| ||
Question Number 213650 by MathematicalUser2357 last updated on 12/Nov/24 | ||
![]() | ||
$$\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{pythagorean}\:\mathrm{theorem}\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} ={c}^{\mathrm{2}} \:\mathrm{exist}. \\ $$ | ||
Answered by MathematicalUser2357 last updated on 12/Nov/24 | ||
![]() | ||
$$ \\ $$$$ \\ $$ | ||