What is assualt in the 4hth degree?
88102
Assault in the following orders are the following things:\n\n1 degree: very serious\n2 degree: somewhat serious\n3 degree: hardly serious\n4 degree: not serious at all
801480
4 is the answer I get
694279
I guess '4'. Do I get an A?
1052621
4.4 degrees celcius (40 Fahrenheit)
283215
[Cu(NH3)4]SO4*H2O (s) + 2HCl(aq) --->\n\nCu(NH3)4 +2 (aq) + H2SO4(aq) + H2O(l) + 2Cl-(aq)
220366
16UT? check your universal time.\n\n----Addition----\n\nI agree with the answer below.\nFor 11º 12' N :  60º 44' W, zone time is -4, so you are trying to calculate azimuth at 12:00pm (16UT), this gives an azimuth answer of 166.1 and altitude of 81.7\n\n255.9º at 12:00pm is too much in east direction.
469262
go to this website\n\nwww.asseenontv.com
714550
degrees = degrees\nminutes = (altitude decimal) * 60\n\nMinutes are to degrees like minutes are to hours.\n\nSo an altitude of 47.5 is 47 degrees and 30 minutes.
754704
You should write NH4(+) otherwise you will not get the oxidation number right.  H is 1 so H4 is 4.\n\nIn order to be left with 1(+) you need 4 - 3 = 1.  So N will be -3.\n\nCheck -3 + 4*1 = +1 OK.
222521
I would guess between 4.3 and 4.4
1143692
A 17h00, soit dans 44mn... non 43 maintenant !\nBon match !
920228
CuSO4•5H20\nThe dot notation denotes the hydrate.
894466
Uh...one and one-fourth degrees below zero.  (Is this a trick question?)
651211
1. 63.75 or 63 3/4\n2. 63\n3. 24.75 or 24 3/4
540818
NH4Cl (note the 4 is a subscript)
710286
the antiderivative of 4 / (t^2 + 1)  IS \n4 arctan(t) + C`
898116
A = b * h / 2\ndA/dT = [b * dh/AT + h * db/dT] / 2\n\nWhen h = 20 and A = 80, b = 8\n\n4 = [8 * dh/dT + 20 * 1] / 2\n4 = 4 * dh/dT + 10\n-6 = 4 * dh/dT\n-3/2 = dh/dT\n\nAs you can see, Gopal has the 4 * dh/dT part correct, but where Gopal has 8 minus 20, it should be 4 minus 10. As a result, Gopal's answer is twice the correct one. Gopal has a lot of good ideas, but is careless with their implementation. You can see that Gopal has unbalanced parentheses and brackets, no doubt the source of the confusion.\n\nAs a rough check, we can ask what the status is one minute after the time given in the problem, when the area is 84 and the base is 9. \n\nh = 2 * 84 / 9\n\nThe altitude computes to 18-2/3, much closer to my rate of change than to Gopal's.