C

9000120304

Level: 
C
The side of a regular hexagonal prism \(ABCDEFA'B'C'D'E'F'\) is \(a = 3\, \mathrm{cm}\) and the height \(v = 8\, \mathrm{cm}\). Find the length of the diagonal \(AD'\).
\(10\, \mathrm{cm}\)
\(\sqrt{73}\, \mathrm{cm}\)
\(\sqrt{82}\, \mathrm{cm}\)
\(2\sqrt{8}\, \mathrm{cm}\)
\(2\sqrt{6}\, \mathrm{cm}\)

9000120305

Level: 
C
The side of a regular hexagonal prism \(ABCDEFA'B'C'D'E'F'\) shown in the picture is \(a = 3\, \mathrm{cm}\) and the height is \(v = 8\, \mathrm{cm}\). Find the angle between the diagonal \(AD'\) and the base plane \(ABC\) (round your result to the nearest degree).
\(53^{\circ }\)
\(37^{\circ }\)
\(45^{\circ }\)
\(61^{\circ }\)
\(72^{\circ }\)

9000106807

Level: 
C
Consider the points \(A = [0;5]\), \(B = [6;1]\), \(C = [7;9]\) and the triangle \(ABC\). Find the direction vector of the line which is the perpendicular bisector of the side \(b\) (i.e. the line through the midpoint of the side \(AC\) which is perpendicular to the segment \(AC\)).
\((4;-7)\)
\((7;4)\)
\((7;9)\)
\((7;-9)\)

9000106808

Level: 
C
Consider the points \(A = [0;5]\), \(B = [6;1]\), \(C = [7;9]\) and the triangle \(ABC\). Find the direction vector of the line which is the bisector of the angle \(ACB\) (i.e. the line which splits the internal angle at the point \(C\) into two angles with equal measures).
\((2;3)\)
\((6;-4)\)
\((7;9)\)
\((7;8)\)

9000106805

Level: 
C
Given points \(A = [0;5]\), \(B = [6;1]\), \(C = [7;9]\), find the direction vector of the line passing through the point \(A\) and the midpoint of the segment \(BC\) (i.e. the median of the triangle \(ABC\) through the vertex \(A\)).
\((1;0)\)
\((1;8)\)
\((1;9)\)
\((6.5;5)\)

9000106903

Level: 
C
The motion with a constant acceleration is described by the relation \(s = \frac{1} {2}at^{2}\). Consequently, the graph which shows the distance as a function of time is part of a parabola. Find the directrix of this parabola, if \(a = 4\, \mathrm{m}/\mathrm{s}^{2}\).
\(s = -\frac{1} {8}\)
\(s = -1\)
\(s = \frac{1} {8}\)
\(s = 1\)

9000106901

Level: 
C
A body is thrown at the initial angle \(\alpha = 45^{\circ }\) and the initial velocity \(v_{0} = 10\, \mathrm{m}/\mathrm{s}\). The trajectory of the body is a parabola. Find the equation of this parabola. Hint: The coordinates of the moving body as functions of time are \[ \begin{aligned}x& = v_{0}t\cdot \cos \alpha , & \\y& = v_{0}t\cdot \sin \alpha -\frac{1} {2}gt^{2}. \\ \end{aligned} \] Consider the standard acceleration due to gravity \(g = 10\, \mathrm{m}/\mathrm{s}^{2}\).
\((x - 5)^{2} = -10\cdot (y - 2.5)\)
\((x - 5)^{2} = 10\cdot (y + 2.5)\)
\(x^{2} = -10\cdot (y - 5)\)
\((x - 5)^{2} = -10\cdot (y + 2.5)\)

9000106902

Level: 
C
Consider a planet traveling around the Sun on an elliptic trajectory. In the perihelion (the point where the planet is nearest to the Sun) is the distance from the planet to the Sun \(4.5\, \mathrm{AU}\). The excentricity of the ellipse is \(0.5\, \mathrm{AU}\). Find the equation for the trajectory of the planet. Use the coordinate system with center in the Sun and \(x\)-axis along the major axis of the ellipse.
\(\frac{(x-0.5)^{2}} {25} + \frac{y^{2}} {24.75} = 1\)
\(\frac{x^{2}} {25} + \frac{(y-0.5)^{2}} {24.75} = 1\)
\(\frac{x^{2}} {25} + \frac{y^{2}} {24.75} = 1\)
\(\frac{(x-0.5)^{2}} {24.75} + \frac{y^{2}} {25} = 1\)