Answer:
- P(t) = 100·2.3^t
- 529 after 2 hours
- 441 per hour, rate of growth at 2 hours
- 5.5 hours to reach 10,000
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
Answer:
4
Step-by-step explanation:
2(10 -6x)=x -8x
20 - 12x = x - 8x
20 = x - 8x + 12x
20 = 5x
20 / 5 = x
x = 4
Step-by-step explanation:
30 minutes x an hour and half past prolly around 1 hr and 30 sec
The solution for equation is x = -6
<em><u>Solution:</u></em>
<em><u>Given equation is:</u></em>

We have to solve the equation
According to Bodmas rule, if an expression contains brackets ((), {}, []) we have to first solve or simplify the bracket followed by of (powers and roots etc.), then division, multiplication, addition and subtraction from left to right
Therefore, solve for brackets in given equation

Solve for terms in left hand side of equation

Move the variables to one side and constants to other side

Thus the solution for equation is x = -6