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vodomira [7]
3 years ago
14

John is comparing the prices of two bags of the

Mathematics
1 answer:
Lady bird [3.3K]3 years ago
8 0

Answer:

So, its better to buy larger bag

Step By Step Explanation:

Let

Smaller bag hold = x ounces

Price of x ounce in smaller bag = $ y

Price of 1 ounce in smaller bag = \frac{y}{x} ....(1

Larger bag will hold = x + 10 ounces

Price of x+10 ounces in larger bag = y + 1.5

Price of 1 ounce in larger bag = \frac{y + 1.5}{x + 10} ...(2

Now equation 1 tells that 1 ounce in smaller costs \frac{y}{x}, and equation 2 tells in ounce in larger bag costs \frac{y + 1.5}{x + 10} .

Now as we can see, we can get equation 2 from equation 1 by increasing numerator by 1.5 and denominator by 10. In this case denominator has increased more than numerator so the value of overall fraction \frac{y + 1.5}{x + 10} would be less than \frac{y}{x} . It means our larger bag has less price per ounce.

So, its better to buy larger bag

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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
aliya0001 [1]

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.

Volume = 500 gallons

Initial Amount of Salt, A(0)=50 pounds

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

Concentration, C(t)=\dfrac{Amount}{Volume}=\dfrac{A(t)}{500}

R_{out}=(concentration of salt in outflow)(output rate of brine)

=(\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

\int(Ae^{\frac{t}{250}})'=\int 6e^{\frac{t}{250}} dt\\Ae^{\frac{t}{250}}=6*250e^{\frac{t}{250}}+C, $(C a constant of integration)\\Ae^{\frac{t}{250}}=1500e^{\frac{t}{250}}+C\\$Divide all through by e^{\frac{t}{250}}\\A(t)=1500+Ce^{-\frac{t}{250}}

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}}

4 0
3 years ago
What is the fraction and decimal for 71%
Nadusha1986 [10]
The fraction is : 71/100
The decimal is : 0.71
3 0
4 years ago
Read 2 more answers
Are these two lines parallel, perpendicular or neither
Deffense [45]
They are Perpendicular
6 0
3 years ago
2 2/5 divided by 6 4/5
Brrunno [24]
0.35294117647 so it would be 0.35
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3 years ago
What is an equation of the line that passes through (2,-3) and is perpendicular to y=x-5
Digiron [165]

y = -x - 1 is the equation of the line that passes through (2,-3) and is perpendicular to y = x - 5

<em><u>Solution:</u></em>

Given that we have to write the equation of the line that passes through (2,-3) and is perpendicular to y = x - 5

<em><u>The equation of line in slope intercept form is given as:</u></em>

y = mx + c ------ eqn 1

Where, "m" is the slope of line and "c" is the y intercept

<em><u>Given equation of line is:</u></em>

y = x - 5

On comparing the above equation with eqn 1,

m = 1

We know that,

Product of slope of line and slope of line perpendicular to given line is equal to -1

Therefore,

1 \times \text{ slope of line perpendicular to given line } = -1\\

Slope of line perpendicular to given line = -1

<em><u>Now we have to find the equation of line with slope -1 and passing through (2, -3)</u></em>

Substitute m = -1 and (x, y) = (2, -3) in eqn 1

-3 = -1(2) + c

-3 = -2 + c

c = -3 + 2

c = -1

<em><u>Substitute m = -1 and c = - 1 in eqn 1</u></em>

y = -x - 1

Thus the equation of line is found

3 0
3 years ago
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