Difference between revisions of "2019 Mock AMC 10B Problems/Problem 3"
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After trying each option we have | After trying each option we have | ||
| − | <cmath></cmath>A) <math>3^\frac{2018}{3}</math> which | + | <cmath></cmath>A) <math>3^\frac{2018}{3}</math> which is irrational as 2018 is not divisible by 3 |
| − | <cmath></cmath>B) <math>3^\frac{2019}{2}</math> which | + | <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 | <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 | + | <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 | <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})</math> | We have <math>\boxed{\bold{E}}</math> <math>(3-\sqrt{2})(3+\sqrt{2})</math> | ||
Latest revision as of 11:47, 5 November 2019
After trying each option we have
![\[\]](http://latex.artofproblemsolving.com/5/7/f/57f1406b65f9f2e6a365b7356da731780b42cceb.png)
A) 
which is irrational as 2018 is not divisible by 3
B) 
which is irrational as 2019 isn't divisible by 2
C) 
which equals 
which is irrational
D) 
equals 
, which is irrational
E) 
which is rational
We have 
