We can only have a right triangle if and only if a^2+b^2 = c^2 is a true equation. The 'c' is the longest side, aka hypotenuse. The legs 'a' and 'b' can be in any order you want.

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For choice A,

a = 2

b = 3

c = sqrt(10)

So,

a^2+b^2 = 2^2+3^2 = 4+9 = 13

but

c^2 = (sqrt(10))^2 = 10

which is not equal to 13 from above. Cross choice A off the list.

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Checking choice B

a = 3

b = 5

c = sqrt(34)

Square each equation

a^2 = 3^2 = 9

b^2 = 5^2 = 25

c^2 = (sqrt(34))^2 = 34

We can see that

a^2+b^2 = 9+25 = 34

which is exactly equal to c^2 above. This confirms the answer.

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Let's check choice C

a = 5, b = 8, c = 12

a^2 = 25, b^2 = 64, c^2 = 144

So,

a^2+b^2 = c^2

25+64 = 144

89 = 144

which is a false equation allowing us to cross choice C off the list.

7 0

3 years ago

PLEASE HELP!!

Ratling [72]

All these cases represent experiments whose outcomes are all equaly likely. What we mean is that the cards have all the same probability of being dealt (1/52), and all the spaces of the spinner have the same probability of being landed into (1/5).

These probabilities are computed as 1 over all the possible outcomes.

(a) If you want a heart card and a five, you want the 5 of hearts. So, only one card would satisfy your request, and thus you have probability 1/52 of dealing the 5 of hearts.

$\dfrac{1}{52}\approx 0.019 = 1.9\%$