Q=1.6 x 10^(19) , C=3.22 x 10^(20) ,K=0.49?
822981
i)Just substitute in the values to get F\nx = 1.1.  Other values are given.  Hence F.\n\nii) Rearrrange the equation\nx = [cosh^-1(QCK/2F-1)]/k\nF = 2.9.  Other values are given, hence x.
741784
(21 g x 52.7 C) + (54.9 g x 31.5 C) / (21 g + 54.9 g)\n\n= 37.4 C
627685
Bring the 225Q and 275Q on one side of the equal sign and the 12500 and 100000 on the other. Then solve for Q.
986663
hmmm ...\n\n(2.5 x 10^3  ) x (.0685) = .17125 x10^3 = 1.7125 x 10^2 g
619470
o.k., in short, d < a < c < b. This is basic material, so I'll give a generously detailed explanation.\n\na) 2 moles N2 * (28 g N2/ mole N2) = 56g N2\n\nb) 1 mole KCl * [(39.1gK +35.5gCl)/ mole KCl] = 74.6 g KCl\n\nc) 9 x 10^23 molecules CO2 * (1 mole CO2/ 6.02 x 10^23)*([12gC + 16gO + 16gO]/ mole CO2) = 66g CO2\n\nd) 2 moles LiF * [(6.94g Li + 19gF)/ mole LiF] = 51.88 g LiF
1081330
I = 60 x e/s\ne = elementary charge\n60 x e/s = 60 x 1.6 x 10^-19 C/s = 9.6 aA\naA = attoampere.\natto =10^-18
944165
F = C * 9/5 + 32\nC = (F - 32) * 5/9\n\nF = 2212 * 9/5 + 32 = 4013.6\n\nHope this helps!
238182
p1    p2\n--- = ---\nq1    q2\n\n 5     2\n--- = ---\n 2    q2\n\nCross multiply,\n\n4/5
943409
Just substitute in the values.\ni) x = 1.5.  Other values are given.  Hence F= 6.62.\nii) Rearrrange the equation\nx = [cosh^-1(QCK/2F-1)]/k\n=4.56\nSubstitute 4.56 back into the original equation to confirm.
226307
= 24 k^2   +  6 k^3 +  18k^2  - 54 k^3\n= 42k^2 - 48 k^3\n= 6k^2 (7 - 8k)\n\nif 6k^2 (7 - 8k)=0\n7 - 8k=0\n8k=7\nk=7/8
532062
If the question is to simplify:\n\nSqrt(144(x^10)(y^12)(z^18)) then the first answer is correct:\n    \n12(x^5)(y^6)(z^9)
1179632
y=kx  \n20=k(6)\n20/6=k\nk=3.33\ny=3.33(9)\ny=30\nso the answer is "E"
652953
I got:\n1. 6x^2 - 36x +48\n2. 6x + 22
299035
just like the person above (bree) says but you can further reduce the 18. \n\nx= 3c sqrt(2)
843058
1) 2.01 * 10^2\n\n2) 1.995 * 10^23\n\n3) 5.35 * 10^-4\n\n4) 3.8418 * 10^-7
759871
These are really hard to solve; you have to square both sides twice.\n\n(sqrt [19y - 20] + 3)^2 = (5sqrt[y-2])^2\n\n19y - 20 + 6 sqrt (19y - 20) + 9 = 25 (y-2)\n\n19y - 11 + 6 sqrt (19y-20) = 25y - 50\n\n(subtract 19y; add 11 to both sides)\n\n6 sqrt(19y - 20) = 6y - 39\n\ndiv both =sides by 3:\n\n2 sqrt(19y-20) = 2y - 13\n\nsquare both sides again:\n\n4 (19y - 20) = (2y - 13)^2 \n\n78y - 80 = 4y^2 - 52y + 169\n\nsubtract 78y; add 80 to both sides\n\n4y^2 - 140y + 249 = 0\n\n\nA huge quadratic to solve, you are on your own from here.  Are you sure you typed in the problem right?
867375
=19g-4h+4-20g+20\n\n= -g-4h+24