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oksian1 [2.3K]
2 years ago
14

Copy the figure below onto a separate sheet of paper. Find the image of the figure under reflections in line m and then line t.

In the box below, describe the location of the new image of point G. Use words like above, below, left, or right of line m and t.

Mathematics
1 answer:
san4es73 [151]2 years ago
4 0

The image of the figure under reflections in line m and then line t is shown below.

<h3>What is a transformation of geometry?</h3>

A spatial transformation is each mapping of feature shapes to itself, and it maintains some spatial correlation between figures.

Reflection does not change the size and shape of the geometry.

The image of the figure under reflections in line m and then line t.

The reflection of the rectangle DEFG will be given in the figure.

More about the transformation of geometry link is given below.

brainly.com/question/22532832

#SPJ1

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3 years ago
Use variation of parameters to find a general solution to the differential equation given that the functions y1 and y2 are linea
Aleks [24]

Recall that variation of parameters is used to solve second-order ODEs of the form

<em>y''(t)</em> + <em>p(t)</em> <em>y'(t)</em> + <em>q(t)</em> <em>y(t)</em> = <em>f(t)</em>

so the first thing you need to do is divide both sides of your equation by <em>t</em> :

<em>y''</em> + (2<em>t</em> - 1)/<em>t</em> <em>y'</em> - 2/<em>t</em> <em>y</em> = 7<em>t</em>

<em />

You're looking for a solution of the form

y=y_1u_1+y_2u_2

where

u_1(t)=\displaystyle-\int\frac{y_2(t)f(t)}{W(y_1,y_2)}\,\mathrm dt

u_2(t)=\displaystyle\int\frac{y_1(t)f(t)}{W(y_1,y_2)}\,\mathrm dt

and <em>W</em> denotes the Wronskian determinant.

Compute the Wronskian:

W(y_1,y_2) = W\left(2t-1,e^{-2t}\right) = \begin{vmatrix}2t-1&e^{-2t}\\2&-2e^{-2t}\end{vmatrix} = -4te^{-2t}

Then

u_1=\displaystyle-\int\frac{7te^{-2t}}{-4te^{-2t}}\,\mathrm dt=\frac74\int\mathrm dt = \frac74t

u_2=\displaystyle\int\frac{7t(2t-1)}{-4te^{-2t}}\,\mathrm dt=-\frac74\int(2t-1)e^{2t}\,\mathrm dt=-\frac74(t-1)e^{2t}

The general solution to the ODE is

y = C_1(2t-1) + C_2e^{-2t} + \dfrac74t(2t-1) - \dfrac74(t-1)e^{2t}e^{-2t}

which simplifies somewhat to

\boxed{y = C_1(2t-1) + C_2e^{-2t} + \dfrac74(2t^2-2t+1)}

4 0
3 years ago
What is the area of a triangle that has a height of 8 feet and a base length of 5 feet?
Lena [83]

Answer:

20 ft²

Step-by-step explanation:

the area (A) of a triangle is calculated using the formula

A = \frac{1}{2} bh ( b is the base and h the height )

here b = 5 and h = 8

A = \frac{1}{2} × 5 × 8 = 20 ft²


7 0
3 years ago
Read 2 more answers
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