Question

Can someone help me answer this question?

Answer 1

To be honest I’m answering the question for coins but that looks hard

Answer 2

Answer: 11/30 (choice H)

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Work Shown:

Part 1

`m \ \ddag = (m)/(m^2-m)\\\\6 \ \ddag = (6)/(6^2-6)\\\\6 \ \ddag = (6)/(36-6)\\\\6 \ \ddag = (6)/(30)\\\\6 \ \ddag = (1)/(5)\\\\`

Explanation: The idea is to replace each copy of m with 6, and then simplify. The same idea will be applied when plugging in m = -5 for part 2 shown in the next section. The last section combines the two results.

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Part 2

`m \ \ddag = (m)/(m^2-m)\\\\(-5) \ \ddag = (-5)/((-5)^2-(-5))\\\\(-5) \ \ddag = (-5)/(25+5)\\\\(-5) \ \ddag = (-5)/(30)\\\\(-5) \ \ddag = -(1)/(6)\\\\`

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Part 3

`6 \ \ddag - \left[(-5)\ \ddag\right] = (1)/(5) - \left(-(1)/(6)\right)\\\\6 \ \ddag - \left[(-5)\ \ddag\right] = (1)/(5) + (1)/(6)\\\\6 \ \ddag - \left[(-5)\ \ddag\right] = (6)/(30) + (5)/(30)\\\\6 \ \ddag - \left[(-5)\ \ddag\right] = (6+5)/(30)\\\\6 \ \ddag - \left[(-5)\ \ddag\right] = \boldsymbol{(11)/(30)}\\\\`

The final answer is 11/30 (choice H)