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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\(\large \color{black}{\begin{align} & \{a,b\}\in \mathbb{Z} ,\ \ \ a,b >2,\ a>b\ \hspace{.33em}\\~\\ & \normalsize \text{which of the following will be always true ?} \hspace{.33em}\\~\\ & a.)\ \dfrac{a+b}{2}>\dfrac{ab}{2} \hspace{.33em}\\~\\ & b.)\ \dfrac{a+b}{2}\geq \dfrac{ab}{2} \hspace{.33em}\\~\\ & c.)\ a^2b^2<a^3b^3\hspace{.33em}\\~\\ & d.)\ \normalsize \text{Both a and c} \hspace{.33em}\\~\\ & e.)\ \normalsize \text{ a, b and c} \hspace{.33em}\\~\\ \end{align}}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0B is confusing, but I thinks it's true The answer is E since we are dealing with positive numbers and bigger than one

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0but in the book it is given as option d.)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Not too good at calc but here it goes... What I did was, was I basically picked random numbers that are greater than 2. (I did multiple trials). Trail 1: a = 8 while b = 6 A. True B. True C. True Trial 2: a = 12 while b = 4 A. True B. True C. True So I would have to agree with @Ahmadnedal that the answer is E. However, I am not in calc therefore I don't serve much of a purpose. Perhaps it is d because of the {a,b} E Z.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for a and b note that \(a+b\) is strictly greater than \(ab\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I understand the reason, some books deal with ineaqalities from different aspects. Although the equality stated in part B will not be achieved, but the argument still holds for any a,b in z that satsifies the constraints given. So I insist on my opinion

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for c you can divide both sides of inequality by \(ab\) which is a positive number and get\[a+b<a^2+ab+b^2\]which is true clearly

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0is the answer option e.) ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It is given that a and b are integers greater than 2 and that a is always greater than b. You can choose a number. Let as say a=4 and b=3 For a)= \[\frac{ 4+3 }{ 2 }>\frac{ 43 }{ 2 }=\frac{ 7 }{ 2 }>\frac{ 1 }{ 2 }\] Similarly at c) \[4^{2}3^{2}<4^{3}3^{3}=7<37\] Therefore both a and c are true.
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