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kakasveta [241]
3 years ago
7

If [infinity]∑n=0cn9n is convergent, does it follow that the following series are convergent? (a) [infinity]∑n=0cn(−3)n

Mathematics
1 answer:
Pepsi [2]3 years ago
7 0

Given: The series ∑cₙ9^n is convergent

To find: The series ∑cₙ(-3)^n is convergent or not.

Solution: If the radius of convergence R the we can conclude that R≥4

So, the series will converge as -3<9.

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Answer:

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6 0
3 years ago
(cotx+cscx)/(sinx+tanx)
Butoxors [25]

Answer:   \bold{\dfrac{cot(x)}{sin(x)}}

<u>Step-by-step explanation:</u>

Convert everything to "sin" and "cos" and then cancel out the common factors.

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\text{Simplify:}\\\\\bigg(\dfrac{cos(x)+1}{sin(x)}\bigg)\div\bigg(\dfrac{sin(x)cos(x)+sin(x)}{cos(x)}\bigg)\\\\\\\text{Multiply by the reciprocal (fraction rules)}:\\\\\bigg(\dfrac{cos(x)+1}{sin(x)}\bigg)\times\bigg(\dfrac{cos(x)}{sin(x)cos(x)+sin(x)}\bigg)\\\\\\\text{Factor out the common term on the right side denominator}:\\\\\bigg(\dfrac{cos(x)+1}{sin(x)}\bigg)\times\bigg(\dfrac{cos(x)}{sin(x)(cos(x)+1)}\bigg)

\text{Cross out the common factor of (cos(x) + 1) from the top and bottom}:\\\\\bigg(\dfrac{1}{sin(x)}\bigg)\times\bigg(\dfrac{cos(x)}{sin(x)}\bigg)\\\\\\\bigg(\dfrac{1}{sin(x)}\bigg)\times cot(x)}\qquad \rightarrow \qquad \dfrac{cot(x)}{sin(x)}

6 0
3 years ago
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8_murik_8 [283]

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$1995.97.

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3 0
2 years ago
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Snezhnost [94]

Answer:

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since 3 x x is 3x and the sign in the bracket is addition you place it there then place 3 x 2.3

being 6.9.

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4 0
3 years ago
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