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3 years ago

Here’s a graph of a linear function. Write the equation that describes that function. Express it in slope intercept form.

Mathematics

1 answer:

worty [1.4K]3 years ago

4 0

<h3>Answer: y = (1/2)x+3</h3>

slope = 1/2, y intercept = 3

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Work Shown:

Pick any two points you want on the line.

I'll pick the two points (0,3) which is the y intercept and also the point (2,4).

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Find the slope of the line through (x1,y1) = (0,3) and (x2,y2) = (2,4)

m = (y2 - y1)/(x2 - x1)

m = (4 - 3)/(2 - 0)

m = 1/2 is the slope

slope = rise/run = 1/2

rise/run = 1/2 means rise = 1 and run = 2.

This means we go up 1 unit each time we move to the right 2 units.

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Since the slope is m = 1/2 and the y intercept is b = 3, we can say

y = mx+b

y = (1/2)x+3

this is the same as y = 0.5x+3 since 1/2 = 0.5

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A line passes through (2,8) and (4,12) which equation best represents the line

Dmitry [639]

Answer:

y=2x+4

Step-by-step explanation:\solving for the slope you get (12-8)/(4-2)=4/2=2

solving for the y intercept you subtract 4 from 8 because 2*2 is 4 and (2,8) is 2 x values away from the y intercept so 8-4=4 so the y intercept is 4

so y=2x+4

4 0

3 years ago

Read 2 more answers

Lim (n/3n-1)^(n-1) n → ∞

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, e :

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

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

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

From the relation 9m = n - 9, we see that m also approaches infinity as n 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:

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