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Sulfer Oxside and Nitrogen oxside
SOx and NOx represent the many different gases obtained by the oxidation of sulfur and nitrogen. Since both sulfur and nitrogen can have a number of valence states, the molecule can continue to be oxidized from: SO, SO2, SO3, and SO4 nitrogen has even more possible oxidation products NO, NO2. N2O (laughing gas) NO3, NO4 , N2O5 et al.
Therefore the use of the variable x is to preclude typing all the possibilities.
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What is - x divided by x squared minus 1?
Do you mean: - x / (x2 - 1)? If so, that's as simplified as you can get. (- x / x2) - 1? If so, that simplifies to (-1/x) - 1. (-x/x)2 - 1? If so, that simplifies to 0. -(x/x)2 - 1? If so, that simplifies to -2. Or perhaps that dash in the question is actually a hyphen, in which case: x / (x2 - 1)? If so, that's also as simplified as you can get. (x / x2) - 1? If so, that simplifies to (1/x) - 1. (x/x)2 - 1? If so, that simplifies to 0.
What 3x 45 and how do you solve for x?
Do you mean 3x = 45 then if so divided both sides by 3 to find the value of x which is 15
How do you factorise x squared minus x 7?
The answer to this depends on what you mean by "x 7" If you mean: x2 -7x, then it can be factored out as x(x - 7) If you mean: x2 - x7, then you can factor it out as: x2(1 - x5) If you mean: x2 - x + 7, then it can not be factored If you mean: (x2 - x)7, then the inner term can be factored, giving you (x[x - 1])7 If you mean something else, then you will need to be more clear with your question.
How do you integrate x power e power x?
{xe^x dx integrate by parts let f(x) = x so f'(x) = 1 and g(x) = e^x so g'(x) = e^x so.. f(x)*g(x) - {(g(x)*f'(x)) dx therefore xe^x - {(e^x * 1) dx so.. xe^x - e^x + C factorize so... (x-1)e^x + C
X squared minus x minus 20?
(X-5)(X+4)=0 so X-5=0 -----X=5 and X=4=0 so X=-4 it factors to (X-5)(X+4) so X=5,-4
