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balu736 [363]
4 years ago
10

Octagon ABCDEFGH and its dilation, octagon A'B'C'D'E'F'G'H', are shown on the coordinate plane below:

Mathematics
2 answers:
True [87]4 years ago
4 0
You can look at one of the points and see what happened to their x and y values. I'm going to take a look at point H.

It went from (2, 6) to (1, 3). This is a dilation factor of 1/2 because notice that you have to multiply the values (2, 6) by 1/2 to get (1, 3). So, your answer is C, and not A.
vovikov84 [41]4 years ago
3 0

its not 1/2 - I got it wrong.

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You are working on a proof. You know that A and B are on the perpendicular bisector of CD. What can you conclude? A) AC = BD B)
MatroZZZ [7]
You can only conclude that AC = AD  

Its C
5 0
3 years ago
Help asap!! There’s a photo
tester [92]

Answer:

C

Step-by-step explanation:

The vertex form for a parabola is a(x-h)^2 + k, where h is the x coordinate of the vertex and k is the y coordinate. a represents the amount that the graph has been "squeezed". Therefore, since the coordinates of the vertex of this parabola are (2,4), and it is "squeezed" by a factor of 2, the correct answer is C. Hope this helps!

4 0
3 years ago
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kicyunya [14]
The answer to the question

6 0
4 years ago
A physics exam consists of 9 multiple-choice questions and 6 open-ended problems in which all work must be shown. If an examinee
IrinaVladis [17]

Answer:

720 ways

Step-by-step explanation:

Generally, combination is expressed as;

                                  ^{n} C_{r} = \frac{n!}{r!(n-r)!}

The question consists of 9 multiple-choice questions and examinee must answer 7 of the multiple-choice questions.

                                   ⇒ ⁹C₇ =\frac{9!}{7!(9-7)!}

                                        =\frac{9!}{7!(2)!}

                                        = 36

The question consists of 6 open-ended problems and examinee must answer 3 of the open-ended problems.

                                    ⇒ ⁶C₃ =\frac{6!}{3!(6-3)!}

                                         =\frac{6!}{3!(3)!}

                                         = 20

Combining the two combinations to determine the number of ways the questions and problems be chosen if an examinee must answer 7 of the multiple-choice questions and 3 of the open-ended problem.

                                        ⁹C₇ × ⁶C₃

                                       = 36 × 20

                                       = 720 ways  

3 0
3 years ago
Could yall help me out :) thanks <3
juin [17]

Answer:

it is blank

Step-by-step explanation:

it is blank. its black.we cant see anything

4 0
3 years ago
Read 2 more answers
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