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nexus9112 [7]
3 years ago
10

Give one example of an equation with variables on both sides that has all real numbers as the solution. Also give an example of

an equation with variables on both sides that has no real solutions.
Mathematics
1 answer:
Ghella [55]3 years ago
6 0
1) 2x + 3 = 2x + 3
This works because if you solve it, you will get either x = x or 3 = 3 which are always true, so x, can have an infinite number of solutions
2) 3(x + 4) = 3x + 11
This has no solution because if you solve it, you're gonna get 12 = 11 and that is NEVER true. Whatever x is, 11 cannot equal 12!
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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
Mashcka [7]

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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Answer:

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Step-by-step explanation:

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Read 2 more answers
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