Their values work out as: a = -2 and b = 4
The values of p and q work out as -2 and 4 respectively thus complying with the given conditions.
Possible values: a = -2 and b = 9 or a = 5/2 and b = -9 Drawing a sketch on graph paper with the information already given helps.
Perpendicular equation: y = ax+14 Slope of line: 2-6/1-b = -1/a Multiply both sides by 1-b: -4 = -1+b/a By trial and improvement: -4 = -1+9/-2 By trial and improvement: -4 = -1-9/2.5 Therefore: a = -2 and b = 9 or a = 2.5 and b = -9
To satisfy the terms of the given equation the values of 'a' and 'b' are -2 and 4 respectively because:- End points: (-2, 2) and (6, 4) Midpoint: (2, 3) Slope: 1/4 Perpendicular slope: -4 Perpendicular equation: y-3 = -4(x-2) => y = -4x+11 or y+4 = 11
Their values work out as: a = -2 and b = 4
The values of p and q work out as -2 and 4 respectively thus complying with the given conditions.
They must be equidistant from the point of bisection which is their midpoint and works out that a = -2 and b = 4 Sketching the equations on the Cartesian plane will also help you in determining their values
Possible values: a = -2 and b = 9 or a = 5/2 and b = -9 Drawing a sketch on graph paper with the information already given helps.
If the points are (b, 2) and (6, c) then to satisfy the straight line equations it works out that b = -2 and c = 4 which means that the points are (-2, 2) and (6, 4)
Perpendicular equation: y = ax+14 Slope of line: 2-6/1-b = -1/a Multiply both sides by 1-b: -4 = -1+b/a By trial and improvement: -4 = -1+9/-2 By trial and improvement: -4 = -1-9/2.5 Therefore: a = -2 and b = 9 or a = 2.5 and b = -9
To satisfy the terms of the given equation the values of 'a' and 'b' are -2 and 4 respectively because:- End points: (-2, 2) and (6, 4) Midpoint: (2, 3) Slope: 1/4 Perpendicular slope: -4 Perpendicular equation: y-3 = -4(x-2) => y = -4x+11 or y+4 = 11
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7x + 10y = 4.5 : 10y = -7x + 4.5 : y = -x.7/10 + 0.45, the gradient of this line is -7/10 Two straight lines are perpendicular if the product of their gradients is -1. Let the equation for the perpendicular line be y = mx + c Then m x -7/10 = -1 : m = 10/7 The equation for the perpendicular line is y = x.10/7 + c If the values of x and y for the point of intersection are provided then these can be substituted in the perpendicular line equation and the value of c obtained. If appropriate, the equation can then be restructured to a format similar to the original equation.
No, these are of different values.
That depends on what it is that you're looking for the tangent of. One generic technique though would be:Find the derivative of the curve who's tangents you're looking at.Calculate the perpendicular slope to that of the line you're given.Solve that derivative for all values giving you that perpendicular slope.Plug those values into the original curve to find a defining point, and write your line functions with those.
y = x This is a line and a function. Function values are y values.