Yes.
solve each equation in terms of y and the slope is the value in front of the x term; the slope of one of them is the negative inverse slope of the other p/q = 3/2
If you have a quadratic, which is factored like (x - P)(x - Q) = 0, so P & Q are solutions for x. Multiplying the binomials gives: x2 - Px - Qx + PQ = 0 ---> x2 - (P+Q)x + PQ = 0, so the negative of the sum is the coefficient of the x term, and the product is the constant term (no variable x).
Let P and Q have coordinates (Px, Py) and (Qx, Qy), respectively (here x and y are intended as subscripts) Note that Py = f(Px) and Qy = f(Qx) Slope m = (vertical difference) / (horizontal difference) m = (Qy-Py) / (Qx-Px)
It is px - 6.
Yes. Suppose x divides y then there exist an integer p such that y = px. Suppose y divides z then there exist an integer q such that z = qy. Therefore z = q*px = qp*x Since p and q are integers then pq is an integer and therefore x divides z. That is to say: if x divides y and y divides z, then x divides z.
Prove all x px or all x qx then all x px or qx
'PX' ??? I don't known what 'PX' means. However, in Statisitcs(Maths), when written as 'P(x)' it means the probability of 'x'.
Yes.
Px equals -0.6 plus 60x-1500
solve each equation in terms of y and the slope is the value in front of the x term; the slope of one of them is the negative inverse slope of the other p/q = 3/2
14 over x
#include <limits.h> int x, *px, y, *py; px= &x; py= &y; *px= INT_MIN; *py= INT_MAX;
whath
int x, *px, **ppx, ***pppx; px = &x; ppx = &px; pppx = &ppx; ***pppx= 12345;
If you have a quadratic, which is factored like (x - P)(x - Q) = 0, so P & Q are solutions for x. Multiplying the binomials gives: x2 - Px - Qx + PQ = 0 ---> x2 - (P+Q)x + PQ = 0, so the negative of the sum is the coefficient of the x term, and the product is the constant term (no variable x).
Let P and Q have coordinates (Px, Py) and (Qx, Qy), respectively (here x and y are intended as subscripts) Note that Py = f(Px) and Qy = f(Qx) Slope m = (vertical difference) / (horizontal difference) m = (Qy-Py) / (Qx-Px)