Help Pleasee

Reflect shape A in the line x = -2

Help PleaseeReflect Shape A In The Line X = -2

Answers

Answer 1

Answer: Check out the diagram below

The reflected image is shown in red.

=================================================

Explanation:

Draw a vertical line through -2 on the x axis. This is the mirror line.

Now focus on the upper right corner of the figure, which is at (-3, -1). Notice how the horizontal distance from this corner point to the mirror line is exactly 1 unit. If we move another 1 unit to the right, then we'll arrive at (-1,-1) which is where the reflected point lands or ends up.

In short, the upper right corner point (-3,-1) reflects over x = -2 to land on (-1,-1)

----------------------

As another example, the upper left corner point (-5, -1) will move exactly 4 spaces to the right to get to the mirror line. Then we move another 4 spaces to the right to get to (2,-1).

So the upper left corner (-5,-1) will ultimately move to (2,-1) after the reflection over x = -2.

Apply these steps to the other corner points and you'll end up with what is shown below.

Take note that a point like A(-5,-1) moves to A'(1,-1), and similar to the other points as well. Also, notice that when going from A to B to C, etc we are moving clockwise. We move counterclockwise when going from A' to B' to C' etc. Reflections always swap the orientation.

Help PleaseeReflect Shape A In The Line X = -2
Answer 2

Transformation involves changing the position of a shape.

The coordinates of the image are:

[tex]\mathbf{A '= (1,1)}[/tex]      [tex]\mathbf{B' = (-1,-1)}[/tex]     [tex]\mathbf{C' = (-1,-4)}[/tex]     [tex]\mathbf{D' = (0,-4)}[/tex]

[tex]\mathbf{E' = (0,-3)}[/tex]      [tex]\mathbf{F = (1,-3)}[/tex]

The vertices of the shape are:

[tex]\mathbf{A = (-5,-1)}[/tex]

[tex]\mathbf{B = (-3,-1)}[/tex]

[tex]\mathbf{C = (-3,-4)}[/tex]

[tex]\mathbf{D = (-4,-4)}[/tex]

[tex]\mathbf{E = (-4,-3)}[/tex]

[tex]\mathbf{F = (-5,-3)}[/tex]

The rule of reflection across line x =-2 is:

[tex]\mathbf{(x,y)=(-x -4,y)}[/tex]

So, we have:

[tex]\mathbf{A '= (5 - 4,-1) = (1,-1)}[/tex]

[tex]\mathbf{B' = (3-4,-1) = (-1,-1)}[/tex]

[tex]\mathbf{C' = (3-4,-4) = (-1,-4)}[/tex]

[tex]\mathbf{D' = (4-4,-4) = (0,-4)}[/tex]

[tex]\mathbf{E' = (4-4,-3) = (0,-3)}[/tex]

[tex]\mathbf{F = (5-4,-3) = (1,-3)}[/tex]

So, the coordinates of the image are:

[tex]\mathbf{A '= (1,-1)}[/tex]

[tex]\mathbf{B' = (-1,-1)}[/tex]

[tex]\mathbf{C' = (-1,-4)}[/tex]

[tex]\mathbf{D' = (0,-4)}[/tex]

[tex]\mathbf{E' = (0,-3)}[/tex]

[tex]\mathbf{F = (1,-3)}[/tex]

See attachment for the image of the transformations

Read more about transformations at:

https://brainly.com/question/11709244

Help PleaseeReflect Shape A In The Line X = -2

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Answer:

[tex](a)\ y = -0.75x + 26[/tex]

[tex](b)\ y = 14.75[/tex]

Step-by-step explanation:

Given

See attachment for scatter plot

Required

Equation of the line of best fit

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[tex]y = -0.75x + 26[/tex]

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[tex]y = -0.75x + 26[/tex] becomes

[tex]y = -0.75*15 + 26[/tex]

[tex]y = -11.25 + 26[/tex]

[tex]y = 14.75[/tex]

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Answer:

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Step-by-step explanation:

Given the function f(x) = 0.5|x – 4| – 3, we are to find the values of x for f(x) = 7

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Step-by-step explanation:

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Step-by-step explanation:

[tex]\frac{x}{3} + \frac{1}{1} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [ \ LCM \ of \ 1 , 3 = 3 \ ]\\\\= \frac{x}{3} + \frac{1 \times 3}{1 \times 3} \\\\= \frac{x}{3} + \frac{3}{3}\\\\=\frac{x+ 3}{3} \\\\[/tex]

Answer:

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Step-by-step explanation:

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=====================================================

Explanation:

The diagrams for problems 7 and 8 are shown below.

------------------------------------

Problem 7

Apply the tangent ratio to find that...

tan(angle) = opposite/adjacent

tan(42) = x/28

28*tan(42) = x

x = 28*tan(42)

x = 25.2113132403396

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Make sure your calculator is in degree mode. One way to check is to type in tan(45) and you should get 1 as a result.

------------------------------------

Problem 8, part 1

To find the angle y, we must use the sine ratio

sin(angle) = opposite/hypotenuse

sin(y) = 18/23

y = arcsin(18/23)

y = 51.5000495907521

y = 52 degrees approximately

------------------------------------

Problem 8, part 2

We can use the pythagorean theorem to find the missing side x

a^2+b^2 = c^2

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Or alternatively, we can apply the tangent ratio on the angle y we found earlier to help find x

tan(angle) = opposite/adjacent

tan(y) = 18/x

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x = 14.3178210636621

x = 14.3

------------------------------------

Problem 9

The double tickmarks show that BE = EC = 10. That means BC = BE+EC = 10+10 = 20.

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20x = 440

x = 440/20

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Answers

Answer:

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Step-by-step explanation:

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Answers

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The tower is approximately 57.34 meters long

Step-by-step explanation:

The given parameters are;

The type of triangle formed by ΔABC = Non-right triangle

The measure of ∠ACB = 55°

The horizontal displacement of the tower, BX = 5 m

The length of BC = 45 m

Therefore, we have;

Triangle ΔABC type = Right triangle

By the tangent to an acute angle, θ, in a right triangle, we have;

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Answer:

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Answers

Step-by-step explanation:

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Answer:

114.5

Step-by-step explanation:

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Answers

Answer:

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Answers

Answer:

6.4 meters per second

Step-by-step explanation:

Sarah the cheetah ran 100 meters at a speed of 16.8 meters per second. An olympian ran the 100-meter dash in 9.6 seconds. How much faster was Sarah the cheetah’s speed, to the nearest tenth of a meter per second?

0.9 meters per second

1.6 meters per second

6.4 meters per second

10.4 meters per second

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the tenth is the first number after the decimal place. To convert to the nearest tenth, look at the number after the tenth (the hundredth). If the number is greater or equal to 5, add 1 to the tenth figure. If this is not the case, add zero

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Answers

Answer:

You wrote the question wrong...

m is not -16 it is -1/6 (that is a big difference)

y=mx+b

y=-1/6x+b

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Step-by-step explanation:

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Answers

Answer:

5x-20=3x(alternate angle)

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x=20/2

x=10

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y=150/6

y=25

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