Parallel to axis ОХ.Вопрос № 9
Solve the system if equations:
(-4; 2; -1).
(4; -2; 1).
(4; -1; 2).
(0; 2; 1).
(-1; 2; 0).
Вопрос № 10
Compute and determine the 2nd order: 
17.
-18.
0.
10.
-10.
Вопрос № 11
The equation of a circumference of radius R = 5 with center at the origin:
х2+у2=25.
х2+у2=16.
у=kx+b.
(х-а)2+(у-b)2=r2.

Вопрос № 12
The equation of a circumference of radius R = 7 with the coordinates of the center: the abscissa a = 3, the ordinate b=-2:
(х-3)2+(у+2)2=49.
х2+у2=16.
у=kx+b.
(х-а)2+(у-b)2=r2.
х2+у2=25.
Вопрос № 13
The point of intersection of the circumference (х-4)2+у2=25:
М(0; 3)
М(-2; -4).
М(-2; -4).
М(0; -4).
М(0; 0).
Вопрос № 14
Coordinates of a center and radius R of the circumference (х-2)2+(у+4)2=25:
О(2;-4); R=5.
О(0; 0); R=5.
О(2;-4); R=25.
О(2;4); R=25.
О(-2;4); R=5.
Вопрос № 15
Coordinates of a center and radius of the circumference х2+у2-25=0:
О(0; 0); R=5.
О(2;-4); R=5.
О(2;-4); R=25.
О(2;4); R=25.
О(-2;4); R=5.
Вопрос № 16
Show the equation of circumference, where the center is situated in point (2;-3) and circumference passes through the point (5;1):
(х-2)2+(у+3)2=25.
х2+у2=16.
у=kx+b.
(х-а)2+(у-b)2=r2.
(х-3)2+(у+2)2=49.
Вопрос № 17
The distance between centers of the circumferences х2+у2=16 and (х+3)2+(у+4)2=25:
5.
4.
3.
25.
16.
Вопрос № 18
Abscissa of the circumference’s point х2+(у+4)2=41 and the point on it with ordinate equals zero:
5.
4.
3.
25.
16.
Вопрос № 19
The curve, specified by equalization (х-а)2+(у-b)2=r2:
Circumference.
Parabola.
Ellipse.
Hyperbola.
Straight line.
Вопрос № 20
Ordinate of the circumference’s point (х+3)2+у2=25, where abscissa equals zero:
4.
5.
3.
25.
16.
Вопрос № 21
A canonical equalization of the ellipse:

(х-а)2+(у-b)2=r2.
у=kx+b.
х2+у2=16.

Вопрос № 22
The curve, set by the equalization :
Ellipse.
Circumference.
Parabola.
Hyperbola.
Straight line.
Вопрос № 23
The point of intersection the hyperbola х2-4у2=16 with the axis of abscissas:
М( ±4; 0).
М( -5; 1).
М( ±5; 0).
М( ±6; 0).
М( ±7; 0).
Вопрос № 24
Coordinates the point М, hyperbola х2-9у2=16 with the ordinate, equals 1:
М( ±5; 1).
М( ±4; 0).
М( ±5; 5).
М( 0; 0).
М( ±5; 25).
Вопрос № 25
Canonical type of hyperbola 64х2-25у2=1600:





Вопрос № 26
Canonical type of the ellipse 9х2+25у2=225:





Вопрос № 27
Equalizations of asymptotes of the hyperbola :

, c>a.
, c<a.
.
.
Вопрос № 28
Equalizations of asymptotes of the hyperbola :
.
.
, c<a.
, c<a.
.
Вопрос № 29
Describe the distance d from origin coordinates to point М(х;у):
;
;
;
;
;
Вопрос № 30
The distance d from origin coordinates to point М(-3; 4):
5;
25;
1;
-7;
-12;
Вопрос № 31
The distance between two points М1(х1;у1)и М2(х2;у2):





Вопрос № 32
The distance between two points М1(8; 3)и М2(0; -3):
10.
0.
11.
100.
-11.
Вопрос № 33
Length of the cutoff АВ with the coordinates А(х1;у1) and В(х2;у2):





Вопрос № 34
Length of the cutoff АВ with the coordinates А(2; 4)и В(5;8):
5;
25;
1;
-7;
-12;
Вопрос № 35
A triangle set by the coordinates of its apices А(1; 1), В(4;1), С(1;5). Length of the side АВ equals:
3;
25;
1;
-7;
-12;
Вопрос № 36
Coordinates of the interval’s midpoint АВ, А(х1;у1)and В(х2;у2):
.
.
.
.
.
Вопрос № 37
Coordinates of the interval’s midpoint АВ, А(1;-1)и В(5;9):
(3; 4).
(1;-1).
(5; 9).
(3; 4).
(6; 8).
Вопрос № 38
A rectangle prescribed by coordinates of its apices А(1; 1), В(3;1), С(1;5). Coordinates of the eg’s midpoint АС:
М(2; 1), N(2;3), P(1;3).
М(1; 1), N(2;3), P(1;5).
М(2; 2), N(3;3), P(1;3).
М(1; 1), N(2;3), P(1;3).
М(2; 1), N(3;1), P(1;5).
Вопрос № 39
A rectangle prescribed by coordinates of its apices А(1; 1), В(8;-5), С(3;5).Point М the midpoint of the leg АС. Length of the median ВМ equals:
10;
6;
7;
8;
9;
Вопрос № 40
Disposition of straight Ах+Ву+С=0, if В=0, С 0:
parallel to axis ОХ;
axis ОХ;
parallel to axis ОУ;
axis ОУ;
passes through the origin coordinates.
Вопрос № 41
Angular coefficient of the straight 2,5у-5х+5=0:
2;
2,5;
-2;
-2,5;
5;
Вопрос № 42
Disposition of stright Ах+Ву+С=0, если А=0, С 0:
parallel to axis ОХ;
axis ОХ;
parallel to axis ОУ;
axis ОУ;
passes through the origin coordinates;
Вопрос № 43
А(2; -3) and В(4;3). Coordinates of the point, divides the interval АВ in two:
(3;0).
(2; -3).
(4; 3).
(-2; 3).
(6; -3).
Вопрос № 44
Equalization of parabola, that symmetrical relative to the axis coordinates:





