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Finger [1]
3 years ago
11

A pet shop has 25 dogs at the start of the year. In January, 5 dogs were sold and 3 dogs were bought. In February, 8 dogs were s

old and the store took delivery of some new dogs. At the start of March, the number of dogs in the pet shop was the same as the beginning of the year. Determine the number of pets bought in February.
Mathematics
2 answers:
MariettaO [177]3 years ago
8 0
The number of dogs the store bought in March was 10.
wolverine [178]3 years ago
8 0
1)Ok we know that at the start of march, they had the same number of dogs as in the start of january which is 25 dogs. 

2) In january 5 dogs were sold so that means that we now only have 20 dogs

3) They bought back 3 new dogs in january so that means they now have 23 dogs

4) In february 8 dogs were sold. So we do 23-8=15 dogs. They now have 15 dogs left. 

5) Now they're buying dogs so that the total number of dogs in their possession is 25 again. 25-15 is the difference needed to get back 25. 
25-15=10 dogs. They bought 10 dogs at the beginning of march.
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The question is incomplete. The complete question is :

The population of a certain town was 10,000 in 1990. The rate of change of a population, measured in hundreds of people per year, is modeled by P prime of t equals two-hundred times e to the 0.02t power, where t is measured in years since 1990. Discuss the meaning of the integral from zero to twenty of P prime of t, d t. Calculate the change in population between 1995 and 2000. Do we have enough information to calculate the population in 2020? If so, what is the population in 2020?

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According to the question,

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Now integrating,

$\int_0^{20}\frac{dP(t)}{dt}dt=\int_0^{20}200e^{0.02t} \ dt$

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$\int1.dP(t)=200e^{0.02t}dt$

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k is change in population.

So in 1995,

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