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Verdich [7]
3 years ago
7

A​ sparkling-water distributor wants to make up 200 gal of sparkling water to sell for $ 5.00 per gallon. She wishes to mix thre

e grades of water selling for ​$10​.00, ​$1​.00, and​ $4.50 per​ gallon, respectively. She must use twice as much of the​ $4.50 water as the $ 1.00 comma water. How many gallons of each should she​ use?​
Mathematics
1 answer:
max2010maxim [7]3 years ago
5 0

Answer:

500

Step-by-step explanation:

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A 400 gallon tank initially contains 100 gal of brine containing 50 pounds of salt. Brine containing 1 pound of salt per gallon
posledela

Answer:

The amount of salt in the tank when it is full of brine is 393.75 pounds.

Step-by-step explanation:

This is a mixing problem. In these problems we will start with a substance that is dissolved in a liquid. Liquid will be entering and leaving a holding tank. The liquid entering the tank may or may not contain more of the substance dissolved in it. Liquid leaving the tank will of course contain the substance dissolved in it. If Q(t) gives the amount of the substance dissolved in the liquid in the tank at any time t we want to develop a differential equation that, when solved, will give us an expression for Q(t).

The main equation that we’ll be using to model this situation is:

Rate of change of <em>Q(t)</em> = Rate at which <em>Q(t)</em> enters the tank – Rate at which <em>Q(t)</em> exits the tank

where,

Rate at which Q(t) enters the tank = (flow rate of liquid entering) x

(concentration of substance in liquid entering)

Rate at which Q(t) exits the tank = (flow rate of liquid exiting) x

(concentration of substance in liquid exiting)

Let y<em>(t)</em> be the amount of salt (in pounds) in the tank at time <em>t</em> (in seconds). Then we can represent the situation with the below picture.

Then the differential equation we’re after is

\frac{dy}{dt} = (Rate \:in)- (Rate \:out)\\\\\frac{dy}{dt} = 5 \:\frac{gal}{s} \cdot 1 \:\frac{pound}{gal}-3 \:\frac{gal}{s}\cdot \frac{y(t)}{V(t)}  \:\frac{pound}{gal}\\\\\frac{dy}{dt} =5\:\frac{pound}{s}-3 \frac{y(t)}{V(t)}  \:\frac{pound}{s}

V(t) is the volume of brine in the tank at time <em>t. </em>To find it we know that at time 0 there were 100 gallons, 5 gallons are added and 3 are drained, and the net increase is 2 gallons per second. So,

V(t)=100 + 2t

We can then write the initial value problem:

\frac{dy}{dt} =5-\frac{3y}{100+2t} , \quad y(0)=50

We have a linear differential equation. A first-order linear differential equation is one that can be put into the form

\frac{dy}{dx}+P(x)y =Q(x)

where <em>P</em> and <em>Q</em> are continuous functions on a given interval.

In our case, we have that

\frac{dy}{dt}+\frac{3y}{100+2t} =5 , \quad y(0)=50

The solution process for a first order linear differential equation is as follows.

Step 1: Find the integrating factor, \mu \left( x \right), using \mu \left( x \right) = \,{{\bf{e}}^{\int{{P\left( x \right)\,dx}}}

\mu \left( t \right) = \,{{e}}^{\int{{\frac{3}{100+2t}\,dt}}}\\\int \frac{3}{100+2t}dt=\frac{3}{2}\ln \left|100+2t\right|\\\\\mu \left( t \right) =e^{\frac{3}{2}\ln \left|100+2t\right|}\\\\\mu \left( t \right) =(100+2t)^{\frac{3}{2}

Step 2: Multiply everything in the differential equation by \mu \left( x \right) and verify that the left side becomes the product rule \left( {\mu \left( t \right)y\left( t \right)} \right)' and write it as such.

\frac{dy}{dt}\cdot \left(100+2t\right)^{\frac{3}{2}}+\frac{3y}{100+2t}\cdot \left(100+2t\right)^{\frac{3}{2}}=5 \left(100+2t\right)^{\frac{3}{2}}\\\\\frac{dy}{dt}\cdot \left(100+2t\right)^{\frac{3}{2}}+3y\cdot \left(100+2t\right)^{\frac{1}{2}}=5 \left(100+2t\right)^{\frac{3}{2}}\\\\\frac{dy}{dt}(y \left(100+2t\right)^{\frac{3}{2}})=5\left(100+2t\right)^{\frac{3}{2}}

Step 3: Integrate both sides.

\int \frac{dy}{dt}(y \left(100+2t\right)^{\frac{3}{2}})dt=\int 5\left(100+2t\right)^{\frac{3}{2}}dt\\\\y \left(100+2t\right)^{\frac{3}{2}}=(100+2t)^{\frac{5}{2} }+ C

Step 4: Find the value of the constant and solve for the solution y(t).

50 \left(100+2(0)\right)^{\frac{3}{2}}=(100+2(0))^{\frac{5}{2} }+ C\\\\100000+C=50000\\\\C=-50000

y \left(100+2t\right)^{\frac{3}{2}}=(100+2t)^{\frac{5}{2} }-50000\\\\y(t)=100+2t-\frac{50000}{\left(100+2t\right)^{\frac{3}{2}}}

Now, the tank is full of brine when:

V(t) = 400\\100+2t=400\\t=150

The amount of salt in the tank when it is full of brine is

y(150)=100+2(150)-\frac{50000}{\left(100+2(150)\right)^{\frac{3}{2}}}\\\\y(150)=393.75

6 0
3 years ago
A salesperson earns a base salary of $4000 per month plus 15% commission on sales. her monthly income f(s) is given by the funct
artcher [175]
9.) Given that the salesperson earns a base salary of $4000 per month plus 15% commission on sales if the salesperson's commission is lowered to 5%, then <span>her monthly income g(s) is given by the function g(s)=4000 +0.05s where s is monthly sales in dollars



10) </span>Given that the salesperson earns a base salary of $4000 per month plus 15% commission on sales if the salesperson's base salary is doubled, then her monthly income h(s) is given by the function g(s)=8000 +0.15s where s is monthly sales in dollars



11) Given that the salesperson earns a base salary of $4000 per month plus 15% commission on sales <span>if the salesperson's base salary is cut in half and her commission is doubled</span>, then her monthly income k(s) is given by the function k(s)=2000 +0.3s where s is monthly sales in dollars.
8 0
3 years ago
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What is the sum of 7452816653 times 23534 and divided by 1423
DIA [1.3K]
The answer is 123,256,912,938.65
4 0
3 years ago
Will give brainliest 100 points Given: Q = 7m + 3n, R = 11 - 2m, S = n + 5, and T = -m - 3n + 8.
gulaghasi [49]

Answer:

-3m-4n+14 (D)

Step-by-step explanation:

R-S+T=

<em>Substitute the Values given</em>

(11-2m)-(n+5)+(-m-3n+8)=

<em>Open Parenthesis</em>

11-2m-n-5-m-3n+8=

<em>Combine Like Terms</em>

-3m-4n+14

6 0
3 years ago
Solve each equation mentally 1/7 × x = 1
Firlakuza [10]

Answer:

7

Step-by-step explanation:

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