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professor190 [17]
3 years ago
9

Fred earned $22.50 in nine hours. how much did he earn after 4 hours? please show all your work

Mathematics
1 answer:
dusya [7]3 years ago
6 0
I believe the answer is 10

22.50 divided by 9 to show how much for 1 hour. That is 2.5 so next 2.5 x 4 = 10

I don’t completely know if this is correct but I believe it is
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Which expressions are equivalent to -3(2+7) A: -6+21 or B:-21+6​
Neporo4naja [7]

<u>Answer:</u>

None

<u>Explanation:</u>

-3(2+7) = -3×9 = -27

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Hence there is no equivalence

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3 years ago
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svlad2 [7]

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qaws [65]

Answer:

Step-by-step explanation:

Let P be the population of the community

So the population of a community increase at a rate proportional to the number of people present at a time

That is

\frac{dp}{dt} \propto p\\\\\frac{dp}{dt} =kp\\\\ [k \texttt {is constant}]\\\\\frac{dp}{dt} -kp =0

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p(t)=p_0e^{kt}---(1)

where p is the present population

p₀ is the initial population

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2p_o=p_oe^{5k}\\\\e^{5k}=2

Apply In on both side to get

Ine^{5k}=In2\\\\5k=In2\\\\k=\frac{In2}{5} \\\\\therefore k=\frac{In2}{5}

Substitute k=\frac{In2}{5}  in p(t)=p_oe^{kt} to get

\large \boxed {p(t)=p_oe^{\frac{In2}{5}t }}

Given that population of a community is 9000 at 3 years

substitute t = 3 in {p(t)=p_oe^{\frac{In2}{5}t }}

p(3)=p_oe^{3 (\frac{In2}{5}) }\\\\9000=p_oe^{3 (\frac{In2}{5}) }\\\\p_o=\frac{9000}{e^{3(\frac{In2}{5} )}} \\\\=5937.8

<h3>Therefore, the initial population is 5937.8</h3>
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Lynna [10]

Answer:

the answer is 19

Step-by-step explanation:

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