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Monica [59]
2 years ago
9

The cost to rent a raft is $7 per person.A raft can hold up to 6 people.There is a $3 launch fee per raft.What is the total cost

for a group of 6?Explain
Mathematics
2 answers:
Romashka-Z-Leto [24]2 years ago
8 0

Answer:the total cost for a group of 6 is $45

Step-by-step explanation:

Let x represent the number of persons that rent the raft.

The cost to rent a raft is $7 per person. This means that if x persons rent the raft, the cost would be 7x

There is a $3 launch fee per raft. This means that the total cost of for x persons would be

7x + 3

A raft can hold up to 6 people.

If a group of 6 persons rent the raft, the total cost of renting the raft would be

7 × 6 + 3 = 42 + 3 = $45

Harman [31]2 years ago
4 0
Answer: $45
Explanation: $7 multiplied by 6 people = 42. Then you add the addition of the $3 launch fee to make the raft go
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Lim (n/3n-1)^(n-1)<br> n<br> →<br> ∞
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Looks like the given limit is

\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1}

With some simple algebra, we can rewrite

\dfrac n{3n-1} = \dfrac13 \cdot \dfrac n{n-9} = \dfrac13 \cdot \dfrac{(n-9)+9}{n-9} = \dfrac13 \cdot \left(1 + \dfrac9{n-9}\right)

then distribute the limit over the product,

\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1} = \lim_{n\to\infty}\left(\dfrac13\right)^{n-1} \cdot \lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1}

The first limit is 0, since 1/3ⁿ is a positive, decreasing sequence. But before claiming the overall limit is also 0, we need to show that the second limit is also finite.

For the second limit, recall the definition of the constant, <em>e</em> :

\displaystyle e = \lim_{n\to\infty} \left(1+\frac1n\right)^n

To make our limit resemble this one more closely, make a substitution; replace 9/(<em>n</em> - 9) with 1/<em>m</em>, so that

\dfrac{9}{n-9} = \dfrac1m \implies 9m = n-9 \implies 9m+8 = n-1

From the relation 9<em>m</em> = <em>n</em> - 9, we see that <em>m</em> also approaches infinity as <em>n</em> approaches infinity. So, the second limit is rewritten as

\displaystyle\lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1} = \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m+8}

Now we apply some more properties of multiplication and limits:

\displaystyle \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m+8} = \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m} \cdot \lim_{m\to\infty}\left(1+\dfrac1m\right)^8 \\\\ = \lim_{m\to\infty}\left(\left(1+\dfrac1m\right)^m\right)^9 \cdot \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)\right)^8 \\\\ = \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)^m\right)^9 \cdot \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)\right)^8 \\\\ = e^9 \cdot 1^8 = e^9

So, the overall limit is indeed 0:

\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1} = \underbrace{\lim_{n\to\infty}\left(\dfrac13\right)^{n-1}}_0 \cdot \underbrace{\lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1}}_{e^9} = \boxed{0}

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Step-by-step explanation:

Great question, it is always good to ask away and get rid of any doubts that you may be having.

I have created an illustration which is attached below in order to help you understand the situation. As we are told the passenger rode the subway from point (0 , 0) to point (-2, 0). Then he rode the subway from point (-2, 0) to the final destination of point (-2 , -10). In order to find the value of the slope we need to use the slope formula which is

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