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3 years ago

What is not a congruence transformation

Mathematics

1 answer:

vodka [1.7K]3 years ago

7 0

Hi there!

There are four different types of transformations:

-Reflections

-Translations

-Rotations

-Dilations

In a reflection, the shape is simply reflected across a line of reflection. It does not change size or shape, so therefore it is a congruence transformation.

In a translation, the shape is moved up, down, left, or right. It does not change size or shape, so therefore it is a congruence transformation.

In a rotation, the shape is rotated around a point of rotation. It does not change size or shape, so therefore it is a congruence transformation.

In a dilation, the shape is made larger or smaller by a scale factor. It does change size, so therefore it is not a congruence transformation.

The transformation that is not a congruence transformation is a DILATION.

Hope this helps!! :)

If there's anything else that I can help you with, please let me know!

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Answer:

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Step-by-step explanation:

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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:

$y_{p}(t) = C$

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

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

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:

$y_{p}(t) = C$

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

$(C)' + 4C = 8$

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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