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3 years ago

Help ASAP (RIGHT ANSWER GETS BRAINLIEST)

Mathematics

2 answers:

Gnoma [55]3 years ago

5 0

Answer:

Dilation by a scale factor of 4 followed by reflection over the y-axis

Step-by-step explanation:

densk [106]3 years ago

5 0

<h2>Answer:</h2>

The set of transformations that has been performed on triangle EFG to form triangle E"F"G" is:

Dilation by a scale factor of 4 followed by reflection over the y-axis

<h2>Step-by-step explanation:</h2>

The coordinates of the vertices of triangle EFG are:

E(-2,1)

F(-2,0)

and G(-3,0)

Now when it is dilated by a scale factor of 4 then the rule of transformation is:

(x,y) → (4x,4y)

Hence,

E(-2,1) → E'(-8,4)

F(-2,0) → F'(-8,0)

and G(-3,0) → G'(-12,0)

Now when this figure is again reflected over the y-axis then the rule of transformation is:

(x,y) → (-x,y)

Hence,

E'(-8,4) → E"(8,4)

F'(-8,0) → F"(8,0)

and G'(-12,0) → G"(12,0)

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3 years ago

Read 2 more answers

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Igoryamba

Answer:

If you want to know what b is then it is 16

Step-by-step explanation:

38 + b = 54

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2 years ago

Complete the given diagram by dragging expressions to each leg of the triangle. Then, correctly complete the equation to derive

Nimfa-mama [501]

The equation to derive the distance d is $\sqrt{(x2-x1)^2+(y2-y1)^2}$ . The lengths of the other legs of the given triangle are (y2 - y1) and (x2 - x1).

<h3>What is the formula for calculating the distance between two points?</h3>

Consider the two points (x1, y1) and (x2, y2)

The formula used for calculating the distance between the two points is

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<h3>Calculation:</h3>

Given that,

The triangle in the graph has vertices (x1, y1), (x2, y2), and (x2, y1)

Since this triangle makes 90°, it is a right-angled triangle.

Hypotenuse = (x1, y1) to (x2, y2), Adjacent = (x1, y1) to (x2,y1), and Opposite = (x2, y1) to (x2, y2).

Consider the length of the hypotenuse = d

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$d = \sqrt{(x2-x1)^2+(y2-y1)^2}$

And the lengths of the other two legs of the given triangle are,

Length of the adjacent side: (x1, y1) to (x2,y1)

= $\sqrt{(x2-x1)^2+(y1-y1)^2}$

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= $\sqrt{0+(y2-y1)^2}$

= $(y2-y1)$

Therefore, the derived distances for the given triangle are:

$d=\sqrt{(x2-x1)^2+(y2-y1)^2}$ , (x2 - x1), and (y2 - y1).

Learn more about the distance between two points here:

brainly.com/question/661229

#SPJ1

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1 year ago

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Step-by-step explanation:

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2 years ago

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andreyandreev [35.5K]

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3 years ago