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siniylev [52]
2 years ago
12

Circle A has center of (4, 5), and a radius of 3 and circle B has a center of (1, 7), and a radius of 9. What steps will help sh

ow that circle A is similar to circle B? (6 points) Translate circle A using the rule (x + 3, y - 2). Сь Dilate circle A by a scale factor of 3. Oc Rotate/circle A 90° about the center, D Reflect circle A over the x-axis,​
Mathematics
2 answers:
Katarina [22]2 years ago
8 0

Answer:

Dilate Circle A by a scale factor of 3

Step-by-step explanation:

Circle A has a radius of 3, Circle B has a radius of 9. By dilating Circle a by a scale factor of 3, it will make it the same size as Circle B, showing that the two circles are similar.

Rotating and reflecting does nothing because both circles are in the same quadrant and rotating a circle does nothing to the placement.

Translating the circle (x+3, y-2) does not put it in the same place as circle B.

Tanzania [10]2 years ago
7 0

Answer:

The other person is correct. B: Dilate circle A by a scale factor of 3.

Step-by-step explanation:

i took the test

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Which two tables represent the same function?
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Answer:

The 1st and the 5th tables represent the same function

Step-by-step explanation:

* Lets explain how to solve the problem

- There are five tables of functions, two of them are equal

- To find the two equal function lets find their equations

- The form of the equation of a line whose endpoints are (x1 , y1) and

  (x2 , y2) is \frac{y-y_{1}}{x-x_{1}}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}

* Lets make the equation of each table

# (x1 , y1) = (4 , 8) and (x2 , y2) = (6 , 7)

∵ x1 = 4 , x2 = 6 and y1 = 8 , y2 = 7

∴ \frac{y-8}{x-4}=\frac{7-8}{6-4}

∴ \frac{y-8}{x-4}=\frac{-1}{2}

- By using cross multiplication

∴ 2(y - 8) = -1(x - 4) ⇒ simplify

∴ 2y - 16 = -x + 4 ⇒ add x and 16 for two sides

∴ x + 2y = 20 ⇒ (1)

# (x1 , y1) = (4 , 5) and (x2 , y2) = (6 , 4)

∵ x1 = 4 , x2 = 6 and y1 = 5 , y2 = 4

∴ \frac{y-5}{x-4}=\frac{4-5}{6-4}

∴ \frac{y-5}{x-4}=\frac{-1}{2}

- By using cross multiplication

∴ 2(y - 5) = -1(x - 4) ⇒ simplify

∴ 2y - 10 = -x + 4 ⇒ add x and 10 for two sides

∴ x + 2y = 14 ⇒ (2)

# (x1 , y1) = (2 , 8) and (x2 , y2) = (8 , 5)

∵ x1 = 2 , x2 = 8 and y1 = 8 , y2 = 5

∴ \frac{y-8}{x-2}=\frac{5-8}{8-2}

∴ \frac{y-8}{x-2}=\frac{-3}{6}=====\frac{y-8}{x-2}=\frac{-1}{2}

- By using cross multiplication

∴ 2(y - 8) = -1(x - 2) ⇒ simplify

∴ 2y - 16 = -x + 2 ⇒ add x and 16 for two sides

∴ x + 2y = 18 ⇒ (3)

# (x1 , y1) = (2 , 10) and (x2 , y2) = (6 , 14)

∵ x1 = 2 , x2 = 6 and y1 = 10 , y2 = 14

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∴ \frac{y-10}{x-2}=\frac{4}{4}======\frac{y-10}{x-2}=1

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∴ (y - 10) = (x - 2)

∴ y - 10 = x - 2 ⇒ add 2 and subtract y in the two sides

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∵ x1 = 2 , x2 = 8 and y1 = 9 , y2 = 6

∴ \frac{y-9}{x-2}=\frac{6-9}{8-2}

∴ \frac{y-9}{x-2}=\frac{-3}{6}======\frac{y-9}{x-2}=\frac{-1}{2}

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∴ 2y - 18 = -x + 2 ⇒ add x and 18 for two sides

∴ x + 2y = 20 ⇒ (5)

- Equations (1) and (5) are the same

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