Вопрос № 9
Date: 2015-10-07; view: 526.
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;
| <== previous lecture | | | next lecture ==> |
| ГЕОМЕТРИЯ. ТЕОРИЯ. | | | Вопрос № 42 |