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Alex777 [14]
3 years ago
7

which postulate or theorem, if any,could be used to prove the triangles congruent? if not enough information is given, write not

enough information

Mathematics
1 answer:
Rus_ich [418]3 years ago
6 0
The ways we can prove the triangles congruent:

SAS (side, angle, side)
ASA (angle, side, angle)
AAS (angle, angle, side) 
SSS (all sides equal)
RHS/HL (Right Angle-Hypotenuse-Side/Hypotenuse-legs)

In these cases:
A. <B = <E
<A = <D
AC=DF

AAS (angle, angle, side) could be used.

B. AB=ED
BC=EF
AC=DF

SSS (all sides equal) could be used.

C.<BEA=<CED (right opposite angles)

As there is only one pair of equal angles we can find,there is not enough information.

D.<B = <E
BC=CE
<ECD < ACE(right opposite angles)

ASA (angle, side, angle) could be used.

Hope it helps!
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At a football stadium, 25% of the fans in attendance were teenagers. If there were 130 teenagers at the football stadium, what w
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Answer:

Total number of people in the stadium = 520

Step-by-step explanation:

Let the total number of people in the stadium = x

Since, 25% of the fans are teenagers, number of teenagers in the stadium

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= \frac{25}{100}\times x

= 0,25x

If the number of teenagers in the stadium = 130

0.25x = 130

x = \frac{130}{0.25}

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Therefore, total number of fans in the stadium = 520.

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Step-by-step explanation:

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3 0
2 years ago
Read 2 more answers
1. State if each scenario involves a permutation or a combination. Then find the number of possibilities:(a) Castel and Joe are
BaLLatris [955]

Answer:

a) This is a permutation, there are 210 ways to do it

b) This is a permutation with 6 possibilities

c) This is a combination with a total of 1081575 possibilities

Step-by-step explanation:

a) Here the order does matter bacause it is not the same that they go to France in the weak trip than they go there for 2 days, as a result, this scenario is a permutation.

You need to pick 3 countries from a total of 7, but the order matters, so the total amount of ways to do this is

{7 \choose 3} * 3! = \frac{7!}{(7-3)!} = 210

b) Here each permutation of 123 will give you a possible (and a different one from the others) lock-combination, so this is again a permutation.

The total amount of ways we have to do the selectio of the lock-combination is the total amount of permutations of a set of 3 elements, in other words, it is 3! = 6.

c) If you only cares about who advances to the finals, then the order doesnt matter here, you just want 8 people out of the 25 available. This is therefore a combination, and the total amount of possible cases are

{25 \choos 8} = 1081575 .

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3 years ago
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