Difference between revisions of "2019 Mock AMC 10B Problems/Problem 3"

Latest revision as of 10:47, 5 November 2019

After trying each option we have $\[\]$ A) $3^\frac{2018}{3}$ which is irrational as 2018 is not divisible by 3 $\[\]$ B) $3^\frac{2019}{2}$ which is irrational as 2019 isn't divisible by 2 $\[\]$ C) $3^2+\sqrt{2}^2+6\sqrt{2}$ which equals $11+6\sqrt{2}$ which is irrational $\[\]$ D) $(2pi)^2$ equals $4pi^2$ , which is irrational $\[\]$ E) $(3-\sqrt{2})(3+\sqrt{2})=9-2=7$ which is rational We have $\boxed{\bold{E}}$ $(3-\sqrt{2})(3+\sqrt{2})$

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 After trying each option we have
-<cmath></cmath>A) <math>3^(2018/3)</math> which isn't rational as 2018 is not divisible by 3<math>
+<cmath></cmath>A) <math>3^\frac{2018}{3}</math> which is irrational as 2018 is not divisible by 3
-<cmath></cmath>B) </math>3^(2019/2)<math> which isn't rational as 2019 isn't divisible by 2</math>
+<cmath></cmath>B) <math>3^\frac{2019}{2}</math> which is irrational as 2019 isn't divisible by 2
-<cmath></cmath>C) <math>3^2+\sqrt{2}^2+6\sqrt{2}</math> which equals <math>11+6\sqrt{2}</math> which is irrational<math>
+<cmath></cmath>C) <math>3^2+\sqrt{2}^2+6\sqrt{2}</math> which equals <math>11+6\sqrt{2}</math> which is irrational
-<cmath></cmath>D) </math>(2pi)^2<math> equals </math>4pi^2<math> which is irrational</math>
+<cmath></cmath>D) <math>(2pi)^2</math> equals <math>4pi^2</math>, which is irrational
-<cmath></cmath>E) <math>(3-\sqrt{2})(3+\sqrt{2})=9-2=7</math> which is rational<math>
+<cmath></cmath>E) <math>(3-\sqrt{2})(3+\sqrt{2})=9-2=7</math> which is rational
-We have </math>\boxed{\bold{E}}<math> </math>(3-\sqrt{2})(3+\sqrt{2})$
+We have <math>\boxed{\bold{E}}</math> <math>(3-\sqrt{2})(3+\sqrt{2})</math>

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Difference between revisions of "2019 Mock AMC 10B Problems/Problem 3"

Latest revision as of 10:47, 5 November 2019