ME =29 EG = 21 MG = 20 what is the tan of E?
26943
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924339
a) y= 10^(tan (πθ))\n= e^(ln10*tan(πθ))\n= e^u where u = ln10*tan(πθ\ndy/dθ =dy/du * du/dθ\n= e^u * ln10*πsec²(πθ)\n= ln10*πsec²(πθ)*e^(ln10*tan(πθ))\n= ln10.πsec²(πθ)*10^(tan (πθ))\n\nb) y= √[tln(t^4)]\n= √[4t*ln(t)]\n= 2√[t*ln(t)]\n= 2u^½ where u = t*ln(t)\ndy/dt = dy/du * du/dt\n= u^-½ * (ln(t) + 1)\n= (ln(t) + 1)/√[tln(t)]\n\nc) y = arctan(arcsin (x^½))\n= arctan u where u = arcsin v where v = x^½\ndy/dx = dy /du * du/dv * dv/dx\n= 1/(1+ u²) * 1/√[1 - v²] * ½x^(-½)\n= 1/[2√(1 - x)*(1 + (arcsin (x^½))²)]\n\nd)xe^y = y - 1\ne^y + xe^y * dy/dx = dy/dx\nSo dy/dy - xe^y * dy/dx = e^y\nie dy/dx (1 - xe^y) = e^y\ndy/dx = e^y/ (1 - xe^y)\n= 1/(e^(-y) - x)
795130
y = tan(πx - 1)\nlet u = πx - 1\nnow, dy/dx = (dy/du)(du/dx)\nd(tan(u)/du = sec^2(u)\ndu/dx = π, so \ndy/dx = πsec^2(u)\ndy/dx = πsec^2(πx - 1)
618692
the correct option is e \n the value of i radian is 180/pi = 57.3248 ( if pi is 3.14)
214671
work = (200N)*(10.0m)*cos(0) + (50.0N)*(35m.0m)*cos(0)\nwork = 2000 Joule + 1750 Joule\nwork = 3750 Joule
428474
m(e) = (2/3) log(e / 10^4.4) energy release = 2 * 10^17?\n\nWhat is the question?  Do you want to know what value of e makes (2/3) log(e / 10^4.4) = 2 * 10^17 ?\n\nThat would mean\nlog(e / 10^4.4) = (3/2) * 2 * 10^17 = 3 * 10^17\ne / 10^4.4 = exp(3 * 10^17)\ne = 10^4.4 exp(3 * 10^17)\n\nwhere exp(x) is just e^x for e = 2.718281828459... the base of natural logarithms, but I write exp(x) because you are already using "e" for something else.
499258
.5kg =  500.0  g\n50mg =    0.05 g\n          2.5  g \n       =======\n        502.55 g
526116
Basically, all you need to know is the initial amount (8.0 grams I assume), and the half life (46 days).  Then you can build an exponential model of the form:\n\nA(t)=A*e^(k*t)  \nA is initial amount\nk is decay constant\nt is time\n\nKnowing that after 46 days, there will be half as much present:\n\n4=8*e^(k*46)\n\nThen solve for k using logarithms.  \n\nln(0.5)=46k\nk=ln(0.5)/46=-0.01507\n\nSubstitute into equation to get time dependent equation:\n\nA(t)=8*e^(-0.01507*t)\n\nNow you want to know how much is left after t=184:\n\nA(184)=8*e^(-0.01507*184)\n\nA(184)=0.5 [grams]\n\nViola....
790092
24g = 4g magnesium + 20g other elements\n1440/24 = 60 so need 60 times the amount\n4g * 60 = 240g magnesium\n20g * 60 = 1200g other elements\n240g + 1200g = 1440g\n\nSo 1200g of other elements
977036
Na: atomic mass 23\nCl: atomic mass 35\nNaCl: atomic mass 23 + 35 = 58\n\n(50 g) / (58 (g / mole)) = 0.862068966 moles\n\n(0.862068966 moles) / (2.5 L) = 0.344827586 moles/L
526884
Another one: 200(1+N/20)
983394
Depends\n\nlim θ→π/2- (ie from below) tanθ = +∞\n\nlim θ→π/2+ (ie from above) tanθ = -∞ \n\nThis is because in the 1st quadrant tanθ >0 and in the second quadrant tanθ <0
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
955015
45.6 grams x 1 mol/323.2 g = 0.141 moles \n\n\nFormula weights \n\nPb= 207.2 x 1 = 207.2\nCr= 52 x 1 =         52.0\nO = 16 x4  =         64.0\n\nPbCrO4 =          323.2
817617
tan(A-B)+tan(B-C)+tan(C-A)\n= (tanA - tanB)/(1 + tanAtanB) + (tanB - tanC)/(1 + tanBtanC) + (tanC - tanA)/(1 + tanAtanC)\n= [(tanA - tanB)(1 + tanBtanC)(1 + tanAtanC) + (tanB - tanC)(1 + tanAtanC)(1 + tanAtanB) + (tanC - tanA)(1 + tanAtanB)(1 + tanBtanC)]/[(1 + tanAtanB)(1 + tanBtanC)(1 + tanAtanC)]\n= 0\nie (tanA - tanB)(1 + tanBtanC)(1 + tanAtanC) + (tanB - tanC)(1 + tanAtanC)(1 + tanAtanB) + (tanC - tanA)(1 + tanAtanB)(1 + tanBtanC) = 0\nOn expanding\n(tanA - tanB)(1 + tanAtanC + tanBtanC + tanAtanBtan²C) \n+ (tanB - tanC)(1 + tanAtanC + tanAtanB + tan²AtanBtanC)\n+ (tanC - tanA)(1 + tanAtanB + tanBtanC + tanAtan²BtanC)\n\n= tanA - tanB + tan²AtanC - tanAtanBtanC + tanAtanBtanC - tan²BtanC + tan²AtanBtan²C - tanAtan²Btan²C \n+ tanB - tanC + tanAtanBtanC - tanAtan²C + tanAtan²B - tanAtanBtanC + tan²Atan²BtanC - tan²AtanBtan²C\n+ tanC - tanA + tanAtanBtanC - tan²AtanB + tanBtan²C - tanAtanBtanC  + tanAtan²Btan²C - tan²Atan²BtanC\n= tan²AtanC - tan²BtanC - tanAtan²C + tanAtan²B - tan²AtanB + tanBtan²C\n=0 \nDivide all by tanAtanBtanC\nSo tanA/tanB - tanB/tanA - tanC/tanB + tanB/tanC - tanA/tanC - tanC/tanA = 0\n\nie tanA/tanB + tanB/tanC + tanC/tanA = tanB/tanA + tanC/tanB = tanA/tanC\n\nWhence tanA/tanB + tanB/tanC + tanC/tanA = cotA/cotB + cotB/cotC + cotC/cotA\n\nSince a triangle then tan(A + B) = tan(π - C) = - tanC\nSo (tanA + tanB)/(1 - tanAtanB) = -tanC\nie tan A + tan B = tanAtanBtanC - tanC\nso tanA + tanB + tanC = tanAtanBtanC
928283
I belive your u is the Greek letter mu, the prefix for micro, meaning 10^(-6).  The term ug would therefore equal 1/1,000,000 gram, which is 1/1,000 of a milligram, abbreviated mg.\n\nSo, 50mg = 50,000 ug
437730
See http://members.aol.com/logan20/elecspon.html for the equation to use.