Answer: Choice B) {3, 5, sqrt(34)}
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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.
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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.
You would do 78/4 which equals 19.5. Hope this helps!
Both the numerator and denominator are continuous at
, which means the quotient rule for limits applies:
Perhaps you meant to write that
instead? In that case, you would have
the answer is 2x2+12z-5 (its also in the picture)
Answer:
for the second questio it is certainly The last choice D
Step-by-step explanation:
