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strojnjashka [21]
3 years ago
13

★PLSSSSS HELPPPPPPPPP★

Mathematics
2 answers:
Vlad1618 [11]3 years ago
6 0

Answer:

The third option

Step-by-step explanation:

This is because you plug in 1 as a test into each of the equations and see if it equals to 12.25. It works for the third option. To verify that you can also plug in different x values and see if they match the y value.

allsm [11]3 years ago
3 0
C - hopes this helps
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Answer:

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Step-by-step explanation:

We are given the following information:

After an antibiotic tablet is taken, the concentration of the antibiotic in the bloodstream is modeled by the function where the time t is measured in hours and C is measured in \mu g/mL

C(t) = 8(e^{(-0.4t)}-e^{(-0.6t)})

Thus, we are given the time interval [0,12] for t.

  • We can apply the first derivative test, to know the absolute maximum value because we have a closed interval for t.
  • The first derivative test focusing on a particular point. If the function switches or changes from increasing to decreasing at the point, then the function will achieve a highest value at that point.

First, we differentiate C(t) with respect to t, to get,

\frac{d(C(t))}{dt} = 8(-0.4e^{(-0.4t)}+ 0.6e^{(-0.6t)})

Equating the first derivative to zero, we get,

\frac{d(C(t))}{dt} = 0\\\\8(-0.4e^{(-0.4t)}+ 0.6e^{(-0.6t)}) = 0

Solving, we get,

8(-0.4e^{(-0.4t)}+ 0.6e^{(-0.6t)}) = 0\\\displaystyle\frac{e^{-0.4}}{e^{-0.6}} = \frac{0.6}{0.4}\\\\e^{0.2t} = 1.5\\\\t = \frac{ln(1.5)}{0.2}\\\\t \approx 2

At t = 0

C(0) = 8(e^{(0)}-e^{(0)}) = 0

At t = 2

C(2) = 8(e^{(-0.8)}-e^{(-1.2)}) = 1.185

At t = 12

C(12) = 8(e^{(-4.8)}-e^{(-7.2)}) = 0.059

Thus, the maximum concentration of the antibiotic during the first 12 hours is 1.185 \mu g/mL at t= 2 hours.

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