Difference between revisions of "2007 CEMC Pascal Problems/Problem 6"
(Created page with "==Problem== The value of <math>\frac{\sqrt{64} + \sqrt{36}}{\sqrt{64 + 36}}</math> is <math> \text{ (A) }\ \frac{7}{5} \qquad\text{ (B) }\ \frac{16}{5} \qquad\text{ (C) }\ \f...") |
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We get <math>\frac{\sqrt{64} + \sqrt{36}}{\sqrt{64 + 36}} = \frac{8 + 6}{\sqrt{100}} = \frac{14}{10}</math> | We get <math>\frac{\sqrt{64} + \sqrt{36}}{\sqrt{64 + 36}} = \frac{8 + 6}{\sqrt{100}} = \frac{14}{10}</math> | ||
| − | Simplifying, we have <math>\frac{14 \div 2}{10 \div 2} = \boxed {\textbf {(A) } frac{7}{5}}</math> | + | Simplifying, we have <math>\frac{14 \div 2}{10 \div 2} = \boxed {\textbf {(A) } \frac{7}{5}}</math> |
~anabel.disher | ~anabel.disher | ||
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<math>\frac{8 + 6}{10}</math> | <math>\frac{8 + 6}{10}</math> | ||
| − | Using the same process as solution 1, we get <math>\boxed {\textbf {(A) } frac{7}{5}}</math>. | + | Using the same process as solution 1, we get <math>\boxed {\textbf {(A) } \frac{7}{5}}</math>. |
~anabel.disher | ~anabel.disher | ||
Latest revision as of 18:46, 7 September 2025
Problem
The value of 
is
Solution 1
We get 
Simplifying, we have 
~anabel.disher
Solution 1.5
We can remember that 
using pythagorean triples. 
and 
, so we have:
Using the same process as solution 1, we get 
.
~anabel.disher
