International Mathematical Competition

To solve the equation, we can use the quadratic formula: (x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}). Plugging in the values of (a = 1, b = -5, c = 6), we get: (x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(6)}}{2(1)}) = \frac{5 \pm \sqrt{25 - 24}}{2} = \frac{5 \pm 1}{2} = 2, 3).

To find the area of a triangle, we can use Heron's formula: (Area = \sqrt{s(s-a)(s-b)(s-c)}), where (s) is the semi-perimeter and (a, b, c) are the lengths of the sides. Plugging in the values of (a = 6 cm, b = 8 cm, c = 10 cm), we get: (s = \frac{6 cm + 8 cm + 10 cm}{2} = 12 cm). Therefore, the area of the triangle is: (Area = \sqrt{12 cm(12 cm - 6 cm)(12 cm - 8 cm)(12 cm - 10 cm)} = 36 cm^2).

To find the derivative of the function, we can use the power rule: (\frac{d}{dx}x^n = nx^(n-1)). Applying this rule to each term of the function, we get: (f'(x) = \frac{d}{dx}(x^3 - 2x^2 + 3x - 4) = 3x^2 - 2(2x) + 3(1) - 0 = 3x^2 - 4x + 3).

To find the equation of the line, we can use the point-slope form: (y - y_1 = m(x - x_1)), where ((x_1, y_1)) is a point on the line and (m) is the slope of the line. Using the two given points, we can find the slope: (m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{7 - 3}{5 - 2} = 2). Plugging in the point ((2, 3)) and the slope (m = 2), we get the equation of the line: (y - 3 = 2(x - 2) = 2x - 4). Therefore, the equation of the line is: (y = 2x - 1).

To find the volume of a sphere, we can use the formula: (Volume = \frac{4}{3}\pi r^3). Plugging in the value of (r = 5 cm), we get: (Volume = \frac{4}{3}\pi (5 cm)^3 = \frac{4}{3}\pi (125 cm^3) = 300\pi cm^3).

To solve the differential equation, we can use the method of separation of variables: (\frac{dy}{dx} = 2x + 1) becomes (dy = (2x + 1)dx). Integrating both sides, we get: (\int dy = \int (2x + 1)dx) which gives (y = x^2 + x + C), where (C) is the constant of integration.

The sum of the first 100 positive integers can be found using the formula: (Sum = \frac{n(n+1)}{2}), where (n) is the number of integers. Plugging in (n = 100), we get: (Sum = \frac{100(100+1)}{2} = \frac{100(101)}{2} = 5050).

To find the area of the region, we need to find the points of intersection of the two curves. Setting (y = x^2) and (y = 2x + 1) equal to each other, we get: (x^2 = 2x + 1) which gives (x^2 - 2x - 1 = 0). Solving for (x), we get (x = 1) and (x = -1). Therefore, the area of the region is: (Area = \int_{-1}^{1} (2x + 1 - x^2) dx = \left[x^2 + x - \frac{x^3}{3}\right]_{-1}^{1} = \left[(1)^2 + (1) - \frac{(1)^3}{3}\right] - \left[(-1)^2 + (-1) - \frac{(-1)^3}{3}\right] = \frac{2}{3}) square units.

To find the equation of the plane, we can use the point-normal form: (\vec{r} \cdot \vec{n} = d), where (\vec{r}) is a vector from the origin to any point on the plane, (\vec{n}) is the normal vector, and (d) is a constant. Plugging in the given values, we get: ((x - 1) \cdot 2 + (y - 2) \cdot (-1) + (z - 3) \cdot 3 = d) which simplifies to: (2x - y + 3z = d). Since the plane passes through the point ((1, 2, 3)), we can plug in these values to find the value of (d): (2(1) - (2) + 3(3) = d) which gives (d = 10). Therefore, the equation of the plane is: (2x - y + 3z = 10).

To find the limit, we can use L'Hopital's rule: (\lim_{x \to 0} \frac{\sin(3x)}{x} = \lim_{x \to 0} \frac{\frac{d}{dx}[\sin(3x)]}{\frac{d}{dx}[x]} = \lim_{x \to 0} \frac{3\cos(3x)}{1} = 3\cos(0) = 3).

To solve the differential equation, we can use the method of characteristic equations: (r^2 - 4r + 4 = 0) which gives (r = 2) as the double root. Therefore, the general solution is: (y = c_1 e^{2x} + c_2 e^{-2x}), where (c_1) and (c_2) are constants.

To find the area of the triangle, we can use the formula: (Area = \frac{1}{2}|x_1 y_2 - x_2 y_1|), where ((x_1, y_1)) and ((x_2, y_2)) are the coordinates of two vertices of the triangle. Plugging in the coordinates of the vertices, we get: (Area = \frac{1}{2}|(0)(2) - (0)(-1)| = \frac{1}{2}|0 - 0| = \frac{1}{2}(0) = 0) square units.

The equation of a circle with center ((h, k)) and radius (r) is: ((x - h)^2 + (y - k)^2 = r^2). Plugging in the values of (h = 2, k = -3, r = 5), we get: ((x - 2)^2 + (y + 3)^2 = 5^2 = 25).

To find the volume of the solid, we can use the formula: (Volume = \pi \int_{a}^{b} [f(x)]^2 dx), where (f(x)) is the function that defines the upper boundary of the region and ([a, b]) are the limits of integration. Plugging in the values of (f(x) = 2 - x) and ([a, b] = [0, 2]), we get: (Volume = \pi \int_{0}^{2} (2 - x)^2 dx = \pi \left[2x - \frac{x^3}{3}\right]_{0}^{2} = \pi \left[(2)(2) - \frac{(2)^3}{3} - (2)(0) - \frac{(0)^3}{3}\right] = \frac{16\pi}{3}) cubic units.