Answer:
Its the third one
Step-by-step explanation:
Plot all of the points onto the graph, and draw a line through all of the points. It most closely resembles the third one
To solve the exercirses which are shown in the figure attached, you must follow the proccedure below:
7) (x)=x³-6x²+8x
(x)=x(x³-6x²+8)
(x)=(x-4)(x-2)x
The lenght is: x
The height is= (x-4)
8) √(2x+8)-6=4
1. You need to clear the variable "x". Then:
√(2x+8)=4+6
√(2x+8)=10
(√2x+8)²=10²
2x+8=100
2x=100-8
x=92/2
x=46
9) l4x+3l=9+2x
1. To solve the left member, you must evaluate two cases: it could be positive,or negative. Then:
2. Negative:
l4x+3l=9+2x
-4x-3=9+2x
-4x-2x=9+3
-6x=12
x=12/-6
x=-2
3. Positive:
l4x+3l=9+2x
4x+3=9+2x
4x-2x=9-3
2x=6
x=6/2
x=3
Answer:
6 1/4
Step-by-step explanation:
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
About 5 or 4.9 is the estimation