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3241004551 [841]

3 years ago

11

1. Solve the System by Graphing Y=X+ 1 y = -x + 5

2 answers:

eimsori [14]3 years ago

7 0

Answer The answer is n

nordsb [41]3 years ago

4 0

Answer:

n

Step-by-step explanation:

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SOLVE. (4a-5)÷(6a²+30a)+(a-1)÷(6a²+30a)

Lady bird [3.3K]

180a^3+900a^3+4a-5
__________________ + a-1/ 6a^a
6a ( a+ 5)

4 0

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

Spinner has six sections of equal shape and size. Each section is either black or white. The table shows the result of an experi

Novosadov [1.4K]

Answer:

The spinner is 3 times more likely to land on white.

Step-by-step explanation:

The spinner landed 57 times on white and 19 times on black.

57 is 3 * 19

Answer: The spinner is 3 times more likely to land on white.

4 0

3 years ago

Read 2 more answers

You order 1.5 pounds of turkey at a deli. You will accept the turkey if its weight is between 1.54 lb and 1.46 lb. What absolute

andrey2020 [161]

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1.46<1.54 hope this helps

7 0

3 years ago

An item was sold at a loss of 20%. Had it been sold for Rs 3000more then the profit would have been 10%. What is the cost price

vagabundo [1.1K]

Answer:

Cost Price = Rs 10000

Step-by-step explanation:

Assume:

Cost of the item = x

Item was sold at a loss of 20%:

Loss = 20% of x = 0.2x

Item sold = x - 0.2x = 0.8x

Item sold at a profit of 10%:

Profit = 10% of x = 0.1x

item sold = x + 0.1x = 1.1x

Solve:

Difference = 1.1x - 0.8x = 0.3x

0.3x = Rs 3000

x = Rs 3000 ÷ 0.3

x = Rs 10000

3 0

2 years ago