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Viktor [21]
4 years ago
9

In the figure, CD = EF and AB = CE. Complete the statements to prove that AB = DF.

Mathematics
2 answers:
myrzilka [38]4 years ago
5 0

Answer:

answers is c

Step-by-step explanation:

CD=EF                  -----> given

CD+ DE= EF+ DE     -----> add DE to both side

CE = CD+DE      

DF= EF+ DE     

CE=DF     -----> transitive property of equality, since CD+DE = EF+ DE 

Since CE=DF and AB=CE (given), then AB=DF 

viva [34]4 years ago
3 0

Answer:

1. Addition Property of Equality

2. Segment addition

3. Substitution Property of Equality.

4. Transitive Property of Equality.

Step-by-step explanation:

Here, given : CD = EF and AB = CE

To Show:  AB = DF

Now, as given in the steps:

1. CD + DE = EF + DE by the (addition) Property of Equality.

As we have added the EQUAL QUANTITY on both sides of the equality.

2.CE = CD + DE and DF = EF + DE by (segment addition).

As here CE and DF are line segments. And the length of  a

Line Segment = Sum of all its parts in which it is divided.

3.CE = DF by the (Addition, subtraction, substitution, transitive)Property of Equality.

Here, as  we know CE  = CD  + DE

but CD  = EF , so SUBSTITUTE EF in place of CD

⇒ CE = EF + ED  = FD (by substitution) Property of Equality.

4.Given, AB = CE and CE = DF implies AB = DF by the (transitive)Property of Equality.

As, x = y , y = z ⇒ x = z is the TRANSITIVE PROPERTY

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1. Let's review the information provided to us for solving the questions:

Power capacity of the wind farms = 2,500 Megawatts or 2.5 Gigawatts

2. Let's resolve the questions a and b:

Part A

Assuming wind farms typically generate 35​% of their​ capacity, how much​ energy, in​ kilowatt-hours, can the​ region's wind farms generate in one​ year?

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Part B

Given that the average household in the region uses about​ 10,000 kilowatt-hours of energy each​ year, how many households can be powered by these wind​ farms?

For calculating the amount of households we divide the total amount of energy the wind farms can generate (7,665'000,000 kilowatt-hours) and we divide it by the average household consumption (10,000 kilowatt-hours)

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