15+x<24 what is the solution to this inequality

Mathematics

1 answer:

ra1l [238]3 years ago

5 0

Move all terms not containing x to the right side of the inequality. x < 9

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about the second one... well, is a "fait accompli" that using the pythagorean theorem, if x = 8 and y = 5, the hypotenuse must be √(8² + 5²) = √(89), which is neither of those choices.

5, 8, 13 are no dice, namely 5² + 8² ≠ 13

25, 64, 17 is are no dice too, because 25² + 17² ≠ 64²

however, 5,12 and 13 are indeed a pythagorean triple

also is 39, 80, 89.

when looking for a pythagorean triple, recall that c² = a² + b².

so the longest leg is the sum of the square of the small ones.

so what you'd do is, check the small legs, square them, add them up, if they're indeed a pythagorean triple, they "must" add up to the longest leg.

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$\Omega=\{2\spadesuit;\ 3\spadesuit;...;A\spadesuit;2\heartsuit ;\ 3\heartsuit;...;A\heartsuit;\ 2\diamondsuit;\ 3\diamondsuit;...;A\diamondsuit;\ 2\clubsuit;\ 3\clubsuit;...;A\clubsuit\}\\\\|\Omega|=52\\\\A=\{3\spadesuit;\ 4\spadesuit;\ 3\diamondsuit;\ 4\diamondsuit;\ 3\clubsuit;\ 4\clubsuit;\ 2\heartsuit;\ 3\heartsuit;\ 4\heartsuit;...;A\heartsuit\}\\\\|A|=2+2+2+13=19\\\\P(A)=\dfrac{|A|}{|\Omega|}\to P(A)=\dfrac{19}{52}$

$Answer:\ \dfrac{19}{52}$

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