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After 2 hours, there are 1,400 mL of fluids remaining in a patient’s IV. The fluids drip at a rate of 300 mL per hour. Let x be the time passed, in hours, and y be the amount of fluid left in the IV, in mL. Write a linear function that models this scenario.

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Answer:

[tex] y(2) = 1400[/tex]

Using this condition we got:

[tex]1400= -300*2 +b[/tex]

And solving for b we got:

[tex] b= 1400+ 600= 2000[/tex]

So then our linear function is given by:

[tex] y = -300x +2000[/tex]

Where y is the amount of fluid left and x the number of hours ellapsing

Step-by-step explanation:

We want to set up a linear function like this one:

[tex]y = mx+b[/tex]

Where y is the amount of fluid left, m the slope and b the initial amount. From the info given we know thatm = -300. And we also have the following condition:

[tex] y(2) = 1400[/tex]

Using this condition we got:

[tex]1400= -300*2 +b[/tex]

And solving for b we got:

[tex] b= 1400+ 600= 2000[/tex]

So then our linear function is given by:

[tex] y = -300x +2000[/tex]

Where y is the amount of fluid left and x the number of hours ellapsing

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