To find the approximate length between points A (0, 0) and B (2, 5), you can use the distance formula: ( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ). Plugging in the coordinates, ( d = \sqrt{(2 - 0)^2 + (5 - 0)^2} = \sqrt{4 + 25} = \sqrt{29} ). Therefore, the approximate length between points A and B is about 5.39 units.
The length of the number (04) is 2 digits, as it consists of the digits 0 and 4. Similarly, the length of the number (50) is also 2 digits, comprising the digits 5 and 0. Therefore, both (04) and (50) have the same length of 2.
To find the length of the line segment AB, you can use the distance formula: ( AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ). For points A(0, 0) and B(6, 3), the calculation is ( AB = \sqrt{(6 - 0)^2 + (3 - 0)^2} = \sqrt{6^2 + 3^2} = \sqrt{36 + 9} = \sqrt{45} = 3\sqrt{5} ). Therefore, the length of AB is ( 3\sqrt{5} ).
Points: (5, 0) and (3, 4) and (-5, 0) Equation works out as: x^2+y^2 = 25 Radius: 5 units in length Centre of circle is at the point of origin (0, 0) on the Cartesian plane.
Circle equation: x^2 +8x +y^2 -9 = 0 Completing the square: (x+4)^2 +y^2 = 25 Center of circle: (-4, 0) Radius of circle: 5 Distance from (-4, 0) to (9, 0) = 13 which will be the hypotenuse of a right triangle Length of tangent line using Pythagoras; theorem: 13^2 -5^2 = 144 Therefore length of tangent line is the square root of 144 = 12 units
Distance between (3, 0) and (0, -6) = sqrt[(3 - 0)2 + (0 - -6)2] = sqrt(32 + 62) = sqrt(45) = 3*sqrt(5) or 6.71 approx.
√[(2 - 0)2 + (4 - 0)2] = √(4 + 16) = √20 ≈ 4.5 or √20 = √(4 x 5) = 2√5
The length of the number (04) is 2 digits, as it consists of the digits 0 and 4. Similarly, the length of the number (50) is also 2 digits, comprising the digits 5 and 0. Therefore, both (04) and (50) have the same length of 2.
The approximate radius is five (5) inches.
To find the length of the line segment AB, you can use the distance formula: ( AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ). For points A(0, 0) and B(6, 3), the calculation is ( AB = \sqrt{(6 - 0)^2 + (3 - 0)^2} = \sqrt{6^2 + 3^2} = \sqrt{36 + 9} = \sqrt{45} = 3\sqrt{5} ). Therefore, the length of AB is ( 3\sqrt{5} ).
Points: (5, 0) and (3, 4) and (-5, 0) Equation works out as: x^2+y^2 = 25 Radius: 5 units in length Centre of circle is at the point of origin (0, 0) on the Cartesian plane.
There are (2^5) bit strings of length 5, as each bit can be either 0 or 1. Therefore, the total number of bit strings is (32).
Circle equation: x^2 +8x +y^2 -9 = 0 Completing the square: (x+4)^2 +y^2 = 25 Center of circle: (-4, 0) Radius of circle: 5 Distance from (-4, 0) to (9, 0) = 13 which will be the hypotenuse of a right triangle Length of tangent line using Pythagoras; theorem: 13^2 -5^2 = 144 Therefore length of tangent line is the square root of 144 = 12 units
Equation of (7, 5) = 4x-3y-13 = 0 Intersection of the straight line equations = (4, 1) Length of line is the square root of the sum of (7-4)2+(5-1)2 =25 Therefore: length of the line from (7, 5) perpendicular to 3x+4y-16 is 5 units
Distance between (3, 0) and (0, -6) = sqrt[(3 - 0)2 + (0 - -6)2] = sqrt(32 + 62) = sqrt(45) = 3*sqrt(5) or 6.71 approx.
1 Point of origin: (7, 5) 2 Equation: 3x+4y-16 = 0 3 Perpendicular equation: 4x-3y-13 = 0 4 Both equations intersect at: (4, 1) 5 Line length is the square root of: (7-4)2+(5-1)2 = 5
If: x^2+y^2+4x+6y-40 = 0 and x-y =10 or as x = 10+y Then: (10+y)^2+y^2+4(10+y)+6y-40 = 0 Thus: 100+20y+y^2+y^2+40+4y+6y-40 = 0 Collecting like terms: 2y^2+30y+100 = 0 or as y^2+15y+50 = 0 When factored: (y+5)(y+10) = 0 So: y = -5 or y = -10 By substitution equations intersect at: (0, -10) and (5, -5) Length of line is the square root of: (5-0)^2 plus (-5--10)^2 = square root of 50 Therefore length of line is: square root of 50 or about 7.071 to three decimal places
About 5 meters (16.4 feet) would be a good estimate of a car length. That's the approximate length of a large family car.