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
#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)
To solve this: Using long division divide the polynomial by (x - 2) and (x - 3). The result of the final subtraction (which will be some expression involving p and q) will be the same as the remainders. When the divisor is a factor, the remainder is 0. Using the results of the final two subtractions gives two simultaneous equations in p and q which can then be solved. Hint on how to do the long division: As you are dividing by x - 2 and x - 3, at each stage find by what you need to multiply the x which when subtracted will remove the highest power of x remaining. Multiply the whole x-2 or the x-3 by this multiplier and subtract; each subtraction will involve x to some power and x to one less than that power. Now you know the method, have a go before reading the solution below: --------------------------------------------------------------------------------------------------------- Divide x³ + px² + qx + 6 by (x - 2) which is a factor ________________x²_+___(p+2)x_+___(2p+q+4) ______------------------------------------------------ (x-2)_|_x³_+__px²_+_______qx_+___________6 ________x³_+_-2x² ________------------ __________(p+2)x²_+_______qx __________(p+2)x²_+_-2(p+2)x __________------------------------- ___________________(2p+q+4)x_+___________6 ___________________(2p+q+4)x_+_-2(2p+q+4) ___________________-------------------------------- __________________________________4p+2q+14 As (x - 2) is a factor, this final subtraction must result in 0 → 4p + 2q + 14 = 0 → 2p + q + 7 = 0 → 2p + q = -7 Divide x³ + px² + qx + 6 by (x - 3) with remainder 3: ________________x²_+___(p+3)x_+___(3p+q+9) ______------------------------------------------------ (x-3)_|_x³_+__px²_+_______qx_+___________6 _________x³_+_-3x² _________----------- ___________(p+3)x²_+_______qx ___________(p+3)x²_+_-3(p+3)x __________------------------------- ____________________(3p+q+9)x_+___________6 ____________________(3p+q+9)x_+_-3(3p+q+9) ____________________-------------------------------- ___________________________________9p+3q+33 As dividing by (x - 3) leaves a remainder of 3 → 9p + 3q + 33 = 3 → 3p + q + 11 = 1 → 3p + q = -10 There are now two simultaneous equations in p and q which can be solved: 2p + q = -73p + q = -10(2) - (1) gives: 3p - 2p + q - q = -10 - (-1) → p = -3 Substituting in (a) gives: 2×-3 + q = -7 → q = -1 → p = -3 & q = -1