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Kipish [7]
3 years ago
8

a roller coaster can take 24 riders in a single trip. 72 people went through the line to ride the roller coaster. how many trips

did the roller coaster make?
Mathematics
1 answer:
bulgar [2K]3 years ago
5 0

Answer:

3 trips

Step-by-step explanation:

We would divide the amount of riders the roller coaster could take in a single trip by the total number of riders to determine how many trips it can make.

max amount of riders/total riders

72/24 = 3

Hence, the rollercoaster made 3 trips.

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Evaluate using L'Hopital's rule:<br><br> <img src="https://tex.z-dn.net/?f=%5Clim_%7Bx%20%5Cto%201%5E%7B%2B%7D%7D%20%28x%20-%201
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Answer:

\displaystyle \lim_{x\to 1^+}(x-1)^\ln(x)}=1

Step-by-step explanation:

We are given:

\displaystyle \lim_{x\to 1^+}(x-1)^\ln(x)

And we want to evaluate it using L'Hopital's Rule.

First, using direct substitution, we will acquire:

=(1-1)^\ln(1)}=0^0

Which is indeterminate.

In order to apply L'Hopital's Rule, we first need to manipulate the expression. We will let:

y=(x-1)^\ln(x)

By taking the natural log of both sides:

\ln(y)=\ln(x)\ln(x-1)

And by taking the limit as x approaches 1 from the right of both functions:

\displaystyle \lim_{x\to 1^+}\ln(y)=\lim_{x\to 1^+}\ln(x)\ln(x-1)

Rewrite:

\displaystyle \lim_{x\to 1^+}\ln(y)= \lim_{x\to 1^+}\frac{\ln(x-1)}{\ln(x)^{-1}}

Using direct substitution on the right will result in 0/0. Hence, we can now apply L'Hopital's Rule:

\displaystyle \lim_{x\to 1^+}\ln(y)= \lim_{x\to 1^+}\frac{1/(x-1)}{-\ln(x)^{-2}(1/x)}

Simplify:

\displaystyle \lim_{x\to 1^+}\ln(y)= \lim_{x\to 1^+}\frac{1/(x-1)}{-\ln(x)^{-2}(1/x)}\Big(\frac{-x\ln(x)^2}{-x\ln(x)^2}\Big)

Simplify:

\displaystyle  \lim_{x\to 1^+}\ln(y)= \lim_{x\to 1^+}-\frac{x\ln(x)^2}{x-1}

Now, by using direct substitution, we will acquire:

\displaystyle \Rightarrow -\frac{1\ln(1)^2}{1-1}=\frac{0}{0}

Hence, we will apply L'Hopital's Rule once more. Utilize the product rule:

\displaystyle \lim_{x\to 1^+}\ln(y)= \lim_{x\to 1^+}-(\ln(x)^2+2x\ln(x))

Finally, direct substitution yields:

\Rightarrow -(\ln(1)^2+2(1)\ln(1))=-(0+0)=0

Thus:

\displaystyle \lim_{x\to 1^+}\ln(y)=0

By the Composite Function Property for limits:

\displaystyle  \lim_{x\to 1^+}\ln(y)=\ln( \lim_{x\to 1^+}y)=0

Raising both sides to e produces:

\displaystyle e^{\ln \lim_{x\to 1^+}y}=e^0

Therefore:

\displaystyle \lim_{x\to 1^+}y=1

Substitution:

\displaystyle \lim_{x\to 1^+}(x-1)^\ln(x)}=1

4 0
2 years ago
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