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Helen [10]
2 years ago
9

Quadrilateral H’ is the image of quadrilateral H after a sequence of transformations. If quadrilateral H’ is congruent to quadri

lateral H, which transformations could have been used ?
A dilation by a scale factor of 2 followed by a reflection
A rotation followed by a dilation by a scale factor of 1/2
A reflection,a translation of 1 unit left, and then a dilation by a scale factor of 3
A translation of 6 units down a rotation and then a reflection
Mathematics
2 answers:
Nataly_w [17]2 years ago
8 0
The last one is correct



A translation of 6 units down a rotation and then a reflection






good luck
Klio2033 [76]2 years ago
7 0

Answer: A translation of 6 units down a rotation and then a reflection

Step-by-step explanation:

  • The rigid transformations preserves the side length and angle measure of a figure such that figure doesn't shrink or get enlarger. It creates congruent figures.
  • There are four rigid transformations such as :

a) reflections, b) rotations, c) translations and d) glide reflection.

  • A dilation changes the size of the image when the scale factor is not equal to 1. It does not creates congruent images.

Hence, if quadrilateral H’ is congruent to quadrilateral H, then the transformation must be :

A translation of 6 units down a rotation and then a reflection.

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Please answer the following questions​
Vlada [557]

Step-by-step explanation:

sorry I can only explain as there are no labels to each diagram

The first diagram is single and can solved using triangular formular given as 1/2 ×base × height

A = 1/2 × 5 × 12

A = 30cm^2..

as for the second one...it consist of 2 diagrams which will be solved separately before adding ...it can simply be done using Pythagoras theorem..

To get the smaller part ...out tita is 45degrees while our adjacent is 4 and opposite is x we are to find x which is the height...

using SOH CAH TOA...

WE HAVE TAN45= opp/adj

Tan45= x/ 4

Tan 45 =1 ...so

1 = x/ 4

and x= 4 ...

so...having our height as 4 and base as 4 ..

Area of smaller triangle become 1/2 × 4 × 4

A = 8cm^2 ...

......SOLVING FOR THE SECOND DIAGRAM ..

WE HAVE the height as ( dotted spot + undotted spot ) = 4 + 4 = 8cm

and our base can be gotten from

Tan45 = opp / adj

1 = 8/x ..

x = 8cm ....so the base is 8 and the height is 8

..

The Area becomes 1/2 × 8×8 = 32cm ...

Total area becomes 32cm + 8cm = 40cm^2

5 0
3 years ago
How do you solve equations that contain like terms
trasher [3.6K]
You add or subtract it for instance: 14x+258-52. You have to subtract 52 from 258 because of the symbol (-) in front of it. Since there is no other numbers with variables 14x is left alone. Hope this helps
6 0
3 years ago
Read 2 more answers
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2/15==x/60

Multiplly
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now divide
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6 0
2 years ago
The temperature at 8:00am is 5 degrees Celsius and increases 2 degrees per hour. Does this equal y=2x+5
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y represents the temperature.

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Please consider marking this answer as Brainliest to help me advance.
5 0
3 years ago
Triangle ABC and DEF are congruent. Find x.
miv72 [106K]

Δ ABC ≡ ΔDEF

⇒ ∠DEF = ∠ABC

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<u>Solve x:</u>

x - 8 = 65

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Answer: x = 73°

4 0
3 years ago
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