Вопрос № 45
Equalization of parabola, symmetrical relatively to axis of abscissas:





Вопрос № 46
Equalization of parabola directrix:





Вопрос № 47
Size of р parabola is calling:
Parameter.
Ordinate.
Abscissa.
Focus.
Directrix.
Вопрос № 48
Coordinates of focus F parabola:

и 



Вопрос № 49
Coordinates of focus F parabola :





Вопрос № 50
Equalization of directrix parabola :





Вопрос № 51
Canonical equalization of hyperbola, where а=5, в=8:





Вопрос № 52
Canonical equalization of ellipse, if а=7, b=5:





Вопрос № 53
Canonical equalization, and distance between the focuses equal 8 and small axle b=3:





Вопрос № 54
Canonical equalization of ellipse, where large axle а=6 , concentricity. =0,5:





Вопрос № 55
What the dimension of matrix С=АВ, if А(m x k), В(k x n):
(m x n).
(k x n).
(n x m).
(m x k).
(n x n).
Вопрос № 56
What the dimension of matrix С=АВ, if А(2 x 3), В(3 x 4):
(2 x 4).
(2 x 3).
(2 x 2).
(3 x 4).
(4 x 3).
Вопрос № 57
What the dimension of matrix С=АВ, if А(3 x 4), В(4 x 1):
(3 x 1).
(3 x 4).
(4 x 4).
(3 x 3).
(1 x 3).
Вопрос № 58
What the dimension of matrix С=АВ, if А(2 x 4), В(4 x 2):
(2 x 2).
(4 x 2).
(2 x 8).
(2 x 4).
(4 x 4).
Вопрос № 59
Find an element С23 matrix С=АВ, if , :
10.
5.
-10.
-5.
-9.
Вопрос № 60
Find an element С12 matrix С=АВ, if , :
-5.
5 .
10.
-10.
-9.
Вопрос № 61
Find an element С33 matrix С=АВ, if , :
2.
-5.
-10.
10.
5.
Вопрос № 62
If the determiner square matrix equals zero, she called:
Singular.
Nonsingular.
Unit.
Inverse.
Diagonal.
Вопрос № 63
If the determiner if square matrix is not equal zero, she called:
Singular.
Nonsingular.
Unit.
Inverse.
Diagonal.
Вопрос № 64
The system of linear equalizations is calling compatible, if:
it has only one solution.
it has at least one solution.
It doesn’t have a solution.
The solutions consist from the whole numbers.
The solutions only positive numbers.
Вопрос № 65
n – unknown quantity, m – quantity of the equalization of the system. What kind of condition contribute application the Kramer’s rule?
m = n.
m £ n.
m ³ n .
m < n.
m > n.
Вопрос № 66
Coordinates of focus F parabola :





Вопрос № 67
If the equalization’ system doesn’t have solution, then the system is calling:
Incompatible .
Compatible.
Determinate.
Heterogeneous.
Homogeneous.
Вопрос № 68
If in the Kramer’s method D=0, Dх¹0, then a system:
Incompatible .
Compatible.
Determinate.
Heterogeneous.
Homogeneous.
Вопрос № 69
How is located the straight Ах+Ву+С=0, if В=0, С¹0:
Parallel to axis ОУ.
Axis ОХ.
Parallel to axis ОХ.
Axis ОУ.
Passes through the origin coordinates.
Вопрос № 70
Formula of the calculus of the distance from point М(х1,у1) to the straight Ах+Ву+С=0:





Вопрос № 71
In what meanings а and в these straights ах-2у-1=0, 6х-4у-в=0 are parallel:
а=3, в 2.
а=2, в=2.
а=3, в=2.
а= -3, в 2.
а=6, в 6.
Вопрос № 72
In what meanings а and в these straights у = ах+2, у = 5х - в are parallel:
а= 5, в -2.
а= -1/5, в -2.
а= -1/5, в= -2.
а= -5, в= -2.
а 5, в= -2.
Вопрос № 73
How is located the straight Ах+Ву+С=0, if A=0
Parallel to the axis ОХ.
Coincides with the axis ОХ.
Date: 2015-12-11; view: 1325
|