$\sqrt{2}$
Solve the simultaneous equation
4y+5x=23
2y+2x=10
Find a number which satisfies this equation:
405 (mod
) = 5
$x^{2}$ + $x^{2}$
40%2
8x- $^{3}$
2x- $^{0.5}$
(-95) + (-4)^3 รท (-8) x (-4)
(a+2)(a-4)
0=2xy - 5 - 2y^4e^{-2xy} + \frac{dy}{dx} \left( 3y^2e^{-2xy} - 2y^3xe^{-2xy} + x^2 \right)
53501843/11
