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mathmath333

  • one year ago

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} & \{a,b\}\in \mathbb{Z} ,\ \ \ a,b >2,\ a>b\ \hspace{.33em}\\~\\ & \normalsize \text{which of the following will be always true ?} \hspace{.33em}\\~\\ & a.)\ \dfrac{a+b}{2}>\dfrac{a-b}{2} \hspace{.33em}\\~\\ & b.)\ \dfrac{a+b}{2}\geq \dfrac{a-b}{2} \hspace{.33em}\\~\\ & c.)\ a^2-b^2<a^3-b^3\hspace{.33em}\\~\\ & d.)\ \normalsize \text{Both a and c} \hspace{.33em}\\~\\ & e.)\ \normalsize \text{ a, b and c} \hspace{.33em}\\~\\ \end{align}}\)

  2. anonymous
    • one year ago
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    B is confusing, but I thinks it's true The answer is E since we are dealing with positive numbers and bigger than one

  3. mathmath333
    • one year ago
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    but in the book it is given as option d.)

  4. anonymous
    • one year ago
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    Not 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 @Ahmad-nedal 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.

  5. anonymous
    • one year ago
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    for a and b note that \(a+b\) is strictly greater than \(a-b\)

  6. anonymous
    • one year ago
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    I 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

  7. anonymous
    • one year ago
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    for c you can divide both sides of inequality by \(a-b\) which is a positive number and get\[a+b<a^2+ab+b^2\]which is true clearly

  8. mathmath333
    • one year ago
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    is the answer option e.) ?

  9. anonymous
    • one year ago
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    Yes

  10. anonymous
    • one year ago
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    it's d

  11. mathmath333
    • one year ago
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    ok thnx

  12. anonymous
    • one year ago
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    It 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)= \[\frac{ 4+3 }{ 2 }>\frac{ 4-3 }{ 2 }=\frac{ 7 }{ 2 }>\frac{ 1 }{ 2 }\] Similarly at c) \[4^{2}-3^{2}<4^{3}-3^{3}=7<37\] Therefore both a and c are true.

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