Everything About Ebay Everything About The Ipod Everything About Golf
|
||||
|
- simultaneous equasion - quadratic rules for equations - 2x squared+3x+4=0 - step by step for solving the quadratic formula - solve two simultaneous quadratic equations - revolving of quadratic equations - how to learn simultaneous equations easily - free quadratic formula solver - quadratic equations mathematics - solving negative simultaneous equations - how to solve quadratic equations y=ax5e2-bx+c - writing an equation into "quadratic form" - basic quadratic projectile equation - "hardest maths questions" - quadradic equation - simulataneous eqations - the hardest quadratic equation - using excel solver for simultaneous equations - math murder mysteries - how to quadratic formula |
Easiest Way Of Working Simultaneous Equations
Solving Quadratic EquationsBut there was more!:I'm looking for the equation of a demand curve, which is usually negative sloping, i.e. y = -x. Well, at least that is how it should look. However, I know that y = -x is a linear equation, and that's not exactly what I need. I guess it could be negatively exponential, but I just don't know. Perhaps it might help to give you a few examples of the points on the curve (points are in x:y format) 1) 1,000:400 I should think that the y-intercept would be around 800. It doesn't matter very much whether the curve touches the x-axis, but I would think that the curve would start approaching it around 8,000 or 9,000. This is actually the curve that I'm trying to find the equation for. Not only that, but then instead of having it in the format of y = ... , I would need it in the format x = ... If you could show me how to find that, I would appreciate it very much. Answer: The curve looks like the top right of a circle as I visualise it. You will find it difficult though to work out the exact equation of the curve. Perhaps this may help put it in the form x=ay^2+bx+c: y=ax^2+bx+c y-c=ax^2+bx You then have to complete the square: y-c=sq( rt(a)x+(b/(2rt(a)) ) - b^2/4a (test it by multiplying it out. I may have made a mistake) y-c+b^2/4a= sq( rt(a)x + (b/2rt(a)) ) rt( y-c+b^2/4a ) =rt(a)x + (b/2rt(a)) rt( y-c+b^2/4a)- (b/2rt(a)) = rt(a)x <rt( y-c+b^2/4a ) - (b/2rt(a)) >/ <rt(a)>= x A bit difficult to follow, so suggest writing it out on paper. Have used sq to mean ^2 at times and obviously rt(x) to mean the square-root of x. May be some mistakes here so apologise in advance if so, but gives the general idea on how to isolate x Problems In The History And Philosophy Of Education: Mathematics[Warning: Long post.] The issue of how to get youth to study math has been ongoing in my lifetime. As soon as I understood the endeavor of education as a problem unto itself, as an object of thought, I learned of math as an important subtopic. How can educators inspire students to a love of math study? I concede that I haven't ~fully~ explored the history of the problem of math education. I don't know when getting students to study math, in the United States or elsewhere, became difficult. error |
|||
|
|
||||