If you mean y = x+5 and x is -1 then y = 4
It is linear. The highest power is 1 (x = x1, y = y1) so it is linear.
This can be done with the equation (x1+x2)/2, (y1+y2)/2 which, when solved, creates a (x,x) solution, or a coordinate pair solution. if you had the points (2,4) and (4,8) you would put x1 (2) plus (+) x2 (4) divided by 2, and 2+4 is 6, and 6/2 is 3, so we know our midpoint x value is 3. Then, we would plug in our 'y' values, so we would have y1 (4) + y2 (8) and 4+8 = 12 and 12/2 is 6, so our solution coordinate ordered pair would be (3,6).
4
(y-y1)=(y2-y1/x2-x1)(x-x1)
The equation for the slope between the points A = (x1, y1) and B = (x2, y2) = (y2 - y1)/(x2 - x1), provided x1 is different from x2. If x1 and x2 are the same then the slope is not defined.
use the formula y-y1=m(x-x1)
The slope of a line between Points 1 and 2 is defined as m = (Y2 - Y1) / (X2 - X1) where coordinates for Point 1 is (X1,Y1) and Point 2 is (X2,Y2). So if the value of Y2-Y1 = 0 then the Slope is zero. If the value of X2 - X1 = zero then the equation is undetermined because you can't solve an equation with a 0 value in the denominator. or m = (Y2 - Y1) / 0
find number of integer solution of X1+x2+x3=24
it equals x1 it equals x1
It is linear. The highest power is 1 (x = x1, y = y1) so it is linear.
An identity is an equation that is always true, for any value of the variable or variables. Here are some examples: x + x = 2x a + b = b + a x1 = x
Suppose you have a differentiable function of x, f(x) and you are seeking the root of f(x): that is, a solution to f(x) = 0.Suppose x1 is the first approximation to the root, and suppose the exact root is at x = x1+h : that is f(x1+h) = 0.Let f'(x) be the derivative of f(x) at x, then, by definition,f'(x1) = limit, as h tends to 0, of {f(x1+h) - f(x1)}/hthen, since f(x1+h) = 0, f'(x1) = -f(x1)/h [approx] or h = -f'(x1)/f(x1) [approx]and so a better estimate of the root is x2 = x1 + h = x1 - f'(x1)/f(x1).
This can be done with the equation (x1+x2)/2, (y1+y2)/2 which, when solved, creates a (x,x) solution, or a coordinate pair solution. if you had the points (2,4) and (4,8) you would put x1 (2) plus (+) x2 (4) divided by 2, and 2+4 is 6, and 6/2 is 3, so we know our midpoint x value is 3. Then, we would plug in our 'y' values, so we would have y1 (4) + y2 (8) and 4+8 = 12 and 12/2 is 6, so our solution coordinate ordered pair would be (3,6).
If this is in the context of finding a root of an equation, the answer is to make some guesses. Find value x1 and x2 such that f(x1) and f(x2) have opposite signs. Then, provided that f is a continuous function over (x1, x2), the bisection method will find its root.
4
(y-y1)=(y2-y1/x2-x1)(x-x1)
Assuming you mean x=1, then it is a solution.