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Illusion [34]
2 years ago
11

Angela has the following coins in her pocket:

Mathematics
1 answer:
GaryK [48]2 years ago
7 0

Answer:

  • There are 10 different combinations

  • The list of different combinations is:

        (10p, 1p), (10p, 50p), (10p, 2p), (10p, 20p), (1p, 50p), (1p, 2p),

        (1p, 20p), (50p, 2p), (50p, 20p), (2p, 20p)

Explanation:

The possible combinations are:

1. Assuming the first coin is 10p:

  • (10p, 1p)
  • (10p, 50p)
  • (10p, 2p)
  • (10p, 20p)

2. Asuming the first coin is 1p

Do not count (1p, 10p) as it is the same combination as (10p, 1p)

  • (1p, 50p)
  • (1p, 2p)
  • (1p, 20p)

3. Assuming the first coin is 50p:

Do not count (50p, 10p) nor (50p, 1p) as they are the same combinations (10p, 50p) and (1p, 50p) counted earlier:

  • (50p, 2p)
  • (50p, 20p)

4. Assuming the first coin is 2p:

The only new combination is:

  • (2p, 20p)

5. All the combinations with 20p have already been listed.

Therefore:

  • There are 4 + 3 + 2 + 1 = 10 different combinations

  • The list of different combinations is:

        (10p, 1p), (10p, 50p), (10p, 2p), (10p, 20p), (1p, 50p), (1p, 2p),

        (1p, 20p), (50p, 2p), (50p, 20p), (2p, 20p)

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Tan 235° = 2tan20°+ tan215°​
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=> tan 55 = 2tan 20 + tan 35

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=> 20  = 55 - 35

taking Tan both sides

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=> Tan 20  = (Tan55 - Tan35) /(1 + Tan55 . Tan35)

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