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lbvjy [14]
3 years ago
9

Juan has c cheesecakes. Clarence has 5 less than 3 times as many cheesecakes as Juan has and Darlene has 2 more than half as man

y cheesecakes as Clarence has. Find an expression in terms of c for the total number of cheesecakes Juan, Clarence, and Darlene have.
Mathematics
1 answer:
mr_godi [17]3 years ago
4 0
Juan has<span> $</span>c<span>$ </span>cheesecakes<span>. </span>Clarence has 5 less than 3 times<span> as </span>many cheesecakes<span> as</span>Juan has<span> and </span>Darlene has 2 more than half<span> as </span>many cheesecakes<span> as </span>Clarence has<span>. </span>Find<span> an </span>expression<span>in </span>terms<span> of $</span>c<span>$ for the </span>total number<span> of </span>cheesecakes Juan<span>, </span>Clarence<span>, and </span>Darlene have<span>. Z K Y. The post below </span>
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Answer:

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So t=0 represents year 1761.

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So we have the doubling time is 5 years.  This means the population will be twice what it was in 5 years.  Let's plug this into:

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Convert to logarithm form:

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Multiply both sides by 1/5:

k=\frac{1}{5}\ln(2)

k=\ln(2^{\frac{1}{5}}) By power rule.

So in the next sentence they actually give us the initial population and we just found k so this is our function for P:

P=39e^{\ln(2^{\frac{1}{5}})t}

So now we plug in 20 to find how many residents there were in 1761:

P=39e^{\ln(2^{\frac{1}{5}})(20)}

This is surely going to the calculator:

P=624

Now if you don't like that, let's try this:

Year 0 we have 39 people.

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Year 10 we have 78(2)=156 people.

Year 15 we have 156(2)=312 people.

Year 20 we have 312(2)=624 people.

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