A bacteria culture initially contains 100 cells and grows at a rate proportional to its size. After an hour the population has increased to 230. (a) Find an expression for the number of bacteria after t hours. (b) Find the number of bacteria after 2 hours. (Round your answer to the nearest whole number.) P(2) = ___bacteria (c) Find the rate of growth after 2 hours. (Round your answer to the nearest whole number.) P'(2) = ___bacteria per hour (d) When will the population reach 10,000? (Round your answer to one decimal place.) t = ___hr

Step-by-step explanation:

It often works well to write an exponential expression as ...

   value = (initial value)×(growth factor)^(t/(growth period))

(a) Here, the growth factor for the bacteria is given as 230/100 = 2.3 in a period of 1 hour. The initial number is 100, so we can write the pupulation function as ...

  P(t) = 100·2.3^t

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(b) P(2) = 100·2.3^2 = 529 . . . number after 2 hours

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(c) P'(t) = ln(2.3)P(t) ≈ 83.2909·2.3^t

  P'(2) = 83.2909·2.3^2 ≈ 441 . . . bacteria per hour

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(d) We want to find t such that ...

  P(t) = 10000

  100·2.3^t = 10000 . . . substitute for P(t)

  2.3^t = 100 . . . . . . . . divide by 100

  t·log(2.3) = log(100)

  t = 2/log(2.3) ≈ 5.5 . . . hours until the population reaches 10,000

step-by-step explanation:

two pairs of adjacent sides are of equal length. this implies that one diagonal divides the kite into congruent triangles, and so the angles between the two pairs of equal sides are equal in measure. it also implies that the diagonals are perpendicular.