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BARSIC [14]
3 years ago
9

Find the distance between the points (0, 10) and (–9, 1).

Mathematics
1 answer:
dimulka [17.4K]3 years ago
5 0

\\ \sf\longmapsto \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

\\ \sf\longmapsto \sqrt{(-9-0)^2+(1-10)^2}

\\ \sf\longmapsto \sqrt{(-9)^2+(-9)^2}

\\ \sf\longmapsto \sqrt{81+81}

\\ \sf\longmapsto \sqrt{162}

\\ \sf\longmapsto 12.42

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Answer:

Option C. f(n) = 16(3/2)⁽ⁿ¯¹⁾

Step-by-step explanation:

To know which option is correct, do the following:

For Option A

f(n) = 3/2(n – 1) + 16

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f(n) = 3/2(1 – 1) + 16

f(n) = 3/2(0) + 16

f(n) = 16

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f(n) = 1.5 + 16

f(n) = 17.5

For Option B

f(n) = 3/2(16)⁽ⁿ¯¹⁾

n = 1

f(n) = 3/2(16)⁽¹¯¹⁾

f(n) = 3/2(16)⁰

f(n) = 3/2 × 1

f(n) = 1

For Option C

f(n) = 16(3/2)⁽ⁿ¯¹⁾

n = 1

f(n) = 16(3/2)⁽¹¯¹⁾

f(n) = 16(3/2)⁰

f(n) = 16 × 1

f(n) = 16

n = 2

f(n) = 16(3/2)⁽ⁿ¯¹⁾

f(n) = 16(3/2)⁽²¯¹⁾

f(n) = 16(3/2)¹

f(n) = 16(3/2)

f(n) = 8 × 3

f(n) = 24

n = 3

f(n) = 16(3/2)⁽ⁿ¯¹⁾

f(n) = 16(3/2)⁽³¯¹⁾

f(n) = 16(3/2)²

f(n) = 16(9/4)

f(n) = 4 × 9

f(n) = 36

For Option D

f(n) = 8n + 8

n = 1

f(n) = 8(1) + 8

f(n) = 8 + 8

f(n) = 16

n = 2

f(n) = 8n + 8

f(n) = 8(2) + 8

f(n) = 16 + 8

f(n) = 24

n = 3

f(n) = 8n + 8

f(n) = 8(3) + 8

f(n) = 24 + 8

f(n) = 32

From the above illustration, only option C describes the sequence.

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