Answer four please!! Thanks
1 answer:
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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:
a. V = (20-x)
b . 1185.185
Step-by-step explanation:
Given that:
- The height: 20 - x (in )
- Let x be the length of a side of the base of the box (x>0)
a. Write a polynomial function in factored form modeling the volume V of the box.
As we know that, this is a rectangular box has a square base so the Volume of it is:
V = h *
<=> V = (20-x)
b. What is the maximum possible volume of the box?
To maximum the volume of it, we need to use first derivative of the volume.
<=> dV / Dx = -3 + 40x
Let dV / Dx = 0, we have:
-3 + 40x = 0
<=> x = 40/3
=>the height h = 20/3
So the maximum possible volume of the box is:
V = 20/3 * 40/3 *40/3
= 1185.185
