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3 years ago

Classify each pair of numbered angles as corresponding, alternate interior, alternate exterior, or none of these.

Mathematics

1 answer:

maksim [4K]3 years ago

8 0

Answer:

The answer in the attached figures

Step-by-step explanation:

we know that

When two lines are crossed by another line, the angles in matching corners are called Corresponding Angles. If the lines are parallel then the corresponding angles are congruent

see the attached figure N 1

When two lines are crossed by another line, the pair of angles on the inner side of each of those two lines but on opposite sides of the transversal, are called Alternate Interior Angles. If the lines are parallel then the alternate interior angles are congruent.

see the attached figure N 2

Vertical Angles are the angles opposite each other when two lines cross

see the attached figure N 3

When two lines are crossed by another line, the pair of angles on the outer side of each of those two lines but on opposite sides of the transversal, are called Alternate Exterior Angles. If the lines are parallel then the alternate exterior angles are congruent.

see the attached figure N 4

When two lines are crossed by another line, the pairs of angles on one side of the transversal but inside the two lines are called Consecutive Interior Angles. If the lines are parallel then the consecutive interior angles are supplementary.

see the attached figure N 5

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harina [27]

Answer:

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

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

$\lim_{t \to \infty} x(t) = x_oe^{\frac{-\alpha y_o}{\beta } }$

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

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

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

Let $e^c = C$

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

Now from the question we are told that

$y(0) = y_o$

Hence

$y(0) = y_o = Ce^{\beta * 0}$

=> $y_o = C$

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

From the question we are told that

$\frac{dx}{dt} = -\alpha xy$

substituting for y

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

=> $\frac{dx}{x} = -\alpha y_oe^{-\beta t} dt$

Now integrating both sides

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

Now taking the exponent of both sides

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

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

Let $e^c = A$

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

Now from the question we are told that

$x(0) = x_o$

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

=> $x_o = K e^{\frac {\alpha y_o }{\beta } }$

divide both side by $(K * x_o)$

=> $K = x_o e^{\frac {\alpha y_o }{\beta } }$

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

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

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

Generally as t tends to infinity , $e^{- \beta t}$ tends to zero

$\lim_{t \to \infty} x(t) = x_oe^{\frac{-\alpha y_o}{\beta } }$

5 0

4 years ago

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grigory [225]

Answer:

Yes

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Step-by-step explanation:

$10$

$3$

$8$

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7 0

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Harlamova29_29 [7]

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