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bezimeni [28]
3 years ago
9

-9, 0, 27/100, 50.5, 1,974. Out of all these numbers what ones are not integers?

Mathematics
1 answer:
tia_tia [17]3 years ago
8 0
All numbers are integers like prime numbers,composite number,even,odd,etc.
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Scorpion4ik [409]

Answer:

You have the correct answer it is the third one.

7 0
3 years ago
Rewrite the expression by favoring out the greatest common factor: 9b+12
malfutka [58]

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3(3b + 4)

the GCF is 3

Step-by-step explanation:

7 0
3 years ago
Is fraction a linear equation​
olya-2409 [2.1K]

Answer:

NO

Step-by-step explanation:

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6 0
3 years ago
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

4 0
3 years ago
Solve<br>-6x^2 + 6 - 2x =x​
Tanzania [10]

Answer:

x1= -1-√17/4

x2= -1+√17/4

3 0
3 years ago
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