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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: õ2+ó2=25. õ2+ó2=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. õ2+ó2=16. ó=kx+b. (õ-à)2+(ó-b)2=r2. õ2+ó2=25. Âîïðîñ ¹ 13 The point of intersection of the circumference (õ-4)2+ó2=25: Ì(0; 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 õ2+ó2-25=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. õ2+ó2=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 õ2+ó2=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)2+ó2=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. õ2+ó2=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
Âîïðîñ ¹ 28 Equalizations of asymptotes of the hyperbola
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 Ì1(õ1;ó1)è Ì2(õ2;ó2):
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(õ1;ó1) and Â(õ2;ó2):
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(õ1;ó1)and Â(õ2;ó2):
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 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 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:
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.
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(õ1,ó1) 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 à=2, â=2. à=3, â=2. à= -3, â à=6, â Âîïðîñ ¹ 72 In what meanings à and â these straights ó = àõ+2, ó = 5õ - â are parallel: à= 5, â à= -1/5, â à= -1/5, â= -2. à= -5, â= -2. à 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: 1316
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