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jeka94
3 years ago
9

Determine whether the sequence converges or diverges. If it converges, give the limit. 60, 10, five divided by three, five divid

ed by eighteen, ...
Mathematics
1 answer:
vodomira [7]3 years ago
6 0
A_n = 60, 10, \frac{5}{3}, \frac{5}{18}

Using geometric sequence form, we can rewrite it as:
A_n = a_1 \cdot r^{n-1}
A_n = 60 \cdot \frac{1}{6}^{n - 1}

Since r < 1, there exists a limit, and thus, it is said to converge.
\lim_{n \to \infty} 60 \cdot (\frac{1}{6})^{n - 1}
= 0

Since the ratio is still positive, as n tends towards infinity, the graph of the sequence will never go below the horizontal. Thus, we say that the limit of the function is zero, which means that as we increase the size of n, eventually after the infinite-th time, it will hit zero.
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4 years ago
1) Eliot and William went hiking. They went at the same speed. Eliot's hike was 8 miles, and William's hike was 12 miles. It too
Vitek1552 [10]

Answer:

1) 2 hours

2) 96 miles

Step-by-step explanation:

1) Eliot and William went hiking. They went at the same speed. Let x mph be their speed.

<u>Elliot:</u>

Distance = 8 miles

Speed = x mph

Time =\dfrac{8}{x} hours

<u>William:</u>

Distance = 12 miles

Speed = x mph

Time =\dfrac{12}{x} hours

It took Eliot 40\ min=\dfrac{40}{60}\ hour=\dfrac{2}{3}\ hour less than William, then

\dfrac{8}{x}+\dfrac{2}{3}=\dfrac{12}{x}\\ \\\dfrac{2}{3}=\dfrac{12}{x}-\dfrac{8}{x}\\ \\\dfrac{2}{3}=\dfrac{4}{x}\\ \\2x=3\cdot 4\ \ [\text{Cross multiply}]\\ \\2x=12\\ \\x=6\ mph

So, William's hike was

\dfrac{12}{6}=2

hours long.

2) <u>Initially:</u>

Distance = 120 miles

Speed = 55 mph

Time =\dfrac{120}{55}=2\dfrac{10}{55}=2\dfrac{2}{11}\ hours

<u>Final:</u>

Speed = 44 mph

Time =2\dfrac{2}{11} hours

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3 years ago
two passenger train traveling in opposite directions meet and pass each other. Each train is 1/18 miles long and is traveling at
xz_007 [3.2K]

9514 1404 393

Answer:

  5 seconds

Step-by-step explanation:

Suppose the front parts of the trains meet at point A. Since both are the same length and traveling the same speed, each will pass point A in time ...

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  time = (1/18 mi)/(40 mi/h) = (1/720 h) × (3600 s)/(1 h) = 5 s

That is, the rear part of each train will be at point A 5 seconds after the front part.

The rear parts will pass each other 5 seconds after the front parts meet.

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3 years ago
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