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2 years ago

Find the sum of the first fifteen terms of the series:

Mathematics

1 answer:

bija089 [108]2 years ago

4 0

Answer:

Step-by-step explanation:

The given series is a geometric series because the consecutive terms differ by a common ratio. The formula for determining the sum of the first n terms, Sn of a geometric sequence is expressed as

Sn = (ar^n - 1)/(r - 1)

Where

n represents the number of term in the sequence.

a represents the first term in the sequence.

r represents the common ratio.

From the information given,

a = 3

r = 6/3 = 12/6 = 2

n = 15

Therefore, the sum of the first 15 terms, S15 is

S15 = (3 × 2^(15) - 1)/2 - 1

S15 = (3 × 32767)/1

S15 = 98301

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What is the value of a?

Annette [7]

Answer:

Step-by-step explanation:

Look at the picture.

The formula of an area of a trapezoid is:

$A=\dfrac{b_1+b_2}{2}\cdot h$

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$h$ - height

We have:

$A=128,\ b_1=a+2,\ b_2=a-2,\ h=8$

Substitute:

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2 years ago

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klasskru [66]

Answer:

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Step-by-step explanation:

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2 years ago

Read 2 more answers

one x-intercept for a parabola is at the point (-6,0) find the other x-intercept for the parabola defined by this equation y=x^2

Tanya [424]

Hello.

First off, you can factor this equation to find the missing x-intercept.

y = x² + 8x + 12
y = (x + 6)(x + 2)

Then, you can use the zero product property to find the x values.

x + 6 = 0 x + 2 = 0
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2 years ago

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jolli1 [7]

The answer is:

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2 years ago

There are 11 candidates for three postions at a restaraunt. One postion is for a cook. The second position is for a food server

LenaWriter [7]

Answer:

165 combinations possible

Step-by-step explanation:

This is a combination problem as opposed to a permutation, because the order in which we fill these positions is not important. We are merely looking for how many ways each of these 11 people can be rearranged and matched up with different candidates, each in a different position each time. The formula can be filled in as follows:

₁₁C₃ = $\frac{11!}{3!(11-3)!}$

which simplifies to

₁₁C₃ = $\frac{11*10*9*8!}{3*2*1(8!)}$

The factorial of 8 will cancel out in the numerator and the denominator, leaving you with

₁₁C₃ = $\frac{990}{6}$

which is 165

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2 years ago