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If a (0 0) and b (8 2) what is the length of ab?
The length of ab can be found by using the Pythagorean theorem. The length of ab is equal to the square root of (0-8)^2 + (0-2)^2 which is equal to the square root of 68. Therefore, the length of ab is equal to 8.24.
What is the length of the line segment that connects points 0 0 and 2 4?
√[(2 - 0)2 + (4 - 0)2] = √(4 + 16) = √20 ≈ 4.5 or √20 = √(4 x 5) = 2√5
How do you calculate the top dimensions of a rectangle if you know its perimeter and bottom area?
Important to note are these formulae: Perimeter_of_rectangle = 2 x (length + width) Area_of_rectangle = length x width So if the perimeter and area are known, then: 2 x (length + width) = perimeter => length + width = perimeter / 2 => length = perimeter / 2 - width length x width = area => (perimeter / 2 - width) x width = area (substituting for length given above) => perimeter / 2 x width - width2 = area => width2 - perimeter / 2 x width + area = 0 which is a quadratic and can be solved either by factorization or by using the formula: width = (perimeter / 2 +/- sqrt(perimeter2 / 4 - 4 x area)) / 2 = (perimeter +/- sqrt(perimeter2 - 16 x area)) / 4 This will provide two values for the width. However, each of these values is the length for the other, so the larger value is the length and the smaller value is the width. Sometimes only 1 value will be found for the width above. In this case, the rectangle is actually a square which means that the length and width are both the same. Examples: 1. perimeter = 6, area = 2 width2 - perimeter / 2 x width + area = 0 => width2 - 6 / 2 x width + 2 = 0 => width2 - 3 x width + 2 = 0 => (width - 2) x (width - 1) = 0 => width = 2 or 1. So the length is 2 and the width is 1. 2. perimeter = 12, area = 9 width2 - perimeter / 2 x width + area = 0 => width2 - 12 / 2 x width + 9 = 0 => width2 - 6 x width + 9 = 0 => (width - 3)2 = 0 => width = 3 So the rectangle is a square with both length and width of 3.
What is the length of the tangent line from the coordinate of 9 0 to a point where it touches the circle of x2 plus 8x plus y2 -9 equals 0?
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
The length and width of a rectangle are consecutive odd integers the area of the rectangle is 63 square units what is the length of the rectangle?
Let x = length; then x + 2 = width which is next odd integer x (x+2) = 63 x^2 + 2x - 63 = 0 (X-7)(x+9) = 0 X = 7 = length; width = 9
What is the length of 3 0 0 -6?
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.
How many strings of length 0 that start with 1 bit or end with 2 bits are possible?
There are no zero-length strings that start with 1 bit or end with 2 bits. In a zero-length string, there are no bits at all.
To check if a particular element present in 2 d array?
for(int i = 0; i < [length][] ; i ++){ for(int j = 0; i < [][length]; i++){ if(array[i][j].equals(object)){ return true; } } } return false; Or something..
The length of a rectangle is 2 feet longer than the with what is the dimensions if the area is 8ft?
Designate the width in feet by w. By the statement of the problem, the length is then w+2. The area of any rectangle is the product of the length and the width, or w(w+2) = 8. Multiplying out and collecting all terms on one side yields w2 + 2w - 8 = 0. The polynomial can be factored into (w + 4)(w - 2) = 0, which is true when w = 2 or -4. The length of a physical rectangle can not be negative; therefore, the width is 2 feet and the length 4 feet.
What is the length of the tangent line from the origin to a point where it touches the circle x2 plus y2 plus 4x minus 6y plus 10 equals 0 on the Cartesian plane?
The tangent of a circle is perpendicular to the radius to the point of contact (Xc, Yc).The point (0, 0), the centre of the circle (Xo, Yo) and the point of contact of the tangent (Xc, Yc) form a right angle triangle.The leg from the point (0, 0) to the point of contact (Xc, Yc) is the required lengthThe leg from the centre of the circle (Xo, Yo) to the point of contact (Xc, Yc) has length equal to the radius (r) of the circleThe hypotenuse is the length between the point (0, 0) and the centre of the circle (Xo, Yo).To solve this:Find the centre (Xo, Yo) of the circle, and its radius r.Use Pythagoras to find the length between the point (0, 0) and the centre of the circle (Xo, Yo)Use Pythagoras to find the length between the point (0, 0) and the point of contact (Xc, Yc) of the tangent - the required length.Hint: a circle with centre (Xo, Yo) and radius r has an equation of the form:(x - Xo)² + (y - Yo)² = r²Have a go at solving it now you know how, before reading the solution below:------------------------------------------------------------------------------Circle:x² + y² + 4x - 6y + 10 = 0→ x² + 4x + y² - 6y + 10 = 0→ (x + (4/2))² - (4/2)² + (y + (-6/2))² - (-6/2)² +10 = 0→ (x + 2)² - 4 + (y - 3)² - 9 + 10 = 0→ (x + 2)² + (y - 3)² = 3 = radius²→ Circle has centre (-2, 3) and radius √3Line from centre of circle (-2, 3) to the given point (0, 0):Using Pythagoras to find length of a line between two points (x1, y1) and (x2, y2):length = √((x2 - x1)² + (y2 - y1)²)To find length between given point (0, 0) and centre of circle (-2, 3)→ length = √((0 - -2)² + (0 - 3)²)= √(2² + (-3)²)= √13Tangent line segment:Using Pythagoras to find length of tangent between point (0, 0) and its point of contact with the circle:centre_to_point² = tangent² + radius²→ tangent = √(centre_to_point² - radius²)= √((√13)² + (√3)²)= √(13 + 3)= √16= 4
How do you use an arc length integral to show the length of the circle of radius r?
The integral from 0 to 2 pi of your constant value r will equal the circumference. This will be equal to 2*pi*r. This can be derived because of the following: Arc length = integral from a to b of sqrt(r^2-(dr/dtheta)^2) dtheta. By substituting the equation r = a constant c, dr/dtheta will equal 0, a will equal 0, and b will equal 2pi (the radians in a circle). By substitution, this becomes the integral from 0 to 2 pi of sqrt(c^2 + 0)dtheta, which leads us back to the original formula.
If the midpoint of a horizontal line segment with a length of 8 is 3 -2 then the coordinates of its endpoints are?
If the midpoint of a horizontal line segment with a length of 8 is (3, -2), then the coordinates of its endpoints are (6, -2) and (0, -4).
