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DanielleElmas [232]
3 years ago
7

Transformations can be thought of as functions that take the points from one object (the pre-image) to the corresponding points

on another object (the image). In the unit, you saw examples of how algebra can be used to rewrite the coordinates of points after they have been transformed. You can also use function notation to rewrite the coordinates. For example, these statements show several different ways you can use function notation for the translation, T, of a point 6 units in the positive x-direction:
T(x, y) = (x + 6, y)
T:(x, y) = (x + 6, y)
T<6, 0>(x, y) = (x + 6, y)
T6, 0(x, y) = (x + 6, y)
Part A
Write functions for each of the following transformations using function notation. Choose a different letter to represent each function. For example, you can use R to represent rotations. Assume that a positive rotation occurs in the counterclockwise direction.

translation of a units to the right and b units up
reflection across the y-axis
reflection across the x-axis
rotation of 90 degrees counterclockwise about the origin, point O
rotation of 180 degrees counterclockwise about the origin, point O
rotation of 270 degrees counterclockwise about the origin, point O
Mathematics
1 answer:
sweet-ann [11.9K]3 years ago
4 0

1. Translation of a units to the right and b units up has rule:

(x,y)→(x+a,y+b).

Then the function is

T(x,y)=(x+a,y+b).

2. Reflection across the y-axis acts in following way:

(x,y)→(-x,y).

Then the function is

R_y(x,y)=(-x,y).

3. Reflection across the x-axis acts in following way:

(x,y)→(x,-y).

Then the function is

R_x(x,y)=(x,-y).

4. Rotation of 90 degrees counterclockwise about the origin, point O acts bu the rule:

(x,y)→(-y,x).

Then the function is

R_{O,90^{\circ}}(x,y)=(-y,x).

5. Rotation of 180 degrees counterclockwise about the origin, point O acts bu the rule:

(x,y)→(-x,-y).

Then the function is

R_{O,180^{\circ}}(x,y)=(-x,-y).

6. Rotation of 270 degrees counterclockwise about the origin, point O acts bu the rule:

(x,y)→(y,-x).

Then the function is

R_{O,270^{\circ}}(x,y)=(y,-x).

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Answer:

The answer to your question is: the first option  - 9/ \sqrt{80}

Step-by-step explanation:

Process

          cosФ = adjacent side / hypotenuse

          -1/9 = adjacent side / hypotenuse

          adjacent side = -1    hypotenuse = 9

          c² = a² + b²

          b² = c² - a²

           b² = 81 - 1

           b = \sqrt{80} \\    b = opposite side

But b is negative because we are in the third quadrant.

          b = -\sqrt{80}

Finally   csc Ф = hypotenuse / opposite side

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Answer:

A. y=3x-10

C. y+6=3(x-15)

Step-by-step explanation:

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The given line is 6x+18y=5

Express this in slope-intercept form y=mx+b, where m is the slope and b is the y-intercept.

6x+18y=5\\18y=-6x+5\\y=-\frac{6}{18}x+\frac{5}{18}\\y=-\frac{1}{3}x+\frac{5}{18}

Therefore, the slope of the line is m=-\frac{1}{3}.

Now, for perpendicular lines, the product of their slopes is equal to -1.

Let us find the slopes of each lines.

Option A:

y=3x-10

On comparing with the slope-intercept form, we get slope as  m_{A}=3.

Now, m\times m_{A}=-\frac{1}{3}\times 3=-1. So, option A is perpendicular to the given line.

Option B:

For lines of the form x=a, where, a is a constant, the slope is undefined. So, option B is incorrect.

Option C:

On comparing with the slope-point form, we get slope as  m_{C}=3.

Now, m\times m_{C}=-\frac{1}{3}\times 3=-1. So, option C is perpendicular to the given line.

Option D:

3x+9y=8\\9y=-3x+8\\y=-\frac{3}{9}x+\frac{8}{9}\\y=-\frac{1}{3}x+\frac{8}{9}

On comparing with the slope-intercept form, we get slope as  m_{D}=-\frac{1}{3}.

Now, m\times m_{D}=-\frac{1}{3}\times -\frac{1}{3}=\frac{1}{9}. So, option D is not perpendicular to the given line.

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