If a+b+c=11 and a^2 + b^2 + c^2 = 81.. find: ab + bc + ca..
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Answered by
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Given a + b + c = 11 and a^2 + b^2 + c^2 = 81.
We know that :
= > (a + b + c)^2 = (a^2 + b^2 + c^2) + 2(ab + bc + ca)
= > (11)^2 = 81 + 2(ab + bc + ca)
= > 121 = 81 + 2(ab + bc + ca)
= > 121 - 81 = 2(ab + bc + ca)
= > 40 = 2(ab + bc + ca)
= > 40/2 = ab + bc + ca
= > 20 = ab + bc + ca.
Therefore, the value of ab + bc + ca = 20.
Hope it helps!
mumpiroy11p9628w:
I didn't understand the first step
Thats the formula!
how did you get (a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab + bc +ca) ??
It is a formula. Learn the important formulas http://www.math-only-math.com/square-of-a-trinomial.html
can u send the link once more
http://www.math-only-math.com/square-of-a-trinomial.html
ok thnx
its ok!
Answered by
0
Answer:
Given a + b + c = 11 and a^2 + b^2 + c^2 = 81.
We know that :
= > (a + b + c)^2 = (a^2 + b^2 + c^2) + 2(ab + bc + ca)
= > (11)^2 = 81 + 2(ab + bc + ca)
= > 121 = 81 + 2(ab + bc + ca)
= > 121 - 81 = 2(ab + bc + ca)
= > 40 = 2(ab + bc + ca)
= > 40/2 = ab + bc + ca
= > 20 = ab + bc + ca.
Therefore, the value of ab + bc + ca = 20.
please mark brainliest
Step-by-step explanation:
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