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hoa [83]
2 years ago
10

Proving circles are similar: Choose any of the transformations that would be a step in proving that Circle A is similar to Circl

e C. Select ALL that apply.
A.
Reflect A over the line y=x

B.
Translate C (x+5,y-4)

C.
Dilate A by 3/2

D.
Dilate C by 3/2

E.
Translate A by (x-5, y+4)
Mathematics
1 answer:
Ulleksa [173]2 years ago
7 0

The transformations that would prove that circles A and C are similar are:

  • A. Reflect A over the line y=x
  • C. Dilate A by 3/2

<h3>How to prove that circle A and circle C are similar?</h3>

The circles are given as:

Circle A and B

Assume the following parameters:

  • The center of circle A is (2,3) with a radius of 2
  • The center of circle B is (3,2) with a radius of 3

To start with;

The circle A must be reflected across the line y = x with the following transformation rule:

(x,y) -> (y,x)

So, we have:

(2,3) -> (3,2)

Next, the radius of A must be dilated by 3/2 as follows:

New Radius = 3/2 * 2 = 3

After the transformations, we have the following parameters:

  • The center of circle A is (3,2) with a radius of 3
  • The center of circle B is (3,2) with a radius of 3

Notice that both circles now have the same center and radius.

Hence, both circles are similar

Read more about similar circles at:

brainly.com/question/9177979

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Which second degree polynomial function f(x) has a lead coefficient of 3 and roots 4 and 1?
9966 [12]

For this case we have that the following function complies with the given conditions:

f (x) = 3x ^ 2 - 15x + 12

To prove it, let's find the roots of the polynomial:

3x ^ 2 - 15x + 12 = 0

By doing common factor 3 we have:

3 (x ^ 2 - 5x + 4) = 0

Factoring the second degree polynomial we have:

3 (x-1) (x-4) = 0

Then, the solutions are:

Solution 1:

x-1 = 0\\x = 1

Solution 2:

x-4 = 0\\x = 4

Answer:

A second degree polynomial function f (x) that has a lead coefficient of 3 and roots 4 and 1 is:

f (x) = 3x ^ 2 - 15x + 12

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dangina [55]

Answer:

Domain {-2,0,2}

Range {-2,0,2}

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Step-by-step explanation:

We are given a relation:

{ (-2,-2) , (0,0) , (2,2) }

Domain can be defined as the all possible values of x for a relation. It is considered as a set of all first values of the ordered pairs of a given relation.

Domain of the given relation is {-2,0,2}

Range can be defined as all possible value of y which corresponds to the values of x in the domain. It is considered as a set of all second values of the ordered pairs of a given relation.

Range of the given relation is {-2,0,2}

A relation is a function if only there is one value of y for each value of x. If in the set of ordered pair of the relation, the value of x gets repeated, then the relation is not a function.

As no values of x are getting repeated, the relation is a function.

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