(9,40,41) is a Pythagorean Triple, farther down the list than teachers usually venture.
Answer: D. 41 cm
There's a subset of Pythagorean Triples where the long leg is one less than the hypotenuse,
a^2+b^2 = (b+1)^2
a^2 + b^2 = b^2 + 2b +1
a^2=2b+1
So we get one for every odd number, since the square of an odd number is odd and the square of an even number is even.
b = (a^2 - 1)/2
a=3, b=(3^2-1)/2=4, c=b+1=5
a=5, b=(5^2-1)/2 =12, c = 13
a=7, b=24, c=25
a=9, b=40, c=41
a=11, b=60, c=61
a=13, b=84, c=85
It's good to be able to recognize Pythagorean Triples when we see them.
Otherwise we'd have to work the calculator:
√(9² + 40²) = √1681 = 41
We are given with a triangle and three medians. The intersection of the two medians is also given which is (4,5). What is asked is the intersection between another pair of medians. Since the medians of a triangle intersect at the centroid of a triangle, the intersection is also
<span>B. (4, 5)</span>
Answer:
Step-by-step explanation:
Given that transitor has a 2% defective rate, need to calculate the probability that the 10th transistor produced is the first with a defect.
The probability function is p(x=k) = (1-p)^(n-1) * p
p = (1-defective rate)^(n-1) * defective rate; n=10
p = (1-0.02)^9 * 0.02 = 0.98^9 * 0.02 = 0.01666
