9000039103
Level:
C
Assuming \(x\in \mathbb{R}\),
\(y\in \mathbb{R}\), solve
the following equation.
\[
(2 + 5\mathrm{i})x + (1 -\mathrm{i})y = 13\mathrm{i} + 8
\]
\(x = 3,\ y = 2\)
\(x = 13,\ y = 8\)
\(x = 8,\ y = 13\)
\(x = 2,\ y = 3\)
C
9000039104
Level:
C
Assuming \(x\in \mathbb{R}\),
\(y\in \mathbb{R}\), solve
the following equation.
\[
(3 - 2\mathrm{i})x + (5 - 7\mathrm{i})y = 1 + 3\mathrm{i}
\]
\(x = 2,\ y = -1\)
\(x = -1,\ y = 2\)
\(x = -2,\ y = -1\)
\(x = -1,\ y = -2\)
9000039109
Level:
C
Assuming \(z\in \mathbb{C}\),
solve the following equation.
\[
2z -\overline{iz} = 1 -\mathrm{i}
\]
\(z = 1 -\mathrm{i}\)
\(z = 1 + \mathrm{i}\)
\(z = \frac{1}
{3} -\frac{1}
{3}\mathrm{i}\)
\(z = -\frac{1}
{3} + \frac{1}
{3}\mathrm{i}\)
9000039110
Level:
C
Assuming \(z\in \mathbb{C}\),
solve the following equation.
\[
\left (1 + \mathrm{i}\sqrt{3}\right )z = 1 -\mathrm{i}\sqrt{3}
\]
\(z = -\frac{1}
{2} -\frac{\sqrt{3}}
{2} \mathrm{i}\)
\(z = \frac{\sqrt{3}}
{2} + \frac{1}
{2}\mathrm{i}\)
\(z = -\frac{1}
{2} + \frac{\sqrt{3}}
{2} \mathrm{i}\)
\(z = -\frac{\sqrt{3}}
{2} + \frac{1}
{2}\mathrm{i}\)
9000036107
Level:
C
There are three information panels \(A\),
\(B\) and
\(C\) in the park. The direct
distance between \(B\)
and \(C\) is
\(150\, \mathrm{m}\). The visual angle of this
distance from the panel \(A\)
is \(55^{\circ }\). The visual angle
of the distance \(AC\)
from the panel \(B\) is
\(39^{\circ }\). Find the direct distance
between the panels \(A\)
and \(B\)
and round your answer to nearest meters.
\(183\, \mathrm{m}\)
\(147\, \mathrm{m}\)
\(195\, \mathrm{m}\)
\(218\, \mathrm{m}\)
9000036108
Level:
C
The center of a spherical balloon is at a height of \(500\, \mathrm{m}\)
height. The visual angle of the balloon is
\(1^{\circ }30'\). The elevation angle of the
center of the balloon is \(42^{\circ }50'\).
Find the diameter of the balloon in meters and round to nearest one decimal.
\(19.3\, \mathrm{m}\)
\(18.2\, \mathrm{m}\)
\(18.9\, \mathrm{m}\)
\(19.5\, \mathrm{m}\)
9000036109
Level:
C
The point \(A\) is
located \(20\, \mathrm{cm}\) from a
mirror and the point \(B\)
is located \(50\, \mathrm{cm}\)
from the same mirror. The direct distance between
\(A\) and
\(B\) (the length of
the segment \(AB\))
is \(70\, \mathrm{cm}\).
Find the angle of incidence of the ray through the point
\(A\) which is reflected
to the point \(B\)
and round your answer to nearest degrees. (The angle of incidence is the angle
between the incident ray and the normal to the mirror.)
\(42^{\circ }\)
\(37^{\circ }\)
\(38^{\circ }\)
\(48^{\circ }\)
9000036110
Level:
C
A tower is observed from two different places \(A\) and \(B\). The direct distance between \(A\) and \(B\) is \(65\, \mathrm{m}\). If we denote the bottom of the tower by \(C\), we get a triangle \(ABC\) in which the measure of \(\measuredangle CAB \) is \(71^{\circ }\) and the measure of \(\measuredangle ABC \) is \( 34^{\circ }\). From the point \(A\) the angle of elevation of the top of the tower is \(40^{\circ }18'\). Find the height of the tower. Suppose that all \(A\), \(B\) and \(C\) are in the same height above sea level and round your answers to nearest meters.
\(32\, \mathrm{m}\)
\(30\, \mathrm{m}\)
\(35\, \mathrm{m}\)
\(38\, \mathrm{m}\)
9000038701
Level:
C
The box is on the slope as in the picture. The angle of the slope is
\(\alpha \).
The forces acting on the box are the force of gravity
\(\vec{F_{G}}\) and the
friction \(\vec{F_{t}}\).
The force of gravity can be replaced by two components
\(\vec{F_{1}}\) and
\(\vec{F_{n}}\). (The force
\(\vec{F_{1}}\) is parallel to the slope
and \(\vec{F_{n}}\) is perpendicular to
the slope.) The friction \(F_{t}\)
is given by the formula \(F_{t} = fF_{n}\),
where \(f\)
is the coefficient of the friction.What is the influence of the increasing angle
\(\alpha \) on
the forces acting on the box?
\(F_{1}\) becomes
bigger and \(F_{t}\)
becomes smaller
both \(F_{1}\)
and \(F_{t}\)
become smaller
\(F_{1}\) becomes
bigger, \(F_{t}\)
does not change
\(F_{1}\) becomes
smaller, \(F_{t}\)
does not change
both \(F_{1}\)
and \(F_{t}\)
become bigger
\(F_{1}\) becomes
smaller and \(F_{t}\)
becomes bigger
9000038702
Level:
C
The box is on the slope as in the picture. The angle of the slope is
\(\alpha \).
The forces acting on the box are the force of gravity
\(\vec{F_{G}}\) and the
friction \(\vec{F_{t}}\).
The force of gravity can be replaced by two components
\(\vec{F_{1}}\) and
\(\vec{F_{n}}\). (The force
\(\vec{F_{1}}\) is parallel to
the slope and \(\vec{F_{n}}\)
is perpendicular to the slope.) Find \(F_{1}\).
\(F_{1} = F_{G}\sin \alpha \)
\(F_{1} = \frac{F_{G}}
{\sin \alpha } \)
\(F_{1} = F_{G}\mathop{\mathrm{tg}}\nolimits \alpha \)
\(F_{1} = \frac{F_{G}}
{\mathop{\mathrm{tg}}\nolimits \alpha } \)
\(F_{1} = F_{G}\cos \alpha \)
\(F_{1} = \frac{F_{G}}
{\cos \alpha } \)