Bacteria grow in a culture at a rate proportional to the amount present. Initially, 300 strands of the bacteria are in the culture and after two hours that number has grown by 20 percent. Find (a) an expression for the approximate number of strands in the culture at any time t and (b) the time needed for the bacteria to double its initial size.

(a) number of strands (n) = time (t) ÷ proportionality constant (k)

(b) The time needed for the bacterial to double its initial size is 3.36 hours.

Explanation:

(a) Let the rate (time) be represented by t and the amount (number) of strands of bacteria be represented by n

t is proportional to n, therefore, t = kn (k is the proportionality constant)

Since t = kn, then, n = t/k

(b) Initial amount of strands = 300

Amount of strands after 2 hours = 300 + (300 × 20/100) = 300 + 60 = 360

k = t/n = 2/360 = 0.0056 hour/strand

Double of the initial size is 600 (300×2 = 600)

Time (t) needed for the bacterial to double its initial size = kn = 0.0056×600 = 3.36 hours

glycolysis

explanation:

glycolysis must occur before any of the other steps can occur.

hope this !

correct answer is option a.

tropopause, stratopause and mesopause have all in common aspect that is point of maximum temperature in its layer of the atmosphere. all layers of atmosphere observe abrupt change in temperature. tropopause is the boundary between troposphere and stratosphere. stratopause is the boundary between stratosphere and mesosphere. mesopause is the boundary between mesosphere and thermosphere. these all boundaries are ranges between altitude of 0 to 80 km.