mathmath333
  • mathmath333
Question
Mathematics
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mathmath333
  • mathmath333
Question
Mathematics
katieb
  • katieb
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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mathmath333
  • mathmath333
\(\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
anonymous
  • anonymous
B is confusing, but I thinks it's true The answer is E since we are dealing with positive numbers and bigger than one
mathmath333
  • mathmath333
but in the book it is given as option d.)

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anonymous
  • anonymous
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.
anonymous
  • anonymous
for a and b note that \(a+b\) is strictly greater than \(a-b\)
anonymous
  • anonymous
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
anonymous
  • anonymous
for c you can divide both sides of inequality by \(a-b\) which is a positive number and get\[a+b
mathmath333
  • mathmath333
is the answer option e.) ?
anonymous
  • anonymous
Yes
anonymous
  • anonymous
it's d
mathmath333
  • mathmath333
ok thnx
anonymous
  • anonymous
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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