2z + 5 - xz
3 + z All variables if they have no visible coefficient have an implied coefficient of 1.
(z + 15)(z + 3)
z2 + 9z + 18 = z2 + 6z + 3z + 18 = z(z+6) + 3(z+6) = (z+6)*(z+3)
z+3 simplified as much as possible
Ten z plus one, or ten z plus ten. That depends on where you had parentheses in the original expression: (10 * z) +1 = 10z + 110 * (z + 1) = 10z + 10
3 + z All variables if they have no visible coefficient have an implied coefficient of 1.
3x + y + z = 63x - y + 2z = 9y + z = 3y + z = 3y = 3 - z (substitute 3 - z for y into the first equation of the system)3x + y + z = 63x + (3 - z) + z = 63x + 3 = 63x = 3x = 1 (substitute 3 - z for y and 1 for x into the second equation of the system)3x - y + 2z = 93(1) - (3 - z) + 2z = 93 - 3 + z + 2z = 93z = 9z = 3 (which yields y = 0)y = 3 - z = 3 - 3 = 0So that solution of the system of the equations is x = 1, y = 0, and z = 3.
int i, x, y, z; x=1; y=5; z=1; for (i=0; i<5; ++i) { printf ("%d %d ", x, y); if (x==3) z=z*(-1); x=x+z; y=y-z; } Output: 1 5 2 4 3 3 4 2 5 1
int x = 1, y = 2, z = 3; int f = x + y + z;
3z2 + 45z + 42 = 3(z2 + 15z + 14) = 3(z + 1)(z + 14).
Z + or - i and z-1
6+z = 3 z = 3-6 z = -3
In Magic: The Gathering, "Grim Downfall" is not a card name. It's possible that you meant "Grim Tutor" or "Doomfall." Please provide more context or specific card names for a detailed combo explanation.
(z + 15)(z + 3)
x = -1, y = 2, z = -3.x - y - z = 03x + y - z = 2x + 2y + z = 0Adding equation 3 to equations 1 and 2 gives new equations 1 & 2:2x + y = 04x + 3y = 2Doubling the [new] equation 1 and subtracting from the [new] equation 2 gives:y = 2Substituting back into [new] equation 1 gives:2x + 2 = 0 ==> x = -1Substituting back into the original equation 1 gives:-1 - 2 - z = 0 ==> z = -3Check by substituting back into original equations:3x + y - z = 3(-1) + (2) - (-3) = -3 + 2 + 3 = 2x + 2y + z = (-1) + 2(2) + (-3) = -1 + 4 - 3 = 0
z = 3.6
z2 + 9z + 18 = z2 + 6z + 3z + 18 = z(z+6) + 3(z+6) = (z+6)*(z+3)