The equation describes a circle with its centre at the origin and radius = √13. Each and every point on that circle is a solution.
Combine the equations together and using the quadratic equation formula it works out that the point of contact is at (5/8, 5/2)
Which of the following is the point-slope equation of the line with a slope equals -4 and a point of -2 3?
(2, -2)
They intersect at the point of: (-3/2, 11/4)
If y = 2x +10 and y^2 = 10x then by forming a single quadratic equation and solving it the point of contact is made at (5/8, 5/2)
no...maximum is 360 degrees
Point of intersection: (5/8, 5/2) Square both sides of the first equation and merge it with the second equation to form a quadratic equation:- 4x2-5x+25/16 = 0 Solving the above using the quadratic equation formula gives x as having two poitive roots of 5/8 and substituting this into the linear equation gives y a value of 5/2.
The equation for a circle of radius r and centre (h, k) is: (x - h)² + (y - k)² = r² If the centre is the origin, the centre point is (0, 0), thus h = k = 0, and this becomes: x² + y² = r²
A linear equation is the equation of a line and that consists of an infnite number of points. What you have, in x = -4 and y = -1/3, is a single point. A single point is not a line and so there cannot be a linear equation (suitable for a line) to represent a point.
Do you mean 0.4y+0.6y = 3y -36? If so then: y = 18
To check this, you substitute the values for x and y into the equation. At (1,2) x equals 1 and y equals 2 so substituting this in we get: x + 2y = 1 + 2(2) = 1 + 4 = 5 And since it equals 5 like in the original equation the point is represented by it.