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3 years ago

If a* = n, theni needs answer

Mathematics

1 answer:

ICE Princess25 [194]3 years ago

8 0

Answer:

a= 2

Step-by-step explanation:

You might be interested in

Please answer the question

Lady bird [3.3K]

Answer:

the first one is

+2 +4 + 8 + 16 +32

he multiply by 2 every time he add numbers so its 65

the second one he make this

-8 -4 -2 -1

every time he Take from numbers he devide by 2

so its 5

the third one

+6 +18 + 54 + 162

every time he add he multiply by 3

ao its 162 +80 =242

the fourth one is

+33 +66 + 132 +264

he multiply by 2 every time he add

so its 267+264=531

the fiveth one is

he put the roots

2²,3²,4²,5²,6²

so the answer is 36

Step-by-step explanation:

A 65

B 5

C 242

D 531

E 36

hopefully its helpful man

6 0

3 years ago

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,

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