_{t \to \infty}y(t)=\infty- 0=\infty

If the tank had infinity capacity, there will be absolutely high(infinite) concentration of salt.

The salt concentration will increase without bound.

6 0
3 years ago
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