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Sergio [31]
3 years ago
14

Don and Marie like to go to a good restaurant. Dinners are $16 to $19 dollars with soup or salad. A beverage is $1.50. Dessert i

s $3.75. If one dinner is $16, the other is $19 and each diner has a beverage and dessert, what will the cost be if a 6% tax is charged and a 15% tip on the pre-tax amount is left?
55.06
54.09
56.08
Mathematics
1 answer:
andreev551 [17]3 years ago
3 0

Hi there!

The correct answer is calculated like so:

16 + 19 + 3 + 3.75 + 3.75 = 45.5

45.5 * 0.06 = 2.73

45.5 * 0.15 = 6.83

45.5 + 2.73 + 6.83 = $55.06

Your friend, ASIAX

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Suppose that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another br
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Answer:

A=1500-1450e^{-\dfrac{t}{250}}

Step-by-step explanation:

The large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved.

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Brine solution with concentration of 2 lb/gal is pumped into the tank at a rate of 3 gal/min

R_{in} =(concentration of salt in inflow)(input rate of brine)

=(2\frac{lbs}{gal})( 3\frac{gal}{min})\\R_{in}=6\frac{lbs}{min}

When the solution is well stirred, it is then pumped out at a slower rate of 2 gal/min.

Concentration c(t) of the salt in the tank at time t

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=(\frac{A(t)}{500})( 2\frac{gal}{min})\\R_{out}=\dfrac{A}{250}

Now, the rate of change of the amount of salt in the tank

\dfrac{dA}{dt}=R_{in}-R_{out}

\dfrac{dA}{dt}=6-\dfrac{A}{250}

We solve the resulting differential equation by separation of variables.  

\dfrac{dA}{dt}+\dfrac{A}{250}=6\\$The integrating factor: e^{\int \frac{1}{250}dt} =e^{\frac{t}{250}}\\$Multiplying by the integrating factor all through\\\dfrac{dA}{dt}e^{\frac{t}{250}}+\dfrac{A}{250}e^{\frac{t}{250}}=6e^{\frac{t}{250}}\\(Ae^{\frac{t}{250}})'=6e^{\frac{t}{250}}

Taking the integral of both sides

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Recall that when t=0, A(t)=50 (our initial condition)

50=1500+Ce^{-\frac{0}{250}}50=1500+Ce^{0}\\C=-1450\\$Therefore the amount of salt in the tank at any time t is:\\A=1500-1450e^{-\dfrac{t}{250}}

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