If those letters are variables or unknowns, then it can be simplified to x2 z
logbase5 of x =z x=5^z
5x3y2z3
(x - y)2 - z2 is a difference of two squares (DOTS), those of (x-y) and z. So the factorisation is [(x - y) + z]*[(x - y) - z] = (x - y + z)*(x - y - z)
y varies jointly with x and z if: when x is held fixed, y varies with z and when z is held fixed, y varies with x. Bothe x and z may vary together.
x(z+1) = xz + x
logbase5 of x =z x=5^z
If x = y and y = z then x = z
There are 8 different subsets. The null set. {x} {y} {z} {x y} {x z} {y z} {x y z}
To determine the value of the expression x-a x-b x-c x-z, we need to consider the factors of the expression. Since each letter of the alphabet (excluding x) is subtracted from x in one of the factors, there are a total of 26 factors. This means that the expression must be expanded to include all 26 letters of the alphabet as factors, resulting in x-a x-b x-c ... x-z. Therefore, the value of the expression is x-a x-b x-c ... x-z.
Commutative x + y = y + x x . y = y . x Associative x+(y+z) = (x+y)+z = x+y+z x.(y.z) = (x.y).z = x.y.z Distributive x.(y+z) = x.y + x.z (w+x)(y+z) = wy + xy + wz + xz x + xy = x x + x'y = x + y where, x & y & z are inputs.
if X(Z) is a Z-transform of x[n] and X(Z) is causal then the initial value theorem states that the lim as z tends to infinity for X(Z) must eqaul x(0).
If z = 3, what is 5 x (6 – z)?
x + 1 = y y + 3 = z z = y + 3 = (x + 1) + 3 = x + 4 Or: x = y - 1 = (z - 3) - 1 = z - 4 Which results in the same: x exceeds z by 4.
5x3y2z3
5 x 5 x z x z
z= zany x= ...
(x - y)2 - z2 is a difference of two squares (DOTS), those of (x-y) and z. So the factorisation is [(x - y) + z]*[(x - y) - z] = (x - y + z)*(x - y - z)