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:
37
Step-by-step explanation:
Use the given functions to set up and simplify
Answer:
x=54
Step-by-step explanation:
2x + 2 = 3x - 52
subtract 2x from both sides, now u have:
2 = x - 52
add 52 from both sides, now u have
54 = x
<h3>
Answer: 0.157</h3>
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Explanation:
Convert the fraction 9/50 to decimal form. You can use either long division or a calculator.
You should find that 9/50 = 0.18 which is the same as 0.180
So the original compound inequality is the same as saying 0.125 < x < 0.180
This tells us that x is between 0.125 and 0.180 where x is not equal to either endpoint. We simply need to pick anything in this interval. It can be anything you want (I recommend to use a number line to help pick a value). One such value is 0.157. There are infinitely many values you can select from.
The number 0.157 is between 0.125 and 0.180, ie 0.125 < 0.157 < 0.180
It's very similar to saying 157 is between 125 and 180, ie 125 < 157 < 180.
Answer: y= -5/2x+3
Explanation: by converting the equation into slope-intercept form you can use the slope and y-intercept given in the equation to graph.
Step-by-step:
5x= -2y+6
2y= -5x+6
Y= -5/2x+3
Slope= -5/2x
Y-intercept= 3
