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olganol [36]
3 years ago
14

I'M OFFERING 20 POINTS-

Mathematics
2 answers:
DochEvi [55]3 years ago
8 0

she is correct even check myself so it is scale factor 3; enlargement

Dafna11 [192]3 years ago
3 0

Answer:

Third option is correct. Scale factor 3 ; enlargement.

Step-by-step explanation:

It is given that the figure A'B'C'D' is a dilation of figure ABCD.

We know that after dilation the corresponding sides of image and preimage are in the same proportion.

The image of AD is A'D'.

From the figure it is noticed that the A(-1,2), D(-1,-1), A'(-3,6) and D'(-3,-3).

Distance formula is

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

AD=\sqrt{(-1+1)^2+(-1-2)^2}=\sqrt{9}=3

A'D'=\sqrt{(-3+3)^2+(-3-6)^2}=\sqrt{81}=9

Scale factor is constant which represents the relation between image and preimage.

k=\frac{A'D'}{AD}

k=\frac{9}{3}

k=3

Therefore the scale factor is 3.

If k>0 it means enlargement and if k<0 it means reduction. Therefore third option is correct.

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Hey there :)

We know the distance formula is:
D = \sqrt{(x2-x1)^2+(y2-y1)^2}

1) Coordinates are: ( -10, -5 ) , ( 8 , 5 )
                                    ↓   ↓       ↓    ↓
                                    x₁ y₁     x₂   y₂
We plug in the values 
D =\sqrt{(8 -(-10))^2+(5-(-5))^2}
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We do the same for the rest

2) Coordinates are ( 10 , 0 ) , ( -2 , 1 )
D = \sqrt{(-2-10)^2+(1-0)^2}
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3) Coordinates are ( 4 , 8 ) , ( 7 , -8 )
D = \sqrt{((7-4)^2+(-8-8)^2}
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4) Coordinates are ( -10, -4 ) , ( 8 , 8 )
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5) Coordinates are ( 6 , -3 ) , ( -8 , 9 )
D = \sqrt{(-8-6)^2 + (9-(-3))^2}
   ≈ 18.4 (nearest tenth) 

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