Answer :
F(3)=-16(3)+70(3)+5
F(3)= 167
, check using your calculator
Answer:
11 hrs 1 min (i think so based on math)
Step-by-step explanation:
If Johnna's reading rate is 150 words per min and there are 364 pages with about 275 words, you need to multiply 364 and 275 to get 100100 and divide that by 150 to get 667.3333333 round to the nearest tenth which will give you 667.3 and divide that by 60 because there are 60 minutes in an hour to get 11.12222222 and round that to the nearest tenth to get 11.1 which means it will take Johnna 11 hours and 1 minute to read the book. I hope this is correct I'm really sorry if it isn't. :P
Answer:
The judicial branch is important because it complements the other two branches. The executive branch defends the country and enforces the law, the legislative passes, amends and repeals laws, and the judicial branch interprets the law and passes punishment.
hope this helps :)
Given Information:
Years = t = 35
Semi-annual deposits = P = $2,000
Compounding semi-annually = n = 2
Interest rate = i = 6.5%
Required Information
Accumulated amount = A = ?
Answer:
Accumulated amount = $515,827
Step-by-step explanation:
The future value of amount earned over period of 35 years and interest rate 6.5% with semi-annual deposits is given by
FV = PMT * ((1 + i/n)^nt - 1)/(i/n))
Where
n = 2
i = 0.065
t = 35
FV = 2000*((1 + 0.065/2)^2*35 - 1)/(0.065/2))
FV = 2,000*(257.91)
FV ≈ $515,827
Therefore, Anthony will have an amount of $515,827 when he retires in 35 years.
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
__
(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
__
(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
