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Irina18 [472]
3 years ago
6

Define a sequence and give an example. Choose the correct answer below. A. A sequence ​{a Subscript n ​} is an ordered list of n

umbers of the form ​{a 1 ​, a 2 ​, a 3 ​, ​..., a Subscript n ​, ​...}. For​ example, {1,​ 3, 5,​ ...} is a sequence. B. A sequence ​{a Subscript n ​} is a finite set of numbers ​{a 1 ​, a 2 ​, a 3 ​, ​..., a Subscript n ​}. For​ example, {5,​ 19, 400,​ 2, 17} is a sequence. C. A sequence ​{a Subscript n ​} is an infinite sum of the form a 1 plus a 2 plus a 3 plus ... equals Summation from k equals 1 to infinity a Subscript k . For​ example, 2 plus 3 plus 4 plus ... equals Summation from k equals 1 to infinity (k plus 1 )is a sequence. D. A sequence ​{a Subscript n ​} is an increasing finite set of numbers ​{a 1 ​, a 2 ​, a 3 ​, ​..., a Subscript n ​} such that a 1 less thana 2less thana 3less than...less thana Subscript n. For​ example, {2,​ 5, 17,​ 19, 400} is a sequence.
Mathematics
1 answer:
Anika [276]3 years ago
5 0

Answer:

The correct option is A

A sequence ​{a Subscript n ​} is an ordered list of numbers of the form ​{a 1 ​, a 2 ​, a 3 ​, ​..., a Subscript n ​, ​...}. For​ example, {1,​ 3, 5,​ ...} is a sequence.

Step-by-step explanation:

The right option is A.

A sequence is an ordered list of number,that could be A.P( arithmetic progression, linear sequence)

Or G.P. ( geometric progression, geometric sequence).

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We can begin by multiplying by its conjugate:

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Simplify using the identity:

1 - cos^2A = sin^2A

\sqrt{\frac{(1-cos^2A)}{(1+cosA)^2}} =\\\\\sqrt{\frac{(sin^2A)}{(1+cosA)^2}} =

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Multiply again by the conjugate to get a SINGLE term in the denominator:

{\frac{sinA}{1+cosA} * {\frac{1-cosA}{1-cosA} =\\

Simplify:

{\frac{sinA(1-cosA)}{1-cos^2A} =

Use the above trig identity one more:

{\frac{sinA(1-cosA)}{sin^2A} =

Cancel out sinA:

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Step-by-step explanation:

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Step-by-step explanation:

I will solve your system by substitution.

(You can also solve this system by elimination.)

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y=−3x+2

Step: Substitute−3x+2foryin5x−y=8:

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