Suppose that the concentration of a bacteria sample is 100000 bacteria per milliliter. If the concentration doubles every 2 hours how long will it take for the concentration to reach 380000 bacteria per milliliter?

3.85 hours

Step-by-step explanation:

We have that the model equation in this case would be of the following type, being "and" the concentration of bacteria:

y = a * e ^ (b * t)

where a and b are constants and t is time.

We know that when the time is 0, we know that there are 100,000 bacteria, therefore:

100000 = a * e ^ (b * 0)

100000 = a * 1

a = 100000

they tell us that when the time is 2 hours, the amount doubles, that is:

200000 = a * e ^ (b * 2)

already knowing that a equals 100,000

e ^ (b * 2) = 2

b * 2 = ln 2

b = (ln 2) / 2

b = 0.3465

Having the value of the constants, we will calculate the value of the time when there are 380000, that is:

380000 = 100000 * (e ^ 0.3465 * t)

3.8 = e ^ 0.3465 * t

ln 3.8 = 0.3465 * t

t = 1.335 / 0.3465

t = 3.85

That is to say that in order to reach this concentration 3.85 hours must pass

r(0,90)(x,y) . t(-5,0)

last option

step-by-step explanation:

∆def to ∆d"e"f"

- rotate 90 counterclockwise degrees

- then translated 5 to the left (x --> x - 5)

answer is the last option