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nataly862011 [7]
2 years ago
15

A tank had 6 liters of water. The tank had a small hole in its base through which water leaked out at a constant rate. The tank

was empty after 4 minutes. If y represents the number of liters of water in the tank and x represents the time in minutes, which graph best represents this situation?

Mathematics
2 answers:
Mariana [72]2 years ago
8 0

Based on the given problem above, if we analyze it, we are going to look for the graph which with coordinates (0,6) which represents starting out with 6 gallons, and ending at (4,0). So, the graph that best represents this situation would be the first graph on the top left. Hope this answers your question. Have a great day!

Anastaziya [24]2 years ago
8 0

Answer:

The graph is attached.

Step-by-step explanation:

We will write an equation in slope-intercept form to represent this situation.  Slope-intercept form is

y = mx+b, where m is the slope and b is the y-intercept.

We know that the y-intercept, the initial value, is 6.  This gives us

y = mx+6.

We do not know the rate of change, or slope, of this situation.  We do, however, know that after 4 minutes there is 0 water left in the tank.  Substituting 0 in for y and 4 in for x, we have

0 = m(4)+6

0 = 4m + 6

Subtract 6 from each side:

0-6 = 4m+6-6

-6 = 4m

Divide both sides by 4:

-6/4 = 4m/4

-1.5 = m

This gives us

y = -1.5x + 6.  This can also be written as

y = -3/2x + 6.

To graph this, we begin with the y-intercept, 6.  We go up to 6 on the y-axis and plot a point.

The slope tells us where to find the next point.  Slope is rise/run; our slope is -3/2, so from our y-intercept, we go down 3 and right 2 to make our next point.  Then we draw a line between the points.

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If you meant, "how many zeros are there in the standard form of 'ten divided by 3,' the answer would be, "none."


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A tank initially contains 60 gallons of brine, with 30 pounds of salt in solution. Pure water runs into the tank at 3 gallons pe
adoni [48]

Answer:

the amount of time until 23 pounds of salt remain in the tank is 0.088 minutes.

Step-by-step explanation:

The variation of the concentration of salt can be expressed as:

\frac{dC}{dt}=Ci*Qi-Co*Qo

being

C1: the concentration of salt in the inflow

Qi: the flow entering the tank

C2: the concentration leaving the tank (the same concentration that is in every part of the tank at that moment)

Qo: the flow going out of the tank.

With no salt in the inflow (C1=0), the equation can be reduced to

\frac{dC}{dt}=-Co*Qo

Rearranging the equation, it becomes

\frac{dC}{C}=-Qo*dt

Integrating both sides

\int\frac{dC}{C}=\int-Qo*dt\\ln(\abs{C})+x1=-Qo*t+x2\\ln(\abs{C})=-Qo*t+x\\C=exp^{-Qo*t+x}

It is known that the concentration at t=0 is 30 pounds in 60 gallons, so C(0) is 0.5 pounds/gallon.

C(0)=exp^{-Qo*0+x}=0.5\\exp^{x} =0.5\\x=ln(0.5)=-0.693\\

The final equation for the concentration of salt at any given time is

C=exp^{-3*t-0.693}

To answer how long it will be until there are 23 pounds of salt in the tank, we can use the last equation:

C=exp^{-3*t-0.693}\\(23/60)=exp^{-3*t-0.693}\\ln(23/60)=-3*t-0.693\\t=-\frac{ln(23/60)+0.693}{3}=-\frac{-0.959+0.693}{3}=  -\frac{-0.266}{3}=0.088

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3 years ago
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