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Agata [3.3K]
4 years ago
8

PLEASE HELP ASAP! SHOW STEPS IF POSSIBLE! I looked all over my text book and couldn't find anything to help me solve these last

three problems, i was really only able to answer one but even then I'm not sure if it is right. Please show steps!!
jack leased a car for 159/month for 36 months. the amount due to when he signed the lease 2399 how much did jack pay during the first year of the lease?
A 1908 (My answer) B 4148 C 4307

Jack leased a car for 159/month for 36 months he paid 2399 at signing how much did he pay during the second year of the lease? A 1908 B 3816 C 6056

what was the total cost of the lease? A 5724 B 7964 C 8123
Mathematics
1 answer:
kicyunya [14]4 years ago
3 0
1. A. 1908. In the first year, (a year is 12 months), he paid 159 a month so 159 times 12 is 1908. (You are correct on this one)

2. A. 1908. In the second year, (again a year is 12 months), he paid 159 a month so 159 times 12 is 1908.

(Btw : there would be a third year 36/12=3. He would also pay another 1908 in the third year. To get your answer for number three either do the stragety I wrote for it or do 1908+1908+1908+2399=8123 or (1908•3)+2399.

3. C. <span>8123. </span>Well 159 times 36 is 5724 plus 2399 and get <span>8123. That is the total cost of the lease.

Hoped I helped!

</span>
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I need help with this problem from the calculus portion on my ACT prep guide
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Given a series, the ratio test implies finding the following limit:

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If r<1 then the series converges, if r>1 the series diverges and if r=1 the test is inconclusive and we can't assure if the series converges or diverges. So let's see the terms in this limit:

\begin{gathered} a_n=\frac{2^n}{n5^{n+1}} \\ a_{n+1}=\frac{2^{n+1}}{(n+1)5^{n+2}} \end{gathered}

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\lim _{n\to\infty}\lvert\frac{a_{n+1}}{a_n}\rvert=\lim _{n\to\infty}\lvert\frac{n5^{n+1}}{2^n}\cdot\frac{2^{n+1}}{\mleft(n+1\mright)5^{n+2}}\rvert=\lim _{n\to\infty}\lvert\frac{2^{n+1}}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^{n+1}}{5^{n+2}}\rvert

We can simplify the expressions inside the absolute value:

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\lim _{n\to\infty}\lvert\frac{2}{5}\cdot\frac{n}{n+1}\rvert=\lim _{n\to\infty}\frac{2}{5}\cdot\frac{n}{n+1}

Now let's re-writte n/(n+1):

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\lim _{n\to\infty}\frac{2}{5}\cdot\frac{1}{1+\frac{1}{n}}=\frac{2}{5}\cdot\frac{1}{1+0}=\frac{2}{5}=0.4

So from the test ratio r=0.4 and the series converges. Then the answer is the second option.

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