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3 years ago

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One day, a king left his castle with a bag of silver coins to wander his kingdom. to the first peasant he met, he gave one-half

of the coins plus two more. A little later, he met another peasant to whom he also gave one-half of his coins plus two more. Walking on, he met a third peasant and again gave half his coins plus two more. FInally, the king went home with two coins left in his bag. How many coins did he have to begin with? Create an algebraic equation. Please explain why.

2 answers:

tresset_1 [31]3 years ago

8 0

Answer:

The king left his castle with 44 silver coins.

Step-by-step explanation:

A king left his castle with a bag of silver coins and he gave his coins to 3 peasants in an order. Finally he went home with two coins left.

To solve this question we calculate it in reverse order.

Finally the king has in his bag = 2 coins

Before it he met third peasant king had =

= (2+2) × 2

= 4×2 = 8

Before the king met third peasant, he had 8 coins.

Before he met Second peasant king had :

= (8+2) × 2

= 10 × 2 = 20

Before he met second peasant king had 20 coins.

Before he met first peasant he had :

= (20+ 2) × 2

= 22 × 2 = 44

Before King met first peasant he had 44 silver coins.

galina1969 [7]3 years ago

6 0

Are there any options on what type of problem u want?

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The result for the given 14 length bit string is-

(a) The bit strings of length 14 which have exactly three 0s is 2184.

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A bit-string would be a binary digit sequence (bits). The size of the value is the amount of bits contained in the sequence.

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The concept used here is permutation;

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To know more about permutation, here

brainly.com/question/12468032

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The correct question is-

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