Mathematics, 31.03.2020 01:49 sophie5988

A car's cooling system has a capacity of 20 quarts. Initially, the system contains a mixture of 3 quarts of antifreeze and 17 quarts of water. Water runs into the system at the rate of 1 gal min , then the homogeneous mixture runs out at the same rate. In quarts, how much antifreeze is in the system at the end of 5 minutes? (Round your answer to two decimal places.)

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Answer from: KBH3298

After 5 minutes, there will be 0.98 quarts of antifreeze and 19.02 quarts of water in the tank.

Since a car's cooling system has a capacity of 20 quarts, and initially, the system contains a mixture of 3 quarts of antifreeze and 17 quarts of water, and water runs into the system at the rate of 1 gal min, then the homogeneous mixture runs out at the same rate, to determine, in quarts, how much antifreeze is in the system at the end of 5 minutes, the following calculation must be performed:

1 quart = 0.25 gallons 3 x 0.25 = 0.75 gallons of antifreeze 17 x 0.25 = 4.25 gallons of water 1 = 0.2 of 5 Minute 1 = 0.75 x 0.8 = 0.6 Minute 2 = 0.6 x 0.8 = 0.48 Minute 3 = 0.48 x 0.8 = 0.384 Minute 4 = 0.384 x 0.8 = 0.3072 Minute 5 = 0.3072 x 0.8 = 0.24576 0.24576 gallons = 0.98 quarts

Therefore, after 5 minutes, there will be 0.98 quarts of antifreeze and 19.02 quarts of water in the tank.

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Answer from: 885122bah

1.10 quarts

Step-by-step explanation:

We are given that

Capacity of system=20 quarts= $\frac{20}{4}=5 gal$

1 gallon=4 quarts

Initially mixture contains

Antifreeze=3 quarts= $\frac{3}{4}=0.75 gal$

Water=17 quarts= $\frac{17}{4}=4.25 gal$

Water runs in to the system=1gal/min

We have to find the antifreeze in the system at the end of 5 minutes.

Concentration inflow=0

Let Concentration outflow=x

Rate of system=Concentration inflow(water inflow rate)-Concentration outflow(water outflow rate)

$\frac{dx}{dt}=0\times 1-\frac{1}{5} x=-\frac{x}{5}$

$\frac{dx}{x}=-\frac{dt}{5}$

Taking integration on both sides

$\int\frac{dx}{x}=-\frac{1}{5}\int dt$

$ln x=-\frac{1}{5}t+C$

x=3 and t=0

ln 3=C

$ln x-ln 3=-\frac{1}{5}t$

$ln\frac{x}{3}=-\frac{1}{5}t$

Using formula

$ln m-ln n=ln\frac{m}{n}$

$\frac{x}{3}=e^{-0.2t}$

$x=3e^{-0.2t}$

Substitute t=5

$x=3e^{-0.2\times 5}=3e^{-1}=1.10$

Hence, in the end of 5 minute ,the antifreeze in the system=1.10 quarts

A car's cooling system has a capacity of 20 quarts. Initially, the system contains a mixture of 3 qu

Answer from: Quest

c: lkllmn.

hope it's you

What additional piece of information do you need to prove that lkn~knm by the sas similarity theorem

Answer from: Quest

(f - g)(x) = 5x^3 - 3

step-by-step explanation:

rearrange these functions vertically, with like terms in separate columns:

f(x) = 5x^3 - 2

g(x) = +1

now subtract, column by column:

(f - g)(x) = 5x^3 - 3

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