anonymous
  • anonymous
which one is larger
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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.
katieb
  • katieb
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anonymous
  • anonymous
\[\sqrt[3]{60}\] \[2+\sqrt[3]{7}\]no machines allowed
anonymous
  • anonymous
\[2=\sqrt[3]{8}\]
anonymous
  • anonymous
\[60=(\frac{ 7+8 }{ 2 })10\]

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anonymous
  • anonymous
\[\sqrt[3]{60}= \sqrt[3]{5}(\sqrt[3]{7+8})\]
anonymous
  • anonymous
almost simlar now to\[\sqrt[3]{8}+\sqrt[3]{7}\] HALT
anonymous
  • anonymous
!!!!!!
anonymous
  • anonymous
is comparing 60 with \[(2+\sqrt[3]{7})^3\]the same idea(tautology)
anonymous
  • anonymous
i think we have to use inequalities \[(2+\sqrt[3]{7})^3=8+3(4)\sqrt[3]{7}+3(2)(\sqrt[3]{7})^2+7\]
anonymous
  • anonymous
\[\sqrt[3]{1}<\sqrt[3]{7}<\sqrt[3]{8}\]
anonymous
  • anonymous
\[15+12a+6a^2\] and 60
klimenkov
  • klimenkov
Interesting question!
anonymous
  • anonymous
\[a=\sqrt[3]{7}\]
anonymous
  • anonymous
we can allow machines
anonymous
  • anonymous
\[a<2\]
anonymous
  • anonymous
\[\frac{ x+y+z }{ 3 } \leq \sqrt[3]{xyz}\] useful ineq
anonymous
  • anonymous
geometric mean
anonymous
  • anonymous
exhausted
klimenkov
  • klimenkov
Try to do like this. How to evaluate \(\sqrt[3]{7}\). \(1<{7}<8 \Rightarrow 1<\sqrt[3]{7}<2\). As 7 is closer to 8 than to 1, I hope \(\sqrt[3]{7}\) is closer to 2. Lets put \(\sqrt[3]{7}=2+\alpha\). Then \(7=8+12\alpha+6\alpha^2+\alpha^3\). Since \(\alpha <1\) we can neglect \(\alpha^2 ,\alpha^3\). So, \(\alpha=-\frac1{12}\). So the approximation for \(\sqrt[3]{7}\) is \(2-\frac1{12}=\frac{23}{12}\). Do this once again, because it is not a good accuracy for this problem.

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