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Kaylis [27]
3 years ago
7

What series of transformations to quadrilateral ABCD map the quadrilateral onto​ quadrilateral EFGH ​ to prove that ABCD≅EFGH ?

Mathematics
1 answer:
iris [78.8K]3 years ago
4 0
For this case we have the following transformation rule:
 (x, y) --------> (x, - y - 2) ----------> (x ', y')
 We are going to apply the following transformation rule for vertex D, for example.
 We have then:
 (4, 1) --------> (4, - 1 - 2) ----------> (4, -3)
 Therefore, the transformation is:
 Reflection on the x axis and vertical displacement 2 units down.
 Answer:
 
a reflection across x-axis, and then a translation 2 units down
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Answer:

Step-by-step explanation:

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8 0
3 years ago
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5m + 6n; m = 2 and n = 4
sp2606 [1]

Answer:

34

Step-by-step explanation:

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Hope this helped and have a good day

6 0
1 year ago
What is the measure of ∠F, to the nearest degree?
juin [17]
Three lines given -- it's a natural for the cos(theta) law. A small hint: I think  the preferred way of doing it is to use the cos(theta) law twice. It will give you a definite answer.

Find G first
g = 6 yd
h = 7 yd
f = 5 yards.

g^2 = h^2 + f^2 - 2*h*f*cos(G)
6^2 = 7^2 + 5^2- 2*7*5*cos(G)
36 = 49 + 25 - 70*Cos(G)
36 = 74 - 70*cos(G)
-48 = - 70 * cos(G) Divide by -70
-38/-70 = cos(G)
0.5429 = cos(G)
cos-1(0.5429) = G
G = 57.12

Now find H
h^2 = g^2 + f^2 - 2*g*f*cos(H)
7^2 = 5^2 + 6^2 - 2*5*6*cos(H)
49 = 25 + 36 - 60cos(H)
49 =61  - 60*cos(H) 
Cos(H) = -12 / - 60
Cos(H) = 0.2
H = cos-1(0.2)
H = 78.46

F  can be found because every triangle has 180 degrees
F + 78.46 + 57.12 = 180
F +  135.58 = 180
F = 180 - 135.58
F = 44.41

A <<<< Answer.
8 0
3 years ago
Read 2 more answers
What is the interquartile range of the data set (WILL MARK BRAINLIEST)
mixas84 [53]

20,25,30,30,31,40,41,49

median=30.5

Q¹= 27.5

Q³=40.5

40.5-27.5= 13 IQR = 13

4 0
2 years ago
Help with 10 and 11 please!
Nady [450]

23/50 * 2 = 46/100

11/20 * 5 = 55/100

Hope this helps! :D

7 0
3 years ago
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