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2 years ago

Find the values of x and y.

Mathematics

2 answers:

Arlecino [84]2 years ago

7 0

Answer:

Step-by-step explanation:

2x + 1 = 85 (Alternate Interior Angles Theorem)

2x = 84

x = 42

(3y + 5) + (2x + 1) = 180 (Linear Pair Theorem)

3y + 5 + (85) = 180

3y + 5 = 95

3y = 90

y = 30

siniylev [52]2 years ago

6 0

Answer:

x = 42, y = 30

Step-by-step explanation:

Alternate angles are equal so

85 = 2x + 1

2x = 84

x = 42°

Linear pair

(3y + 5) + (2x + 1) = 180

3y + 5 + 85 = 180

3y = 90

y = 30°

So I think the answer is x = 42° and y = 30°.

Hope it helps.

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Answer:

$y(t) = y_o e^{\beta t}$

$x(t) = x_o e^{\frac{-\alpha y_o }{\beta }[e^{-\beta t} - 1] }$

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Step-by-step explanation:

From the question we are told that

$\frac{dy}{y} = -\beta dt$

Now integrating both sides

$ln y = \beta t + c$

Now taking the exponent of both sides

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=> $y(t) = e^{\beta t} e^c$

Let $e^c = C$

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Now from the question we are told that

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Hence

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=> $y_o = C$

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From the question we are told that

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substituting for y

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Now integrating both sides

$lnx = \alpha \frac{y_o}{\beta } e^{-\beta t} + c$

Now taking the exponent of both sides

$x(t) = e^{\alpha \frac{y_o}{\beta } e^{-\beta t} + c}$

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Let $e^c = A$

=> $x(t) =K e^{\alpha \frac{y_o}{\beta } e^{-\beta t} }$

Now from the question we are told that

$x(0) = x_o$

$x(0)=x_o =K e^{\alpha \frac{y_o}{\beta } e^{-\beta * 0} }$

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divide both side by $(K * x_o)$

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=> $x(t)= x_o e^{\frac{-\alpha * y_o }{\beta} + \frac{\alpha y_o}{\beta } e^{-\beta t} }$

=> $x(t) = x_o e^{\frac{\alpha y_o }{\beta }[e^{-\beta t} - 1] }$

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