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vekshin1
2 years ago
15

Find the 10th term of the arithmetic sequence -x + 9, 4x + 14, 9x + 19, ...

Mathematics
1 answer:
rusak2 [61]2 years ago
5 0

Answer:

44x + 54

Step-by-step explanation:

The nth term of an arithmetic sequence is given by :

Tn = a + (n - 1)d

d = common difference = T2 - T1

d = T2 - T1

a = first term = - x + 9

d = 4x + 14 - (-x + 9)

d = 4x + 14 + x - 9

d = 5x + 5

The 10th term:

T(10) = (-x + 9) + (10 - 1)*(5x + 5)

T(10) = - x + 9 + 9(5x + 5)

T(10) = - x + 9 + 45x + 45

T(10) = 44x + 54

Hence, the 10th term is 44x + 54

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The terminal side of Ø is in quadrant II and cos ø = -5/13 What is sin ø?
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the value of sin ∅ is 12/ 13

<h3>Quadrants and the "cast" Rule:</h3>
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From the given question,

We have,  cos ∅= 5/13

From the trigonometric identities, we have that

sin^2\alpha  + cos^2\alpha  = 1

Then , let's substitute the value of cos ∅

sin^2\alpha +(\frac{-5}{13} )^2 = 1

Let make sin the subject of formula and find the squares of the fraction

sin²∅ = 1 - \frac{25}{169}

Find the LCM

sin²∅ = \frac{169 - 25}{169}

Find the difference

sin²∅ = \frac{144}{169}

Find the square root

sin∅ = \sqrt{\frac{144}{159} }

sin∅ = \frac{12}{13}

In quadrant II , sin is positive, so we have

sin ∅ = 12/ 13

Thus, the value of sin ∅ is 12/ 13

Learn more about quadrant  here:

brainly.com/question/863849

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Step-by-step explanation:

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