Which of the following sets could be the sides of a right triangle? a. {2, 3, share root 10 } b. {3, 5, square root 34} c. {5, 8, 12}

Choice B) {3, 5, sqrt(34)}

Explanation:

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.

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.

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.

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.

i = 2 / - 5

step-by-step explanation:

evaluation of ( 2 - 4 ) ( p + q ) when p = 2 and q = 5 i, by putting values of p and q,                                                                    

step-by-step explanation: