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adelina 88 [10]

4 years ago

14

Which equation could be used to calculate the sum of the geometric series? One-third + two-ninths + StartFraction 4 Over 27 EndF

raction + StartFraction 8 Over 81 EndFraction + StartFraction 16 Over 243 EndFraction S 5 = StartFraction one-third (1 minus (two-thirds) Superscript 5 Baseline) Over (1 minus two-thirds) EndFraction S 5 = StartFraction two-thirds (1 minus (one-third) Superscript 5 Baseline) Over (1 minus one-third) EndFraction S = StartFraction one-third Over (1 minus two-thirds) EndFraction S = StartFraction two-thirds Over (1 minus one-third) EndFraction

2 answers:

Troyanec [42]4 years ago

5 0

Answer:

A

Step-by-step explanation:

edge 2020

kondaur [170]4 years ago

4 0

Answer:

S 5 = StartFraction one-third (1 minus (two-thirds) Superscript 5 Baseline) Over (1 minus two-thirds) EndFraction

Step-by-step explanation:

Given the geometric series:

1/3+2/9+4/27+8/81+16/243

First we must know that the series is a finite series with just 5terms.

Before we can know the formula to calculate sum of the first five terms of the series, we must determine its common ratio (r) first.

r = (2/9)÷1/3 = 4/27÷2/9= 8/81÷4/27

r = 2/9 × 3/1

r = 2/3

Similarly;

r = 4/27×9/2

r = 2/3

Since all values of r is the as them the common ratio is 2/3.

If r< 1 in geometric series, then the formula for finding its sum is applicable

Sn = a(1-rⁿ)/1-r

a is the first term = 1/3

r is the common ratio = 2/3

n is the number of terms = 5

Substituting the values in the formula we have:

S5 = 1/3{1-(2/3)^5}/1-2/3

This gives the requires equation

S 5 = StartFraction one-third (1 minus (two-thirds) Superscript 5 Baseline) Over (1 minus two-thirds) EndFraction

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Solve these linear equations in the form y=yn+yp with yn=y(0)e^at.

WINSTONCH [101]

Answer:

a) $y(t) = y_{0}e^{4t} + 2$ . It does not have a steady state

b) $y(t) = y_{0}e^{-4t} + 2$ . It has a steady state.

Step-by-step explanation:

a) $y' -4y = -8$

The first step is finding $y_{n}(t)$ . So:

$y' - 4y = 0$

We have to find the eigenvalues of this differential equation, which are the roots of this equation:

$r - 4 = 0$

$r = 4$

So:

$y_{n}(t) = y_{0}e^{4t}$

Since this differential equation has a positive eigenvalue, it does not have a steady state.

Now as for the particular solution.

Since the differential equation is equaled to a constant, the particular solution is going to have the following format:

$y_{p}(t) = C$

So

$(y_{p})' -4(y_{p}) = -8$

$(C)' - 4C = -8$

C is a constant, so (C)' = 0.

$-4C = -8$

$4C = 8$

$C = 2$

The solution in the form is

$y(t) = y_{n}(t) + y_{p}(t)$

$y(t) = y_{0}e^{4t} + 2$

b) $y' +4y = 8$

The first step is finding $y_{n}(t)$ . So:

$y' + 4y = 0$

We have to find the eigenvalues of this differential equation, which are the roots of this equation:

$r + 4 =$

$r = -4$

So:

$y_{n}(t) = y_{0}e^{-4t}$

Since this differential equation does not have a positive eigenvalue, it has a steady state.

Now as for the particular solution.

Since the differential equation is equaled to a constant, the particular solution is going to have the following format:

$y_{p}(t) = C$

So

$(y_{p})' +4(y_{p}) = 8$

$(C)' + 4C = 8$

C is a constant, so (C)' = 0.

$4C = 8$

$C = 2$

The solution in the form is

$y(t) = y_{n}(t) + y_{p}(t)$

$y(t) = y_{0}e^{-4t} + 2$

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Masteriza [31]

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Alex_Xolod [135]

Answer:

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Step-by-step explanation:

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irina [24]

Answer:

A.6.2i-4.2 j

Step-by-step explanation:

We are given that

$u=9i-6j$

$v=-3i-2j$

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We have to find the projection of w on to u.

The project of vector a on b

= $\frac{a\cdot b}{\mid b\mid^2}b$

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Using the formula

The projection of w on to u

$=\frac{u\cdot w}{\mid u\mid^2}(u)=\frac{81}{117}(9i-6j)$

The projection of w on to u=6.2i-4.2j

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... 3 = x

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