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Sonja [21]
3 years ago
11

Use complete sentences to describe the transformation that maps ABCD onto its image.

Mathematics
1 answer:
Lapatulllka [165]3 years ago
6 0

General Idea:

When a point or figure on a coordinate plane is moved by sliding it to the right or left or up or down, the movement is called a translation.

Say a point P(x, y) moves up or down ' k ' units, then we can represent that transformation by adding or subtracting respectively 'k' unit to the y-coordinate of the point P.

In the same way if P(x, y) moves right or left ' h ' units, then we can represent that transformation by adding or subtracting respectively 'h' units to the x-coordinate.

P(x, y) becomes P'(x\pm h, y\pm k). We need to use ' + ' sign for 'up' or 'right' translation and use ' - ' sign for ' down' or 'left' translation.

Applying the concept:

The point A of Pre-image is (0, 0). And the point A' of image after translation is (5, 2). We can notice that all the points from the pre-image moves 'UP' 2 units and 'RIGHT' 5 units.

Conclusion:

The transformation that maps ABCD onto its image is translation given by (x + 5, y + 2),

In other words, we can say ABCD is translated 5 units RIGHT and 2 units UP to get to A'B'C'D'.

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If 5x−4y=22 and 3x+6y=30, what is the value of x−5y
amm1812

Answer:

-4 See Below:

Step-by-step explanation:

When you have a system of equations, you need to solve for one variable, and subsitute that variable into the other equation to solve for the other variables value.

3x +6y = 30 In this case, it would be easier to solve for x.

3x= 30-6y  Divide both sides by 3

x= 10-2y

Now we substitute for x in the other equation.

5x- 4y = 22

5(10-2y)-4y=22

50-10y-4y=22

50-14y=22 Add 14y to both sides

50= 14y + 22 Subtract 22 from both sides

28 = 14y     Divide by 14 on both sides.

y = 2

Substitute y=2 and solve for x.

x = 10- 2y

x= 10-2*2

x= 10-4

x= 6

Now we know y=2 and x=6, so we can substitute them into x-5y to find its value.

x - 5y

= 6-5*2

=6-10

= -4

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What is partial derivative of z=(2x+3y)^10 with respect to x,y?
maw [93]
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\dfrac\partial{\partial y}(2x+3y)^{10}=10(2x+3y)^9\times3=30(2x+3y)^9

Or, if you actually did want the second order derivative,

\dfrac{\partial^2}{\partial y\partial x}(2x+3y)^{10}=\dfrac\partial{\partial y}\left[20(2x+3y)^9\right]=180(2x+3y)^8\times3=540(2x+3y)^8

and in case you meant the other way around, no need to compute that, as z_{xy}=z_{yx} by Schwarz' theorem (the partial derivatives are guaranteed to be continuous because z is a polynomial).
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Answer:

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Step-by-step explanation:

You have to subtract 180 and 117 which is 63.

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