Without an equality sign it is not an equation and so therefore its value can't be determined.
0
Either a point or the y-intercept.If the y-intercept is known (call it b), and, calling the slope m, the equation is y = mx + b.If a sample point (x0, y0) is known, then the equation is y - y0 = m(x - x0).
The equation of a line with slope m through a point (x0, y0) has equation: y - y0 = m(x - x0) Thus the equation of the line with slope 2 through the point (4, -3) has equation: y - (-3) = 2(x - 4) → y + 3 = 2x - 8 → y = 2x - 11
linear (A+)
A function is an equation (a relation) which has only one y-value for every x-value. If a single x-value has more than one y-value, the equation is no longer called a function.
If you mean: x+y = 30 and the value of y is 15 then the value of x is also 15
The general equation for a linear approximation is f(x) ≈ f(x0) + f'(x0)(x-x0) where f(x0) is the value of the function at x0 and f'(x0) is the derivative at x0. This describes a tangent line used to approximate the function. In higher order functions, the same concept can be applied. f(x,y) ≈ f(x0,y0) + fx(x0,y0)(x-x0) + fy(x0,y0)(y-y0) where f(x0,y0) is the value of the function at (x0,y0), fx(x0,y0) is the partial derivative with respect to x at (x0,y0), and fy(x0,y0) is the partial derivative with respect to y at (x0,y0). This describes a tangent plane used to approximate a surface.
The equation of a sphere with radius r, centered at (x0 ,y0 ,z0 ) is (x-x0 )+(y-y0 )+(z-z0 )=r2
Either a point or the y-intercept.If the y-intercept is known (call it b), and, calling the slope m, the equation is y = mx + b.If a sample point (x0, y0) is known, then the equation is y - y0 = m(x - x0).
It is simply y0/y.
Hard to tell because of problems with the browser. If the question is about "y = 0" the answer is YES.
Assuming you want the equation of the straight line between the two points (x0, y0) and (x1, y1), the equation is: y - y0 = m(x - x0) where m is the gradient between the two points: m = (y1 - y0) ÷ (x1 - x0) Note: if the two x coordinates are equal, that is x0 = x1, then the equation of the line is x = x0.
The equation of a line with slope m through a point (x0, y0) has equation: y - y0 = m(x - x0) Thus the equation of the line with slope 2 through the point (4, -3) has equation: y - (-3) = 2(x - 4) → y + 3 = 2x - 8 → y = 2x - 11
linear (A+)
The general equation of a line through point (x0, y0) with gradient m is given by: y - y0 = m(x - x0) The gradient m between two points (x0, y0) and (yx1, y1) is given by: m = change_in_y/change_in_x = (y1 - y0)/(x1 - x0) → line through points (1, 5) and (3, 17) is given by: y - 5 = ((17 - 5)/(3 - 1))(x - 1) → y - 5 = (12/2)(x - 1) → y - 5 = 6(x - 1) → y - 5 = 6x - 6 → y = 6x - 1
The equation of a circle with centre (x0, y0) and radius r is given by: (x - x0)² + (y - y0)² = r² The circle with centre (3, -5) and passing through the point (6, -7) has radius: use Pythagoras: radius = distance from centre (x0, y0) to point (x1, y1) = √((x1 - x0)² + (y1 - y0)²) → radius = √((6 - 3)² + (-7 - -5)²) = √(3² + (-2)²) = √(9 + 4) = √13 Thus the equation is: (x - 3)² + (y - -5)² = (√13)² → (x - 3)² + (y + 5)² = 13 Which can be expanded to give: x² - 6x + 9 + y² +10y + 25 = 13 → x² - 6x + y² + 10y + 21 = 0
The equation of a circle with centre (x0, y0) and radius r is given by: (x - x0)² + (y - y0)² = r² For the circle with centre (-1, -7) and radius 10 this gives: (x - -1) + (y - -7)² = 10² → (x + 1)² + (y + 7)² = 100 This can be expanded and rearranged to give: x² +2x + y² + 14y - 50 = 0
A circle with centre (x0, y0) and radius r has the equation of:(x -x0)² + (y - y0)² = r²By writing the equation of any circle in this form its centre and radius can be determined.To completely lie within a quadrant, the centre of the circle must be more than r away from the y- and x-axes:In the first quadrant if: x0 > r and y0 > rIn the second quadrant if: x0 < -r and y0 > rIn the third quadrant if: x0 < -r and y0 < -rIn the fourth quadrant if: x0 > r and y0 < -rIf either x0 or y0 (or both) is exactly r away from the y- or x-axis then the circle is on boundary between quadrants, and if either x0 or y0 (or both) is less than r away from the y- or x-axis, then the circle is in more than one boundary.f x0 < r from the y-axis then the circle is in quadrants I and II, or y0 < r from the x-axis then the circle is in quadrants III and IV; if both less than r away from their respective axes, the the circle is in all four quadrants.