-1
1/h-8 = (1/h)-8 = h8
(2h-3)(h+1) = 0 h = 3/2 or h = -1
f(x) = x/2Then the differential is lim h->0 [f(x+h) - f(x)]/h= lim h->0 [(x+h)/2 - x/2]/h= lim h->0 [h/2]/h= lim h->0 [1/2] = 1/2f(x) = x/2Then the differential is lim h->0 [f(x+h) - f(x)]/h= lim h->0 [(x+h)/2 - x/2]/h= lim h->0 [h/2]/h= lim h->0 [1/2] = 1/2f(x) = x/2Then the differential is lim h->0 [f(x+h) - f(x)]/h= lim h->0 [(x+h)/2 - x/2]/h= lim h->0 [h/2]/h= lim h->0 [1/2] = 1/2f(x) = x/2Then the differential is lim h->0 [f(x+h) - f(x)]/h= lim h->0 [(x+h)/2 - x/2]/h= lim h->0 [h/2]/h= lim h->0 [1/2] = 1/2
There are 8 permutations of three coins ... T T T T T H T H T T H H H T T H T H H H T H H H ... counting heads and sorting by count, you get ... 0 - T T T 1 - T T H 1 - T H T 1 - H T T 2 - T H H 2 - H T H 2 - H H T 3 - H H H ... so, the probability of each possible number of heads is 0: 1 in 8, 1: 3 in 8, 2: 3 in 8, and 3: 1 in 8.
d/dh(h^-1) = -1(h^-2) = -(h^-2)
multiplication is point to point and convolustion is point to multi-point ex multiplication-- s[n]=x[n].h[n] s[0]=[x[0].h[0] s[1]=[x[1].h[1] s[2]=[x[2].h[2] . . . .. s[n-1]=[x[n-1].h[n-1] convollustion s[n]=x[n]*h[n] s[0]=[x[0].h[0]+x[0].h[1]+x[0].h[2]+.......+x[0].h[n-1] s[1]=[x[1].h[0]+x[1].h[1]+x[1].h[2]+.......+x[1].h[n-1] s[2]=[x[2].h[2]+x[2].h[1]+x[2].h[2]+.......+x[2].h[n-1] . . . s[n-1]=[x[n-1].h[0]+x[n-1].h[1]+x[n-1].h[2]+.......+x[n-1].h[n-1].
-1
H-1. It accounts for 99.9885%
1/h-8 = (1/h)-8 = h8
The formula for H plus 1 plus H minus 1 is 2H. This is because the +1 and -1 charges cancel each other out, leaving only the H atom.
(2h-3)(h+1) = 0 h = 3/2 or h = -1
f(x) = x/2Then the differential is lim h->0 [f(x+h) - f(x)]/h= lim h->0 [(x+h)/2 - x/2]/h= lim h->0 [h/2]/h= lim h->0 [1/2] = 1/2f(x) = x/2Then the differential is lim h->0 [f(x+h) - f(x)]/h= lim h->0 [(x+h)/2 - x/2]/h= lim h->0 [h/2]/h= lim h->0 [1/2] = 1/2f(x) = x/2Then the differential is lim h->0 [f(x+h) - f(x)]/h= lim h->0 [(x+h)/2 - x/2]/h= lim h->0 [h/2]/h= lim h->0 [1/2] = 1/2f(x) = x/2Then the differential is lim h->0 [f(x+h) - f(x)]/h= lim h->0 [(x+h)/2 - x/2]/h= lim h->0 [h/2]/h= lim h->0 [1/2] = 1/2
There are 8 permutations of three coins ... T T T T T H T H T T H H H T T H T H H H T H H H ... counting heads and sorting by count, you get ... 0 - T T T 1 - T T H 1 - T H T 1 - H T T 2 - T H H 2 - H T H 2 - H H T 3 - H H H ... so, the probability of each possible number of heads is 0: 1 in 8, 1: 3 in 8, 2: 3 in 8, and 3: 1 in 8.
All isotopes of hydrogen have 1 proton and 1 electron; the isotope H-1 (protium) hasn't a neutron. the mass is 1.Other isotopes are:- H-2 (deuterium): 1 p, 1n, 1e; mass 2.- H-3: 1 p, 2n, 1e; mass 3.- H-4: 1 p, 3n, 1e; mass 4.- H-5: 1 p, 4n, 1e; mass 5.- H-6: 1 p, 5n, 1e; mass 6.- H-7: 1 p, 6n, 1e; mass 7.H-1, H-2 and H-3 are natural isotopes; H-1 and H-2 are stable isotopes.Isotopes from H-3 to H-7 are radioactive and unstable.
P(H and H and H and H)= 1/2 X 1/2 X 1/2 X 1/2 = 1/16
The oxidation number for H is almost always 1+.