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mathmath333
 one year ago
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mathmath333
 one year ago
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mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0\(\large \color{black}{\begin{align} & \normalsize \text{If }\ \large a^2+b^2+c^2=1\ \hspace{.33em}\\~\\ & \normalsize \text{then which of the following cannot be a value of }\ \large (ab+bc+ac)\ ? \hspace{.33em}\\~\\ & a.)\ 0 \hspace{.33em}\\~\\ & b.)\ \dfrac12 \hspace{.33em}\\~\\ & c.)\ \dfrac{1}{4} \hspace{.33em}\\~\\ & d.)\ 1 \hspace{.33em}\\~\\ \end{align}}\)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2\[(a + b + c)^2 = 1 + 2 k \Rightarrow 2k + 1 \ge 0 \Rightarrow k \ge 1/2\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2\[(a + b + c)^2 = \underbrace{a^2 + b^2 + c^2}_{\large = ~ 1} + 2 \underbrace{(a b + b c + ca)}_{\large = ~k }\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Since LHS is \(\ge 0\), RHS is also \(\ge 0\).

alekos
 one year ago
Best ResponseYou've already chosen the best response.0why would (a+b+c)^2 be an odd number? 2k+1

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0it not the odd number he used the formula \((a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ac)\)
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