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If the slope of a line is already known what else is needed in order to create an equation?
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).
Which equation describes the line with a slope of 2 and contains the point (4-3)?
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
Is this equation linear or nonlinear y0?
linear (A+)
How do you identify if the given equation is function or not?
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.
What is the value of x in the equation x y 30 when y 15?
If you mean: x+y = 30 and the value of y is 15 then the value of x is also 15
Equation for linear approximation?
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.
What is the equation of sphere?
The equation of a sphere with radius r, centered at (x0 ,y0 ,z0 ) is (x-x0 )+(y-y0 )+(z-z0 )=r2
If the slope of a line is already known what else is needed in order to create an equation?
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).
What is y0 divided by y?
It is simply y0/y.
Is y0 a linear equation?
Hard to tell because of problems with the browser. If the question is about "y = 0" the answer is YES.
How can you find an equation line between two pair of points?
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.
Which equation describes the line with a slope of 2 and contains the point (4-3)?
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
Is this equation linear or nonlinear y0?
linear (A+)
Which equation represents the line that passes through points (1 5) and (3 17)?
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
What is the equation of a circle whose centre is at 3 -5 and passes through the point of 6 -7 on the Cartesian plane?
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
What is the equation of a circle whose centre is at -1 -7 and has a radius of 10 on the Cartesian plane?
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
What is the equation of the circle with center -2 3 and radius r 5?
Equation of a circle with centre (x0, y0) and radius r is (x - x0)^2 + (y - y0)^2 = r^2 → circle is: (x - -2)^2 + (y - 3)^2 = 5^2 → (x + 2)^2 + (y - 3)^2 = 5^2 → x^2 + 4x + 4 + y^2 - 6y + 9 = 25 → x^2 + 4x + y^2 - 6y - 12 = 0
