Question 391127
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \frac{5}{8}\ +\ \frac{1}{4}\ +\ \frac{2}{3}\ =\ \frac{15}{24}\ +\ \frac{6}{24}\ +\ \frac{16}{24}\ =\ \frac{37}{24}]


therefore:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \frac{37}{24}\ +\ m\ +\ n\ =\ 3]


hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ m\ +\ n\ =\ 3\ -\ \frac{37}{24}\ =\ \frac{72}{24}\ -\ \frac{37}{24}\ =\ \frac{35}{24}]


Select any two real number values you like for *[tex \Large m] and *[tex \Large n] so long as their sum is *[tex \Large \frac{35}{24}].


For example, *[tex \Large m\ =\ \frac{5}{4}] and *[tex \Large n\ =\ \frac{5}{24}] would work.


But so would *[tex \Large m\ =\ \frac{2435}{24}] and *[tex \Large n\ =\ -100],



John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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