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Parallel to axis ÎÕ.

Âîïðîñ ¹ 9

Solve the system if equations:

(-4; 2; -1).

(4; -2; 1).

(4; -1; 2).

(0; 2; 1).

(-1; 2; 0).

Âîïðîñ ¹ 10

Compute and determine the 2nd order:

17.

-18.

0.

10.

-10.

Âîïðîñ ¹ 11

The equation of a circumference of radius R = 5 with center at the origin:

õ22=25.

õ22=16.

ó=kx+b.

(õ-à)2+(ó-b)2=r2.

Âîïðîñ ¹ 12

The equation of a circumference of radius R = 7 with the coordinates of the center: the abscissa a = 3, the ordinate b=-2:

(õ-3)2+(ó+2)2=49.

õ22=16.

ó=kx+b.

(õ-à)2+(ó-b)2=r2.

õ22=25.

Âîïðîñ ¹ 13

The point of intersection of the circumference (õ-4)22=25:

Ì(0; 3)

Ì(-2; -4).

Ì(-2; -4).

Ì(0; -4).

Ì(0; 0).

Âîïðîñ ¹ 14

Coordinates of a center and radius R of the circumference (õ-2)2+(ó+4)2=25:

Î(2;-4); R=5.

Î(0; 0); R=5.

Î(2;-4); R=25.

Î(2;4); R=25.

Î(-2;4); R=5.

Âîïðîñ ¹ 15

Coordinates of a center and radius of the circumference õ22-25=0:

Î(0; 0); R=5.

Î(2;-4); R=5.

Î(2;-4); R=25.

Î(2;4); R=25.

Î(-2;4); R=5.

Âîïðîñ ¹ 16

Show the equation of circumference, where the center is situated in point (2;-3) and circumference passes through the point (5;1):

(õ-2)2+(ó+3)2=25.

õ22=16.

ó=kx+b.

(õ-à)2+(ó-b)2=r2.

(õ-3)2+(ó+2)2=49.

Âîïðîñ ¹ 17

The distance between centers of the circumferences õ22=16 and (õ+3)2+(ó+4)2=25:

5.

4.

3.

25.

16.

Âîïðîñ ¹ 18

Abscissa of the circumference’s point õ2+(ó+4)2=41 and the point on it with ordinate equals zero:

5.

4.

3.

25.

16.

Âîïðîñ ¹ 19

The curve, specified by equalization (õ-à)2+(ó-b)2=r2:

Circumference.

Parabola.

Ellipse.

Hyperbola.

Straight line.

Âîïðîñ ¹ 20

Ordinate of the circumference’s point (õ+3)22=25, where abscissa equals zero:

4.

5.

3.

25.

16.

Âîïðîñ ¹ 21

A canonical equalization of the ellipse:

(õ-à)2+(ó-b)2=r2.

ó=kx+b.

õ22=16.

Âîïðîñ ¹ 22

The curve, set by the equalization :

Ellipse.

Circumference.

Parabola.

Hyperbola.

Straight line.

Âîïðîñ ¹ 23

The point of intersection the hyperbola õ2-4ó2=16 with the axis of abscissas:

Ì( ±4; 0).

Ì( -5; 1).

Ì( ±5; 0).

Ì( ±6; 0).

Ì( ±7; 0).

Âîïðîñ ¹ 24

Coordinates the point Ì, hyperbola õ2-9ó2=16 with the ordinate, equals 1:

Ì( ±5; 1).

Ì( ±4; 0).

Ì( ±5; 5).

Ì( 0; 0).

Ì( ±5; 25).

Âîïðîñ ¹ 25

Canonical type of hyperbola 64õ2-25ó2=1600:

Âîïðîñ ¹ 26

Canonical type of the ellipse 9õ2+25ó2=225:

Âîïðîñ ¹ 27

Equalizations of asymptotes of the hyperbola :

, c>a.

, c<a.

.

