Balkan Mathematical Olympiad

Using the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 1$, $b = -4$, and $c = 3$, we get $x = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(3)}}{2(1)} = \frac{4 \pm \sqrt{16 - 12}}{2} = \frac{4 \pm \sqrt{4}}{2} = \frac{4 \pm 2}{2} = 1$ or $3$. Therefore, the value of $x$ is $1$.

The area of a triangle is given by the formula $A = \frac{1}{2}bh$, where $b$ is the base and $h$ is the height. Substituting the given values, we get $A = \frac{1}{2}(8)(6) = 24$ cm$^2$.

Adding $5$ to both sides of the inequality, we get $2x < 12$. Dividing both sides by $2$, we get $x < 6$. Therefore, the solution to the inequality is $x < 6$.

The sum of the first $n$ positive integers is given by the formula $S_n = \frac{n(n+1)}{2}$. Substituting $n = 100$, we get $S_{100} = \frac{100(101)}{2} = 5050$. Therefore, the sum of the first $100$ positive integers is $5050$.

When flipping a coin, there are two possible outcomes: head or tail. Since both outcomes are equally likely, the probability of getting a head is $1/2 = 0.5$.

The equation of a line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the formula $y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$. Substituting the given points, we get $y - 3 = \frac{7 - 3}{5 - 2}(x - 2) = \frac{4}{3}(x - 2)$. Simplifying, we get $y - 3 = \frac{4}{3}x - \frac{8}{3} = \frac{4}{3}x - \frac{24}{9} = \frac{4}{3}x - 8$. Therefore, the equation of the line is $y = \frac{4}{3}x - 8 + 3 = \frac{4}{3}x - 5$.

The volume of a cube is given by the formula $V = a^3$, where $a$ is the side length. Substituting the given value, we get $V = 4^3 = 64$ cm$^3$.

Rewriting the equation in exponential form, we get $2^4 = x + 3$. Simplifying, we get $16 = x + 3$. Subtracting $3$ from both sides, we get $x = 13$. Therefore, the value of $x$ is $13$.

The derivative of a function $f(x)$ is given by the formula $f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$. Applying this formula to the given function, we get $f'(x) = \lim_{h \to 0} \frac{(x + h)^3 - 2(x + h)^2 + 3(x + h) - 4 - (x^3 - 2x^2 + 3x - 4)}{h} = \lim_{h \to 0} \frac{x^3 + 3x^2h + 3xh^2 + h^3 - 2(x^2 + 2xh + h^2) + 3x + 3h - 4 - x^3 + 2x^2 - 3x + 4}{h} = \lim_{h \to 0} \frac{3x^2h + 3xh^2 + h^3 - 4xh - 4h^2 + 3h}{h} = \lim_{h \to 0} \frac{h(3x^2 + 3xh + h^2 - 4x - 4h + 3)}{h} = \lim_{h \to 0} (3x^2 + 3xh + h^2 - 4x - 4h + 3) = 3x^2 - 4x + 3$.

The equation of a circle with center $(h, k)$ and radius $r$ is given by the formula $(x - h)^2 + (y - k)^2 = r^2$. Substituting the given values, we get $(x - 2)^2 + (y + 3)^2 = 5^2 = 25$. Therefore, the equation of the circle is $(x - 2)^2 + (y + 3)^2 = 25$.

The area of a regular hexagon with a side length $s$ is given by the formula $A = \frac{3\sqrt{3}}{2}s^2$. Substituting the given value, we get $A = \frac{3\sqrt{3}}{2}(6)^2 = 36\sqrt{3}$ cm$^2$.

Using the inverse sine function, we get $x = \sin^{-1}\left(\frac{1}{2}\right) = 30\degree$.

When flipping two coins, there are four possible outcomes: HH, HT, TH, TT. Since each outcome is equally likely, the probability of getting two heads is $1/4 = 0.25$.

The equation of the tangent line to a curve $y = f(x)$ at the point $(x_1, y_1)$ is given by the formula $y - y_1 = f'(x_1)(x - x_1)$. Finding the derivative of the given function, we get $f'(x) = 2x - 2$. Substituting the point $(1, 0)$, we get $f'(1) = 2(1) - 2 = 0$. Therefore, the equation of the tangent line is $y - 0 = 0(x - 1) = -x + 1$. Simplifying, we get $y = -x + 2$.