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lidiya [134]
2 years ago
8

If m = 2, find the value of m+3.

Mathematics
2 answers:
Natasha2012 [34]2 years ago
7 0

Answer:

5

Step-by-step explanation:

m+3=

(m=2)+3=

(2)+3=

2+3=

=5

steposvetlana [31]2 years ago
5 0

Answer:

5

Step-by-step explanation:

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ABC is dilated by a factor of 1/2 to produce A'B'C'
Anton [14]

Answer:

B

Step-by-step explanation:

34 x 1/2 = 17 (Line A'B')

28 x 1/2 = 14 (Angle A')

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3 years ago
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How do you solve this whole thing please?
NemiM [27]

Answer:

Step-by-step explanation:

Let's look at the first two, and hopefully you'll be able to figure the rest out:

1. Answer below

d = rt

The problem asks us to solve for r, so that means we need to get r on one side of the equation by itself. To do so, we will need to divide both sides of the equation by t:

\frac{d}{t} = \frac{rt}{t}

\frac{d}{t} = r

2. Answer below

B = T - Lc

The problem asks us to solve for T, so that means we need to get T on one side of the equation by itself. To do so, we will need to add Lc to both sides of equation:

B + Lc = T - Lc + Lc

B + Lc = T

7 0
4 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

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3 years ago
Graph the function <br><br> x+y=0
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Step-by-step explanation:

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