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VLD [36.1K]
3 years ago
8

A game is played with tokens according to the following rule. In each round, the player with the most tokens gives one token to

each of the other players and also places one token in the discard pile. The game ends when some player runs out of tokens. Evelyn, Amelia and Ava start with 17, 16 and 15 tokens, respectively. How many rounds will there be in the game?​
Mathematics
1 answer:
Lelu [443]3 years ago
6 0

The number of rounds that can be played is given by the product of three

and the least number of tokens a player has at the start.

<h3>Response;</h3>
  • <u>45 rounds</u>

<h3>Reasons for arriving at the above value;</h3>

A round is when the player with the most tokens gives one token to each of the other players, and one placed in the discard pile, we have;

<h3>First three round</h3>

17 - 3 = 14

16 + 1 = 17

15 + 1 = 16

17 - 3 = 14

14 + 1 = 15

16 + 1 = 17

17 - 3 = 14

14 + 1 = 15

15 + 1 = 16

The above values follow a number series such that we have;

<h3>Fourteenth three rounds</h3>

4 - 3 = 1

2 + 1 = 3

3 + 1 = 4

4 - 3 = 1

1 + 1 = 2

3 + 1 = 4

4 - 3 = 1

2 + 1 = 3

1 + 1 = 2

<h3>Fifteenth three rounds</h3>

3 - 3 = 0

1 + 1 = 2

2 + 1 = 3

3 - 3 = 0

0 + 1 = 1

2 + 1 = 3

3 - 3 = 0

0 + 1 = 1

1 + 1 = 2

Therefore, after the fifteenth three rounds, which is 15 × 3 = <u>45 rounds</u> Ava runs out of token, ending the game.

Learn more about number series here:

brainly.com/question/4163549

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Step-by-step explanation:

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Solve the matrix equation for a, b, c, and d. [1 2] [a b] [6 5][3 4] [c d]= [19 8]
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Answer:

The answer is "\bold{\left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}7&-2\\ -\frac{1}{2}&\frac{7}{2}\end{array}\right]}".

Step-by-step explanation:

\bold{\left[\begin{array}{cc}1&2\\3&4\end{array}\right] \left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}6&5\\ 19&8\end{array}\right]}

Solve the L.H.S part:

\left[\begin{array}{cc}1&2\\3&4\end{array}\right] \left[\begin{array}{cc}a&b\\c&d\end{array}\right]\\\\\\\left[\begin{array}{cc}a+2c&b+2d\\3a+4c&3b+4d\end{array}\right]

After calculating the L.H.S part compare the value with R.H.S:

\left[\begin{array}{cc}a+2c&b+2d\\3a+4c&3b+4d\end{array}\right]= \left[\begin{array}{cc}6&5\\ 19&8\end{array}\right]} \\\\

\to a+2c =6....(i)\\\\\to b+2d =5....(ii)\\\\\to 3a+4c =19....(iii)\\\\\to 3b+4d = 8 ....(iv)\\\\

In equation (i) multiply by 3 and subtract by equation (iii):

\to 3a+6c=18\\\to 3a+4c=19\\\\\text{subtract}... \\\\\to 2c = -1\\\\\to  c= - \frac{1}{2}

put the value of c in equation (i):

\to a+ 2 (- \frac{1}{2})=6\\\\\to a- 2 \times \frac{1}{2}=6\\\\\to a- 1=6\\\\\to a =6 +1\\\\\to a = 7\\

In equation (ii) multiply by 3 then subtract by equation (iv):

\to 3b+6d=15\\\to 3b+4d=8\\\\\text{subtract...}\\\\\to 2d = 7\\\\\to d= \frac{7}{2}\\

put the value of d in equation (iv):

\to 3b+4 (\frac{7}{2})=8\\\\\to 3b+4 \times \frac{7}{2}=8\\\\\to 3b+14=8\\\\\to 3b =8-14\\\\\to 3b = -6\\\\\to b= \frac{-6}{3}\\\\\to b= -2

The final answer is "\bold{\left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}7&-2\\ -\frac{1}{2}&\frac{7}{2}\end{array}\right]}".

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