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Korvikt [17]
3 years ago
5

Need help on this ! ‼️

Mathematics
1 answer:
mihalych1998 [28]3 years ago
5 0

Answer:

I don't know this one

Step-by-step explanation:

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What is the decimal equivalent of the -68/255
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Find the general solution of the following ODE: y' + 1/t y = 3 cos(2t), t > 0.
Margarita [4]

Answer:

y = 3sin2t/2 - 3cos2t/4t + C/t

Step-by-step explanation:

The differential equation y' + 1/t y = 3 cos(2t) is a first order differential equation in the form y'+p(t)y = q(t) with integrating factor I = e^∫p(t)dt

Comparing the standard form with the given differential equation.

p(t) = 1/t and q(t) = 3cos(2t)

I = e^∫1/tdt

I = e^ln(t)

I = t

The general solution for first a first order DE is expressed as;

y×I = ∫q(t)Idt + C where I is the integrating factor and C is the constant of integration.

yt = ∫t(3cos2t)dt

yt = 3∫t(cos2t)dt ...... 1

Integrating ∫t(cos2t)dt using integration by part.

Let u = t, dv = cos2tdt

du/dt = 1; du = dt

v = ∫(cos2t)dt

v = sin2t/2

∫t(cos2t)dt = t(sin2t/2) + ∫(sin2t)/2dt

= tsin2t/2 - cos2t/4 ..... 2

Substituting equation 2 into 1

yt = 3(tsin2t/2 - cos2t/4) + C

Divide through by t

y = 3sin2t/2 - 3cos2t/4t + C/t

Hence the general solution to the ODE is y = 3sin2t/2 - 3cos2t/4t + C/t

3 0
3 years ago
Which equation could be used to calculate the sum of the geometric series? One-third + two-ninths + StartFraction 4 Over 27 EndF
kondaur [170]

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

4 0
3 years ago
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