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, D_х¹0, 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