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How do you compute quadratic equation using java?
What do you mean by "compute"? Do you want to graph it? Factor it? Calculate it's function given a set of points that lie on it? If you're looking to compute the function given three points that fall on the parabola, then I have just the code for you. If you're given three points, (x1, y1), (x2, y2) and (x3, y3), then you can compute the coefficients of your quadratic equation like this: a = (y1 * (x2 - x3) + y2 * (x3 - x1) + y3 * (x1 - x2)) / (x1 * x1 * (x2 - x3) + x2 * x2 * (x3 - x1) + x3 * x3 * (x1 - x2)) b = (y1 - y2) / (x1 - x2) - a * (x1 + x2); c = y1 - (x1 * x1) * a - x1 * b; You now can calculate the y co-ordinate of any point given it's x co-ordinate by saying: y = a * x * x + b * x + c;
What does the derivative of a function mean?
The derivative of a function is another function that represents the slope of the first function, slope being the limit of delta y over delta x at any two points x1,y1 and x2,y2 on the graph of the function as delta x approaches zero.
How do you determine if an equation represents a linear function?
In general, it is linear if it is of the form a1x1 + a2x2 + ... + anxn + a0 = 0where a1, a2, ... an and a0 are constants, and x1, x2, ... , xn are variable.If there are only two variables then usually x1 = x and x2 = y.
Why does line parallel to the x-axis will intersect the graph of a function once at most if the function has an inverse?
Suppose you have a function y = f(x) which has an inverse. Therefore there exists a function g(y) such that g(y) = x whenever y = f(x). Now suppose a line parallel to the x axis, y = k (some constant), intersects the graph of y = f(x) at more than one point: say x1 and x2. That means that k = f(x1) and k = f(x2). Now, in the context of the function g, this means that [from the first intersection] g(k) = x1 and [from the first intersection] g(k) = x2 But the function g cannot map k to two different points. That is the contradiction which precludes the possibility of a horizontal line intersecting an invertible function more than once.
How matrix X1 X2 equals matrix X1 X2?
i think its pretty much the same thing because matrix X1 X2 IS ACTUALLY X1 X2
Define one-to-one function?
the function (x),sory I can`t use the sign of the function because it is not available. the function of (x)=4x+4 is one to one function assume function(x1)= function(x2) then 4(x1)+4 =4(x2)+4 4(x1)=4(x2) (x1)=(x2) hence,the function is one to one
What is the inverse function of x5?
x1/5
What function formula can you use to add the numbers in Excel in the range X1 to X10?
=SUM(X1:X10)
How do you find the slope of a function?
Need two points. m = slope. (X1, Y1) and (X2, Y2) m = Y2 - Y1/X2 - X1 ==============Or, if function is in this form...... Y =mX + b ======== Read off of function, or get function is this form.
How much is the resistance when the function switch is set to x1?
x1 is single ohms. If the dial reads 40, the resistance is 40 ohms.
Which is the syntax for passing a structure member as an argument to a function?
struct example { int fld1, fld2; } x1; printf ("fld1=%d, fld2=%d\n", x1.fld1, x1.fld2);
How do you pass a structure member as a argument to a function?
struct example {int fld1, fld2;};struct example x1;printf ("fld1=%d, fld2=%d\n", x1.fld1, x1.fld2);
What is the proof of newton raphson iterative equation?
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).
How do you prove decreasing function theorem?
this is the increasing function theorem, hope it helps "If F'(x) >= 0 , and all x's are and element of [a,b], Then F is increasing on [a,b]" use Mean Value Theorem (M.V.T) Let F'(x)>=0 on some interval Let x1< x2 (points from that interval) by M.V.T there is a point C which is an element of [x1,x2] such that F(x2)-F(x1) / X2- X1 = F'(C) this implies: F(x2)-F(x1) = F'(C) X [x2-x1] F'(C)>=0 [x2-x1]>0 therefore: F(x2)>=F(x1) Therefore: F is increasing on that interval.
How do you compute quadratic equation using java?
What do you mean by "compute"? Do you want to graph it? Factor it? Calculate it's function given a set of points that lie on it? If you're looking to compute the function given three points that fall on the parabola, then I have just the code for you. If you're given three points, (x1, y1), (x2, y2) and (x3, y3), then you can compute the coefficients of your quadratic equation like this: a = (y1 * (x2 - x3) + y2 * (x3 - x1) + y3 * (x1 - x2)) / (x1 * x1 * (x2 - x3) + x2 * x2 * (x3 - x1) + x3 * x3 * (x1 - x2)) b = (y1 - y2) / (x1 - x2) - a * (x1 + x2); c = y1 - (x1 * x1) * a - x1 * b; You now can calculate the y co-ordinate of any point given it's x co-ordinate by saying: y = a * x * x + b * x + c;
An equation of a nonlinear function and provide two inputs to evaluate?
y=x2+3 x1=1 x2=2 y(x1) = 1*1+3 = 4 y(x2) = 2*2+3 = 7 x2/x1 = 2, While y2/y1 = 7/4 !=2, and thus the function is nonlinear.
What is the distance formula in 3D?
The formula: distance=sqrt(((x1-x2)*(x1-x2))+((y1-y2)*(y1-y2))+((z1-z2)*(z1-z2))) In DarkBASIC it's: function distance3D(x1,y1,z1,x2,y2,z2) x=x1-x2 y=y1-y2 z=z1-z2 result=sqrt((x*x)+(y*y)+(z*z)) endfunction result In classic BASIC I think it's: FUNCTION distance3D(x1,y1,z1,x2,y2,z2) x=x1-x2 y=y1-y2 z=z1-z2 result=SQRT((x*x)+(y*y)+(z*z)) RETURN result END FUNCTION
