the price of a certain commodity in dollars/units at time t is given by p=8+4e^-2t+te^-2t?
360360
The average value of a function is given by\n\n(1/(b-a)) S f(x) dx, evaluated from a to b.\n\nSo for this question, the average will be given by\n\n(1/(4-0)) S 8+4e^-2t+te^-2t dt = (1/4)[8t - 2e^(-2t) + S te^-2t dt]\n\nLet u = e^(-t)\ndu =  -e^(-t) dt\n\nSo, S te^-2t dt = - S u^2 du\n= (-u^3)/3\n= -e^(-3t)/3\n\nSo your answer then becomes\n\n(1/4)[8t - 2e^(-2t) - e^(-3t)/3] = 2t - (1/2)e^(-2t) - e^(-3t)/12\n\nEvaluating from 0 to 4,\n= 2(4) - (1/2)e^(-2(4)) - e^(-3(4))/12 - [2(0) - (1/2)e^(-2(0)) - e^(-3(0))/12]\n= 8 - (1/2)e^(-8) - e^(-12)/12 + 1/2 + 1/12\n= 8 5/12 - (1/2)e^(-8) - e^(-12)/12\n~ $8.42
882807
33.5 * p = 18.75\np = 18.75 / 33.5 ~ 0.5697 OR 56.97%\n\nOr did you want to use A = Pe^rt ?\n\n18.75 = 33.5e^5r\nr = ln(18.75 / 33.5) / 5 ~ -0.1161 OR -11.61%
1081089
[ (50000 x 2.4)+(25000 x .8) ] / (50000+250000)=1.8\n\nThe remainder of the info in the problem is not necessary.
189924
just add\n\n$02E2\n+ $85\n-----\nyou tell me...\njust remember to carry the one after hitting 14 (or E)
965285
rate b = rate a / sqrt (mwtb/mwta)\n          = [rate a * sqrt mwta]/sqrt mwtb\n          = 0.0235 * sqrt64 / sqrt 352\n          = 0.188/18.7624 = 0.01 = 10 x 10^-3...aprrox the same answer\n she got that 9.88 by taking m wt UF6  as 362....lol\nplug it in to see...\nbtw, your algebra was wrong....
249879
Take $30.00 and subtract $.10 ($29.90)\nThen figure 80% of that, and the cost is the answer ($23.92)
317034
A = P(1+i)^n\nP = 1000\ni = .08/4 = .02\nn = 8*4 = 32\n\nA = 1000(1+.02)^32\n= 1884.54
575490
I- + 2e- -------> I2\n\n4P + 3e-  -------------> P4\n\nthere ya go :)
235611
8388608 would be the last value.\n16777215 would be the total value in cent.\n\n= $167772.15
839084
Copper prices (LME) in US$/mt\napril, 28 = 7,285.00\naugust, 31 = 7,647.00\nnovember, 9 = 7,220.00\n\nfuture?\nIn two year, the forecast is =~ 6,220.00
437832
Homework.\n\nSuppose a1*x+a2*Tx+..+am*T^(m-1)x=0.\n\nApply T^(m-1) to get a1=0.\nApply T^(m-2) to get a2=0, etc.
966454
No. The first would be $0.008. The second would be $0.80
84594
if x is the original price\n216=x-.1x\n216=.9x\ndivide 216 by .9 this gives you x
835734
-0.070, -0.001, 0.000
22859
Subract the two terms on the left giving:\n-2t < 16\n\nAdd 2t to both sides giving\n0 < 16 + 2t\n\nSubtract 16 from both sides\n\n-16 < 2t\n\nDivide both sides by 2\n\n-8 < t
398683
Use log you will find\n\nor put\n\nx=3.597285024\n\nIts done
751901
OK, we know the change in price going from 2 quarts to 4 quarts - that is 3.09-1.65=$1.44, so we pay 0.72 per quart.\n\nWe know that the equation is linear, so the formula must be\np = A + 0.72q\nwhere p is the price, q is the number of quarts and A is a constant.\nWe can then plug in a price and number of quarts to work out a value for A:\n3.09 = A + 4 x 0.72\n3.09 = A + 2.88\n0.21 = A\n\ngiving us a formula:\np = 0.21 + 0.72q\n\nIt is worth checking that this formula works for our other price and number of quarts:\n1.65 = 0.21 + 0.72 x 2\n1.65 = 1.65\nwhich it does.