I attached the question in the image below

Mathematics

2 answers:

djyliett [7]3 years ago

8 0

Answer:

45°

Step-by-step explanation:

$tan^{-1}(1)$ = 45°

Alexeev081 [22]3 years ago

3 0

Answer:

$\huge\boxed{\theta=45^o\ \vee\ \theta=225^o}$

Step-by-step explanation:

$\tan\theta=1$

$\bold{METHOD\ 1}\\\\\text{Use the table in the attachment}\\\\\tan45^o=1\to\theta=45^o\ \vee\ \theta=45^o+180^o=225^o\\\\\bold{METHOD\ 2}\\\\\tan\theta=1\to\tan^{-1}1=\theta\to\theta=45^o\ \vee\ \theta=225^o$

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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,

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