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Mariana [72]
3 years ago
12

Find the common ratio for the geometric sequence defined by the formula: an=40(2–√)n−1

Mathematics
1 answer:
Maksim231197 [3]3 years ago
3 0

Answer:

b is the answer in your questions

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Divide (x^3-20x+16)/(x-4)
Lorico [155]
\frac{( x^{3}-20x+16) }{(x-4)}

Use the rational root theorem:
a_{o}=16, a_{n}=1

The dividers of a_{o}: 1,2,4,8,16 
The dividers of a_{n}: 1

Therefore, check the following rational numbers: +-\frac{1,2,4,8,16}{1}

\frac{4}{1} is a root of the expression, so factor out x-4

Compute \frac{ x^{3}-20x+16 }{x-4} to get the rest of the equation.

\frac{(x-4)( x^{2} +4x-4)}{x-4}

Cancel the common factor:
x^{2}+4x-4 is the final answer
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3 times the amount of x+2? it is worth 10 points and
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3(x+2) is the expression
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In an arithmetic sequence, a17 = -40 and
Viktor [21]

Answer:

Tn = 2Tn-1 - Tn-2

Step-by-step explanation:

Before we can generate the recursive sequence, we need to find the nth term of the given sequence.

nth term of an AP is given as:

Tn = a+(n-1)d

If a17 = -40

T17 = a+(17-1)d = -40

a+16d = -40 ...(1)

If a28 = -73

T28 = a+(28-1)d = -73

a+27d = -73 ...(2)

Solving both equations simultaneously using elimination method.

Subtracting 1 from 2 we have:

27d - 16d = -73-(-40)

11d = -73+40

11d = -33

d = -3

Substituting d = -3 into 1

a+16(-3) = -40

a - 48 = -40

a = -40+48

a = 8

Given a = 8, d = -3, the nth term of the sequence will be

Tn = 8+(n-1) (-3)

Tn = 8+(-3n+3)

Tn = 8-3n+3

Tn = 11-3n

Given Tn = 11-3n and d = -3

Tn-1 = Tn - d... (3)

Tn-1 = 11-3n +3

Tn-1 = 14-3n

Tn-2 = Tn-2d...(4)

Tn-2 = 11-3n-2(-3)

Tn-2 = 11-3n+6

Tn-2 = 17-3n

From 3, d = Tn - Tn-1

From 4, d = (Tn - Tn-2)/2

Equating both common difference

(Tn - Tn-2)/2 = Tn - Tn-1

Tn - Tn-2 = 2(Tn - Tn-1)

Tn - Tn-2 = 2Tn-2Tn-1

2Tn-Tn = 2Tn-1 - Tn-2

Tn = 2Tn-1 - Tn-2

The recursive formula will be

Tn = 2Tn-1 - Tn-2

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Find the volume of the cone shown below.
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I'm Not Completely sure but the closest Answer seems to be C.
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