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Andrei [34K]
3 years ago
11

4(t+25) = (t+50) - 4 (0.15t)

Mathematics
1 answer:
LenKa [72]3 years ago
6 0

Answer:

4(t+25) = (t+50) - 4 (0.15t)

4t + 100 = t + 50 - 0.6t

4t + 100 = 50 + 0.4t

4t - 0.4t + 100 = 50 + 0.4t - 0.4t

3.6t + 100 = 50

3.6t + 100 - 100 = 50 - 100

3.6t = -50

3.6t / 3.6 = -50 / 3.6

t = - 13.9

Step-by-step explanation:

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Find a particular solution to y" - y + y = 2 sin(3x)
leonid [27]

Answer with explanation:

The given differential equation is

y" -y'+y=2 sin 3x------(1)

Let, y'=z

y"=z'

\frac{dy}{dx}=z\\\\d y=zdx\\\\y=z x

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This is a type of linear differential equation.

Integrating factor

     =e^{\int (x-1) dx}\\\\=e^{\frac{x^2}{2}-x}

Multiplying both sides of equation (1) by integrating factor and integrating we get

\rightarrow z\times e^{\frac{x^2}{2}-x}=\int 2 sin 3 x \times e^{\frac{x^2}{2}-x} dx=I

I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx -\int \frac{2\cos 3x e^{\fra{x^2}{2}-x}}{3} dx\\\\I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx-\frac{2I}{3}\\\\\frac{5I}{3}=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx\\\\I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{5}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{5} dx

8 0
3 years ago
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Answer:

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Quesnel Lake:Lake Baikal

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Quesnel Lake:Great Slave Lake

506  \div 614.172 = .82

8 0
2 years ago
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GarryVolchara [31]
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8 0
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