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shusha [124]
3 years ago
9

Find the slope of the line that contains the points (-9, -3) and (-19, -3).

Mathematics
1 answer:
NNADVOKAT [17]3 years ago
7 0

Answer:

Slope (m) = 0

Step-by-step explanation:

We will use this formula to find the slope⤵⤵⤵

\frac{y _{2} - y _{1}  }{x _{2} - x _{1}  }

(Sorry if it's a bit small btw)

Now we plug in the values accordingly:

-3 - (-3) = -3 + 3 = 0

-19 - (-9) = -19 + 9 = -10

Slope = 0/-10

So we have 0/-10, which means our slope is actually just 0.

(This also means that the line on the graph will be horizontal.)

Hope this helps you :)

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Write the equation that modeled the sequence 63,-21,7,-7/3,7/9
Ilya [14]

Answer:

an = 63(-1/3)^(n-1)

Step-by-step explanation:

This is a geometric sequence with first term 63 and common ratio -1/3.

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an = 63(-1/3)^(n-1).

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3 years ago
Does the series converge or diverge? If it converges, what is the sum? Show your work. ∑ ∞ n = 1 − 4 ( − 1 / 2 ) n − 1
Mekhanik [1.2K]

Answer:

Step-by-step explanation:

Given the series,

∑ ∞ n = 1 − 4 ( − 1 / 2 ) n − 1

I think the series is summation from n = 1 to ∞ of -4(-1/2)^(n-1)

So,

∑ − 4 ( − ½ )^(n − 1). From n = 1 to ∞

There are different types of test to show if a series converges or diverges

So, using Ratio test

Lim n → ∞ (a_n+1 / a_n)

Lim n → ∞ (-4(-1/ 2)^(n+1-1) / -4(-1/2)^(n-1))

Lim n → ∞ ((-4(-1/2)^(n) / -4(-1/2)^(n-1))

Lim n → ∞ (-1/2)ⁿ / (-1/2)^(n-1)

Lim n→ ∞ (-1/2)^(n-n+1)

Lim n→ ∞ (-1/2)^1 = -1/2

Since the limit is less than 0, then, the series converge...

Sum to infinity

Using geometric progression formula

S∞ = a / 1 - r

Where

a is first term

r is common ratio

So, first term is

a_1 = -4(-½)^1-1 = -4(-½)^0 = -4 × 1

a_1 = -4

Common ratio r = a_2 / a_1

a_2 = 4(-½)^2-1 = -4(-½)^1 = -4 × -½ = 2

a_2 = 2

Then,

r = a_2 / a_1 = 2 / -4 = -½

S∞ = -4 / 1--½

S∞ = -4 / 1 + ½

S∞ = -4 / 3/2 = -4 × 2 / 3

S∞ = -8 / 3 = -2⅔

The sum to infinity is -2.67 or -2⅔

Check attachment for better understanding

3 0
3 years ago
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