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1 answer:
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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.
Answer:
Step-by-step explanation:
We know that the standard quadratic equation is ax^2+bx+c=0
Let's compare all the given equation to it and , find discriminant.
1. a=2, b= -7, c=-9
So it has 2 real number solutions.
2. a=1, b=-4, c=4
So it has only 1 real number solution.
3. a=4, b=-3, c=-1
So it has 2 real number solutions.
4. a=1, b=-2, c=-8
So it has 2 real number solutions.
5. a=3, b=5, c=3
Thus it does not has real solutions.