Mathematics, 15.11.2019 04:31, joshajgaa

After an antibiotic tablet is taken, the concentration of the antibiotic in the bloodstream is modelled by the function c(t)=8(e^(-0.4t)-e^(-0.6t)) where the time t is measured in hours and c is measured in mewg/ml.

what is the maximum concentration of the antibiotic during the first 12 hours?

Answers: 2

Show answers

Answers

answered: lays20001

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,

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

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.

answered: Guest

the clues that signal the reader should change tone are the punctuation marks, the grammatical signs. for example, the quotation marks at the beginning of something someone else said literally or the exclamation marks.

the tone of the first line of dialogue until "gettysburg" is a kind tone, a tone of advise. the narrator is trying to the other person in doing something he or she obviously is finding hard to do by giving a piece of advise and bringing up a memory of a successful similar case.

the clue that the reader understand how to read the word "bang" is the exclamation mark. it gives the word a surprise tone, a strong accent.

the best tone for reading the word "bang" is an exciting tone, a surprise one, even a loud one.

the words that should be read with a formal tone are the ones that give factual information. the sentence: mister lincoln couldn't think of anything to say at the gettysburg" gives information about an event and it needs to be read formally, also, when the narrator wants to transmit calmness, a formal and slow tone is needed, because people also transmit messages with the vibrations of our voices and tones.

answered: Guest

the transverse axis of a hyperbola is the line that contains the two vertices and the two focuses.

in this example (hyperbola of equation x^2/2 - y^2/4 = 1), the transverse axis is the x-axis .

step-by-step explanation: