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A farmer has two water tanks that drain at a rate of 2.5 gallons per minute. He is considering replacing the existing tanks with new ones, either Model S or Model T. Information about the new tanks is shown below. All tanks, old and new, hold 100 gallons of water and drain at a constant rate.
Select all options that correctly represent the situation.
The equation, y = 4x + 86, represents the Model T tank.
The existing tanks drain the slowest at 2.5 gallons per minute.
The equation, y = -3x + 100, represents the Model S tank.
The equation, y = 2.5x + 100, represents the existing tanks.
The Model S tank drains the fastest at 3 gallons per minute.
The Model T tank drains the fastest at 3.5 gallons per minute.
Answers
Answer:
(c) The equation, y = -3x + 100, represents the Model S tank.
(f) The Model T tank drains the fastest at 3.5 gallons per minute.
Step-by-step explanation:
Given
See attachment for models S and T
First, we calculate the equation of both models
Model S
[tex](x_1,y_1) = (0,100)\\\\(x_2,y_2) = (6,82)[/tex]
Calculate drainage rate (m)
[tex]m = \frac{y_2 - y_1}{x_2 -x_1}[/tex]
[tex]m = \frac{82-100}{6-0}[/tex]
[tex]m = \frac{-18}{6}[/tex]
[tex]m = -3[/tex] -- This implies that model S drains at 3 gallons per minutes
The equation is:
[tex]y = m(x - x_1) + y_1[/tex]
[tex]y = -3(x - 0)+100[/tex]
[tex]y = -3(x)+100[/tex]
[tex]y = -3x+100[/tex]
Model T
[tex](x_1,y_1) = (0,100)\\\\(x_2,y_2) = (4,86)[/tex]
Calculate drainage rate (m)
[tex]m = \frac{y_2 - y_1}{x_2 -x_1}[/tex]
[tex]m = \frac{86-100}{4-0}[/tex]
[tex]m = \frac{-14}{4}[/tex]
[tex]m = -3.5[/tex] -- This implies that model T drains at 3.5 gallons per minutes
The equation is:
[tex]y = m(x - x_1) + y_1[/tex]
[tex]y = -3.5(x - 0)+100[/tex]
[tex]y = -3.5(x)+100[/tex]
[tex]y = -3.5x+100[/tex]
From the calculations above, the following options are true:
(c) and (f)
It is required to find which statements are true.
The statements that are true are
The equation, y = -3x + 100, represents the Model S tank.
The Model T tank drains the fastest at 3.5 gallons per minute.
Equation of a line is
[tex]y=mx+c[/tex]
where,
[tex]y=\text{Gallons of water left}[/tex]
[tex]x=\text{Minutes passed}[/tex]
[tex]m=\text{Slope}=\dfrac{\Delta y}{\Delta x}[/tex]
[tex]c=\text{Y intercept}[/tex]
The equation of the line for model S is
[tex]y-100=\dfrac{82-100}{6-0}(x-0)\\\Rightarrow y-100=-3x\\\Rightarrow y=-3x+100[/tex]
The option C is true.
The equation of the line for model T is
[tex]y-100=\dfrac{86-100}{4-0}(x-0)\\\Rightarrow y-100=-3.5x\\\Rightarrow y=-3.5x+100[/tex]
The slope or rate of drain is 3.5 gallons per minute.
So, option F is true.
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