Recent content by WORLD-HEN

why is this a "calculus-algebra" question?

Try sending your paper to some one well known, and if he/she feels that your paper is worth it, theyll help you publish it.

It would be the line perpendicular to the line joining the two centres at a distance of 3 units from the point where the line intersects the circumference of either circle.

solve what?

The exact same way. -5(3-3)=0 (-5)(3) + (-3)(-5) = 0 -15 + (-3)(-5) = 0 -3 x -5 = 15 More generally -a(b-b) = 0 -ab + (-b)(-a) = 0 (-b)(-a) = ba

Sorry, haven't used latex much before

yes it is :)

According to classical theory, the force of gravity would be \frac{G m_1 m_2} {R^2} R being the distance from the centre of gravity. Since you want to find the force of gravity on a body elevated from the surface of the Earth by h ,assuming the Earth is spherical, the total distance...

Hi, me with my really old book again. This time , a novel way of turning expressions into partial fractions. It would be best if I show you the examples in the book : \frac{3x^2 +12x +11} {(x+1)(x+2)(x+3)} To express this fraction in the form \frac{A} {x+1} + \frac{B} {x+2} +...

If I understand correctly, you want me to prove that negative times negative is positive, right? -a(a-a) = 0 This statement is true since (a-a) = 0 Using the distributive law : -a(a-a)=0 -a^2 + (-a X -a) = 0 -a X -a = a^2 So, substituting 4 for a gives -4 X -4 = 4^2...

What if the value of h increases with time. Maybe the universe started classically with h=0 and becomes a quantum universe. crazy idea.

If the singularity is infinitely dense, then the gravitational force of attraction at the singularity must be infinite right? This means that the force of attraction between the quarks is over come(since infinite force is what is neaded to break the force). That means that quarks can exist...

Yup, after you pump up your basics, you are going to enjoy courant. But I am warning you, you might get addicted.

Actually , you can extend it to all polynomials. for a cubic equation with factors a,b,c The second derivative = 2!(a + b + c) for a biquadratic expression with factors a,b,c,d The thirds derivative = 3!(a + b + c + d) Similarly, for a n-nomial the (n-1)th derivative = (n-1)!( a +...

No, it was written in 1940, and printed in 1960