−12x+14=−34−12x+14=−34

1 answer:

7 0

<span>Simplifying -12x + 14 = -34 + -12x + 14 + -34 Reorder the terms: 14 + -12x = -34 + -12x + 14 + -34 Reorder the terms: 14 + -12x = -34 + 14 + -34 + -12x Combine like terms: -34 + 14 = -20 14 + -12x = -20 + -34 + -12x Combine like terms: -20 + -34 = -54 14 + -12x = -54 + -12x Add '12x' to each side of the equation. 14 + -12x + 12x = -54 + -12x + 12x Combine like terms: -12x + 12x = 0 14 + 0 = -54 + -12x + 12x 14 = -54 + -12x + 12x Combine like terms: -12x + 12x = 0 14 = -54 + 04 = -54 Solving 4,

In short responce,I'm sorry I took forever,this was hard,Your answer is C,x=4

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After an antibiotic tablet is taken, the concentration of the antibiotic in the bloodstream is modelled by the function C(t)=8(e

Alexxx [7]

Answer:

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

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,

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