When X is 1, regardless of the base p.
p is obviously zero because any number x plus 0 equals itself.
x-a is a factor of the polynomial p(x),if p(a)=0.also,if x-a is a factor of p(x), p(a)=0.
There is no "you" in P = V x I
x2 + 3px + p = 0 x2 + p(3x + 1) = 0 p(3x+1) = -x2 p = -x2/(3x+1) So p can have any value at all. In fact, around x = -1/3, p goes asymptotically to + and - infinity.
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p is obviously zero because any number x plus 0 equals itself.
If p, q, r, ... are the roots of the equations, then (x-p), (x-q), (x-r), etc are the factors (and conversely).
p must equal zero p = 0 (7 x 0) + 8.4 = 8.4 0 + 8.4 = 8.4
p = -log[H+] = 12.4
x-a is a factor of the polynomial p(x),if p(a)=0.also,if x-a is a factor of p(x), p(a)=0.
There is no "you" in P = V x I
p x = 20 x5 - 20 x4 + 24 x2 q x = 4 x2 p = 20 x4 - 20 x3 + 24 x, if xis not equal to 0 p / qx = ( 20 x4 - 20 x3 + 24 x) / (4 x2) = (5 x4 - 5 x3 + 6 x) / x2 So p / qx = 5 x2 - 5 x + 6/x if x is not equal to 0
x2 + 3px + p = 0 x2 + p(3x + 1) = 0 p(3x+1) = -x2 p = -x2/(3x+1) So p can have any value at all. In fact, around x = -1/3, p goes asymptotically to + and - infinity.
P = Cx 'P' = the product 'C' = any integer
p(x)=-log([x])So... pH=-log([H+])pH = 1.4
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5.8