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rewona [7]
3 years ago
5

How to find A1 in a geometric series for which sn= 105, r=-2 and n=6

Mathematics
1 answer:
inna [77]3 years ago
6 0
Here we're discussing the "nth partial sum of a geom series."

                                                        1 - r^n
The formula for that is s(n) = a(1) --------------- 
                                                          1-r

We can subst. the given values and see where that takes us:

                   1 - (-2)^6        1 - 64
105 = a(1)*---------------- = ----------*a(1)  =  -21*a(1)
                     1-(-2)                3

                                               105
Solving this for a(1):  a(1) = --------- = -5    (answer)
                                               -21
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find the volume of the solid formed by revolving the region bounded by the graphs of y = 4x - x^2 and f(x) = x^2 from [0,2] abou
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Answer:

v =  \frac{32\pi}{3}

or

v=33.52

Step-by-step explanation:

Given

f(x) = 4x - x^2

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[a,b] = [0,2]

Required

The volume of the solid formed

Rotating about the x-axis.

Using the washer method to calculate the volume, we have:

\int dv = \int\limit^b_a \pi(f(x)^2 - g(x)^2) dx

Integrate

v = \int\limit^b_a \pi(f(x)^2 - g(x)^2)\ dx

v = \pi \int\limit^b_a (f(x)^2 - g(x)^2)\ dx

Substitute values for a, b, f(x) and g(x)

v = \pi \int\limit^2_0 ((4x - x^2)^2 - (x^2)^2)\ dx

Evaluate the exponents

v = \pi \int\limit^2_0 (16x^2 - 4x^3 - 4x^3 + x^4 - x^4)\ dx

Simplify like terms

v = \pi \int\limit^2_0 (16x^2 - 8x^3 )\ dx

Factor out 8

v = 8\pi \int\limit^2_0 (2x^2 - x^3 )\ dx

Integrate

v = 8\pi [ \frac{2x^{2+1}}{2+1} - \frac{x^{3+1}}{3+1} ]|\limit^2_0

v = 8\pi [ \frac{2x^{3}}{3} - \frac{x^{4}}{4} ]|\limit^2_0

Substitute 2 and 0 for x, respectively

v = 8\pi ([ \frac{2*2^{3}}{3} - \frac{2^{4}}{4} ] - [ \frac{2*0^{3}}{3} - \frac{0^{4}}{4} ])

v = 8\pi ([ \frac{2*2^{3}}{3} - \frac{2^{4}}{4} ] - [ 0 - 0])

v = 8\pi [ \frac{2*2^{3}}{3} - \frac{2^{4}}{4} ]

v = 8\pi [ \frac{16}{3} - \frac{16}{4} ]

Take LCM

v = 8\pi [ \frac{16*4- 16 * 3}{12}]

v = 8\pi [ \frac{64- 48}{12}]

v = 8\pi * \frac{16}{12}

Simplify

v = 8\pi * \frac{4}{3}

v =  \frac{32\pi}{3}

or

v=\frac{32}{3} * \frac{22}{7}

v=\frac{32*22}{3*7}

v=\frac{704}{21}

v=33.52

8 0
3 years ago
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