.

Âîïðîñ ¹ 28

Equalizations of asymptotes of the hyperbola :

.

.

, c<a.

, c<a.

.

Âîïðîñ ¹ 29

Describe the distance d from origin coordinates to point Ì(õ;ó):

;

;

;

;

;

Âîïðîñ ¹ 30

The distance d from origin coordinates to point Ì(-3; 4):

5;

25;

1;

-7;

-12;

Âîïðîñ ¹ 31

The distance between two points Ì111)è Ì222):

Âîïðîñ ¹ 32



The distance between two points Ì1(8; 3)è Ì2(0; -3):

10.

0.

11.

100.

-11.

Âîïðîñ ¹ 33

Length of the cutoff ÀÂ with the coordinates À(õ11) and Â(õ22):

Âîïðîñ ¹ 34

Length of the cutoff ÀÂ with the coordinates À(2; 4)è Â(5;8):

5;

25;

1;

-7;

-12;

Âîïðîñ ¹ 35

A triangle set by the coordinates of its apices À(1; 1), Â(4;1), Ñ(1;5). Length of the side ÀÂ equals:

3;

25;

1;

-7;

-12;

Âîïðîñ ¹ 36

Coordinates of the interval’s midpoint ÀÂ, À(õ11)and Â(õ22):

.

.

.

.

.

Âîïðîñ ¹ 37

Coordinates of the interval’s midpoint ÀÂ, À(1;-1)è Â(5;9):

(3; 4).

(1;-1).

(5; 9).

(3; 4).

(6; 8).

Âîïðîñ ¹ 38

A rectangle prescribed by coordinates of its apices À(1; 1), Â(3;1), Ñ(1;5). Coordinates of the eg’s midpoint ÀÑ:

Ì(2; 1), N(2;3), P(1;3).

Ì(1; 1), N(2;3), P(1;5).

Ì(2; 2), N(3;3), P(1;3).

Ì(1; 1), N(2;3), P(1;3).

Ì(2; 1), N(3;1), P(1;5).

Âîïðîñ ¹ 39

A rectangle prescribed by coordinates of its apices À(1; 1), Â(8;-5), Ñ(3;5).Point Ì the midpoint of the leg ÀÑ. Length of the median ÂÌ equals:

10;

6;

7;

8;

9;

Âîïðîñ ¹ 40

Disposition of straight Àõ+Âó+Ñ=0, if Â=0, Ñ 0:

parallel to axis ÎÕ;

axis ÎÕ;

parallel to axis ÎÓ;

axis ÎÓ;

passes through the origin coordinates.

Âîïðîñ ¹ 41

Angular coefficient of the straight 2,5ó-5õ+5=0:

2;

2,5;

-2;

-2,5;

5;

Âîïðîñ ¹ 42

Disposition of stright Àõ+Âó+Ñ=0, åñëè À=0, Ñ 0:

parallel to axis ÎÕ;

axis ÎÕ;

parallel to axis ÎÓ;

axis ÎÓ;

passes through the origin coordinates;

Âîïðîñ ¹ 43

À(2; -3) and Â(4;3). Coordinates of the point, divides the interval ÀÂ in two:

(3;0).

(2; -3).

(4; 3).

(-2; 3).

(6; -3).

Âîïðîñ ¹ 44

Equalization of parabola, that symmetrical relative to the axis coordinates:

Âîïðîñ ¹ 45

Equalization of parabola, symmetrical relatively to axis of abscissas:

Âîïðîñ ¹ 46

Equalization of parabola directrix:

Âîïðîñ ¹ 47

Size of ð parabola is calling:

Parameter.

Ordinate.

Abscissa.

Focus.

Directrix.

