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anyanavicka [17]
3 years ago
7

(12y)+12 y=15 do I multiply 12 and 15 then add 12

Mathematics
2 answers:
PSYCHO15rus [73]3 years ago
8 0
12y + 12.....when y = 15
12(15) + 12.....multiplication comes before addition in the order of operations...so multiply first.
so yes....u r correct...multiply 12 and 15, then add 12
My name is Ann [436]3 years ago
7 0
Yes you were right cause you have to do what is inside of the parenthesis first so that would be 180 then u would add 12.

your final answer would be 192

hope this was enough help
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In 1859, 24 rabbits were released into the wild in Australia, where they had no natural predators. Their population grew exponen
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Answer:

a) P' = P

   P(t) = 24e^{0.693t} where t is step of 6 months

b) 7.7 years

c)1064.67 rabbits/year

Step-by-step explanation:

The differential equation describing the population growth is

\frac{dP}{dt} = P

Where t is the range of 6 months, or half of a year.

P(t) would have the form of

P(t) = P_0e^{kt}

where P_0 = 24 is the initial population

After 6 month (t = 1), the population is doubled to 48

P(1) = 24e^k = 48

e^k = 2

k = ln(2) = 0.693

Therefore P(t) = 24e^{0.693t}

where t is step of 6 months

b. We can solve for t to get how long it takes to get to a population of 1,000,000:

24e^{0.693t} = 1000000

e^{0.693t} = 1000000 / 24 = 41667

0.693t = ln(41667) = 10.64

t = 10.64 / 0.693 = 15.35

So it would take 15.35 * 0.5 = 7.7 years to reach 1000000

c. P' = P_0ke^{kt}

We need to resolve for k if t is in the range of 1 year. In half of a year (t = 0.5), the population is 48

24e^{0.5k) = 48

0.5k = ln2 = 0.693

k = 1.386

Therefore, P' = 1.386*24e^{1.386t}

At the mid of the 3rd year, where t = 2.5, we can calculate P'

P' = 1.386*24e^{1.386*2.5} = 1064.67 rabbits/year

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