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Dmitry [639]
2 years ago
9

Jerome was curious if \triangle ABC△ABCtriangle, A, B, C was congruent to \triangle FED△FEDtriangle, F, E, D, so he tried to map

one triangle onto the other using transformations: Jerome concluded: "It's not possible to map \triangle ABC△ABCtriangle, A, B, C onto \triangle FED△FEDtriangle, F, E, D using a sequence of rigid transformations, so the triangles are not congruent." What error did Jerome make in his conclusion?\
Mathematics
1 answer:
Lunna [17]2 years ago
8 0

Answer: C

Step-by-step explanation: I just did it on Khan Academy.

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You need to do 2y+y+10+50=180. Then you'd combine like terms to make 3y+60=180 then to get rid of the sixty subtract it from 180 to make 3y=120 then divide by three to get y by itself so 120 divided by three is 40 which makes y whisk forty. Hope this helped.
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If 2m = 4x and 2w = 8x, what is m in terms of w?
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What is 6×9×(-5) equal to?
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6*9*(-5) = -270

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Determine which equations below, when combined with the equation 3x-4y=2, will form a system with no solutions. Choose all that
svetoff [14.1K]
Two equations will not have solution if they are parallel and have different y-intercepts. Parallel lines have the same slope. In a slope-intercept form, the equation of the line can be expressed as,       
 
        y = mx + b 

where m is slope and b is the y-intercept.

Given: 3x - 4y = 2
Slope-intercept: y = 3x/4 - 1/2

A. 2y = 1.5x - 2
Slope-intercept: y = 3x/4 - 1

B. 2y = 1.5x - 1
Slope-intercept:  y = 3x/4 - 1/2

C. 3x + 4y = 2
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2 years ago
Find the smallest relation containing the relation {(1, 2), (1, 4), (3, 3), (4, 1)} that is:
professor190 [17]

Answer:

Remember, if B is a set, R is a relation in B and a is related with b (aRb or (a,b))

1. R is reflexive if for each element a∈B, aRa.

2. R is symmetric if satisfies that if aRb then bRa.

3. R is transitive if satisfies that if aRb and bRc then aRc.

Then, our set B is \{1,2,3,4\}.

a) We need to find a relation R reflexive and transitive that contain the relation R1=\{(1, 2), (1, 4), (3, 3), (4, 1)\}

Then, we need:

1. That 1R1, 2R2, 3R3, 4R4 to the relation be reflexive and,

2. Observe that

  • 1R4 and 4R1, then 1 must be related with itself.
  • 4R1 and 1R4, then 4 must be related with itself.
  • 4R1 and 1R2, then 4 must be related with 2.

Therefore \{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(4,1),(4,2)\} is the smallest relation containing the relation R1.

b) We need a new relation symmetric and transitive, then

  • since 1R2, then 2 must be related with 1.
  • since 1R4, 4 must be related with 1.

and the analysis for be transitive is the same that we did in a).

Observe that

  • 1R2 and 2R1, then 1 must be related with itself.
  • 4R1 and 1R4, then 4 must be related with itself.
  • 2R1 and 1R4, then 2 must be related with 4.
  • 4R1 and 1R2, then 4 must be related with 2.
  • 2R4 and 4R2, then 2 must be related with itself

Therefore, the smallest relation containing R1 that is symmetric and transitive is

\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(2,1),(2,4),(3,3),(4,1),(4,2),(4,4)\}

c) We need a new relation reflexive, symmetric and transitive containing R1.

For be reflexive

  • 1 must be related with 1,
  • 2 must be related with 2,
  • 3 must be related with 3,
  • 4 must be related with 4

For be symmetric

  • since 1R2, 2 must be related with 1,
  • since 1R4, 4 must be related with 1.

For be transitive

  • Since 4R1 and 1R2, 4 must be related with 2,
  • since 2R1 and 1R4, 2 must be related with 4.

Then, the smallest relation reflexive, symmetric and transitive containing R1 is

\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(2,1),(2,4),(3,3),(4,1),(4,2),(4,4)\}

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2 years ago
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