Âîïðîñ ¹ 48

Coordinates of focus F parabola:

è

Âîïðîñ ¹ 49

Coordinates of focus F parabola :

Âîïðîñ ¹ 50

Equalization of directrix parabola :

Âîïðîñ ¹ 51

Canonical equalization of hyperbola, where à=5, â=8:

Âîïðîñ ¹ 52

Canonical equalization of ellipse, if à=7, b=5:

Âîïðîñ ¹ 53

Canonical equalization, and distance between the focuses equal 8 and small axle b=3:

Âîïðîñ ¹ 54

Canonical equalization of ellipse, where large axle à=6 , concentricity. =0,5:

Âîïðîñ ¹ 55

What the dimension of matrix Ñ=ÀÂ, if À(m x k), Â(k x n):

(m x n).

(k x n).

(n x m).

(m x k).

(n x n).

Âîïðîñ ¹ 56

What the dimension of matrix Ñ=ÀÂ, if À(2 x 3), Â(3 x 4):

(2 x 4).

(2 x 3).

(2 x 2).

(3 x 4).

(4 x 3).

Âîïðîñ ¹ 57

What the dimension of matrix Ñ=ÀÂ, if À(3 x 4), Â(4 x 1):

(3 x 1).

(3 x 4).

(4 x 4).

(3 x 3).

(1 x 3).

Âîïðîñ ¹ 58

What the dimension of matrix Ñ=ÀÂ, if À(2 x 4), Â(4 x 2):

(2 x 2).

(4 x 2).

(2 x 8).

(2 x 4).

(4 x 4).

Âîïðîñ ¹ 59

Find an element Ñ23 matrix Ñ=ÀÂ, if , :

10.

5.

-10.

-5.

-9.

Âîïðîñ ¹ 60

Find an element Ñ12 matrix Ñ=ÀÂ, if , :

-5.

5 .

10.

-10.

-9.

Âîïðîñ ¹ 61

Find an element Ñ33 matrix Ñ=ÀÂ, if , :

2.

-5.

-10.

10.

5.

Âîïðîñ ¹ 62

If the determiner square matrix equals zero, she called:

Singular.

Nonsingular.

Unit.

Inverse.

Diagonal.

Âîïðîñ ¹ 63

If the determiner if square matrix is not equal zero, she called:

Singular.

Nonsingular.

Unit.

Inverse.

Diagonal.

Âîïðîñ ¹ 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.

Âîïðîñ ¹ 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.

Âîïðîñ ¹ 66

Coordinates of focus F parabola :

Âîïðîñ ¹ 67

If the equalization’ system doesn’t have solution, then the system is calling:

Incompatible .

Compatible.

Determinate.

Heterogeneous.

Homogeneous.

Âîïðîñ ¹ 68

If in the Kramer’s method D=0, Dõ¹0, then a system:

Incompatible .

Compatible.

Determinate.

Heterogeneous.

Homogeneous.

Âîïðîñ ¹ 69

How is located the straight Àõ+Âó+Ñ=0, if Â=0, ѹ0:

Parallel to axis ÎÓ.

Axis ÎÕ.

Parallel to axis ÎÕ.

Axis ÎÓ.

Passes through the origin coordinates.

Âîïðîñ ¹ 70

Formula of the calculus of the distance from point Ì(õ11) to the straight Àõ+Âó+Ñ=0:

Âîïðîñ ¹ 71

In what meanings à and â these straights àõ-2ó-1=0, 6õ-4ó-â=0 are parallel:

à=3, â 2.

à=2, â=2.

à=3, â=2.

à= -3, â 2.

à=6, â 6.

Âîïðîñ ¹ 72

In what meanings à and â these straights ó = àõ+2, ó = 5õ - â are parallel:

à= 5, â -2.

à= -1/5, â -2.

à= -1/5, â= -2.

à= -5, â= -2.

à 5, â= -2.

Âîïðîñ ¹ 73

How is located the straight Àõ+Âó+Ñ=0, if A=0

Parallel to the axis ÎÕ.

Coincides with the axis ÎÕ.


Date: 2015-12-11; view: 1188


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