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melamori03 [73]
3 years ago
6

Jordan will hike a 10 mile trail at a rate of 2 mi/h. Write a linear equation to represent the distance of the trail that Jordan

still has to hike after x hours. What does the y-intercept of the equation represent?
Mathematics
1 answer:
Lesechka [4]3 years ago
7 0

Answer:

  y = 10 -2x; 10 is the length of the trail

Step-by-step explanation:

Jordan's remaining distance starts at 10 miles, and decreases 2 miles each hour:

  y = -2x +10

The y-intercept is the length of the trail.

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If BOTH equations are in slope-intercept form then the--------------? method would be best, but the------------? method would al
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Answer:

Step-by-step explanation:

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5 0
3 years ago
If the number of bacteria in a colony doubles every 8 hours and there is currently a population of 9,315 bacteria, what will the
Liono4ka [1.6K]

It is given that the bacteria in a colony doubles every 8 hours.

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The present population of the bacteria is 9315.

After 8 hours, the bacteria becomes double. So, the number of bacteria becomes 9315 x 2 = 18630.

Again after 8 hours, the bacteria becomes 18630 x 2 = 37260.

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8 0
1 year ago
Lim (n/3n-1)^(n-1)<br> n<br> →<br> ∞
n200080 [17]

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}

7 0
3 years ago
(a) Starting with the geometric series [infinity] xn n = 0 , find the sum of the series [infinity] nxn − 1 n = 1 , |x| &lt; 1.
Mila [183]

Let <em>f(x)</em> be the sum of the geometric series,

f(x)=\displaystyle\frac1{1-x} = \sum_{n=0}^\infty x^n

for |<em>x</em>| < 1. Then taking the derivative gives the desired sum,

f'(x)=\displaystyle\boxed{\dfrac1{(1-x)^2}} = \sum_{n=0}^\infty nx^{n-1} = \sum_{n=1}^\infty nx^{n-1}

5 0
2 years ago
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