simply Rt=(1)/1/R1 + 1/R2 can't even begin to understand......please help..........?
609281
The simplest formula for the effective resistance of 2 resistors in parallel is:\n\n1/Rt = 1/R(1) + 1/R(2)\n\nwhere Rt is the effective resistance and R(1) and R(2) are the resistance of the two resistors in paralel. You see when you place two resistors in parallel, the effective resistance of the two is always lower than either one resistor by itself. You can think about it this way: the resistor is like a narrow door which the electricity must pass through, so the higher the resistance is, the narrower the opening. When you just have one resistor, it is like only allowing the electricity to pass through one door. However, when you have two resistors in parallel, you opened a second door right besides the frist one for the electricity to pass through. Therefore the overall flow of electricity is increased and the effective resistance of the two resistors in parallel is lower.\n\nWhen you put two resistors in series, on the other hand, is like putting a narrow door behind another narrow door, so it's harder for the electricity to get through both doors and the effective resistance goes up.\n\nR(s) = R(1) + R(2)
930308
x=[2+/-rt(4+60)]/2\n=[2+/-8]/2\n=-3 or 5\nso the factors are\n(x+3)(x-5)
898860
First one: d= - 1\n\nEDIT: the rest I'm too lazy to do. Do your own homework! =D
639100
A = Pe^(rt)\nP = 25000, r = .08 and A = 50000 (since you are doubling and 25000*2 = 50000)\n50000 = 25000e^(.08t)\nDivide both sides by 25000\n2 = e^(.08t)\nTake the ln of each side\nln(2) = .08t\nDivide each side by .08\nln(2)/.08 = t\nt = 8.664339757
200519
+/- 1\n\nThe answer is not 1, because -1 * -1 = 1 also.
458677
There's a very explicit algorithm which explains\nhow to handle number, operators (including left\nor right precedence), functions, etc. on the \nWikipedia site for RPN (see reference #1).\n\nMostly, it's very straight-forward.  If it's an operand\n(ie. a number), then push it on the stack.  If it's an\noperator, then pop off as many operands are needed and\nevaluate.\n\nBut do look at the full algorithm for the details.\n\nOh yeah, for a full-blown implementation, see the source\ncode to the Unix "dc" (desk calculator) program.  The\nSolaris source is at reference #2.
828677
∫sin(ln(t)) dt\n\n= ∫sin(ln(t)) d(t)/dt dt\n\n= tsin(ln(t)) - ∫t * 1/t cos(ln(t)) dt \n\n= tsin(ln(t)) - ∫ cos(ln(t)) dt \n\n= tsin(ln(t)) - ∫ cos(ln(t)) d(t)/dt dt \n\n= tsin(ln(t)) - (tcos(ln(t)) -∫ t * 1/t *-sin(ln(t)) dt)\n\n= tsin(ln(t)) - tcos(ln(t)) -∫sin(ln(t)) dt\n\nSo 2∫sin(ln(t)) dt = tsin(ln(t)) - tcos(ln(t))\n\nSo ∫sin(ln(t)) dt = t/2(sin(ln(t)) - cos(ln(t)))
732315
This one's a little tricky.\n\nAllow me to use "t" instead of "delta"\n\n\nint ((cos(2t))^3)dt\n\nSince (cos(2t))^2 = 1 - (sin(2t))^2\n\nSubstitute.  this becomes:\n\nint ((1 - (sin(2t))^2) * cos(2t))dt\n\nSubstitute u = sin(2t).  Then du = cos(2t)dt.\nSo dt = du / cos(2t)\nThe key here is when you plug this in, the cos(2t) values cancel.\nSo we get:\n\nint (1 - u^2)du\n= u - (1/3)u^3 + C\n= sin(2t) - ((sin(2t))^3)/3 + C
491746
If u could give the entire question i could solve it for u and i'm not jokin. its actually quite easy.Its basically\n\n(w+3)/4w * 2w/(w-3).So wat ud have is\n\n1/2( (w+3)/(w-3) )\n\nits xactly like sayin 2/4 = (2/1) / (4/1)
960662
ry +s = tx - m\nYou must leave y alone for indicate its value in terms of other variables.\n\nFirst of all you move the s to the other side or the = as shown: ry =tx- m - s (you change the sign of s)\n\nThen you move the r as follows: y=(tx-m-s)/r\n\nAnd thats it \n\nYou have a restriction because the r has to be different than 0 to avoid a division by zero.  So restriction is r different from zero.
1102516
by any chance is this a question from your maths homework? lol :D\n\nI cant work it out myself because to put it simple, im crap at maths.
963851
The question doesn't make sense: what is T(n/4) supposed to mean when n = 2 or 3 or 5 ...??
1109416
3/x  + 1    the way you put it..
836547
f(x)=rtx+C\nf(1)=1+C=2\nso C=1\nf(1/4)=rt(1/4)+1=3/2
1065775
If a(N)=Sum(r^n, 0=>N) and -1<r<1...\n\n Is something that looks like 1/(1-r^N)
948100
t-(4/t)=3\ncan be rewritten as\nt^2 - 3t - 4 = 0\n\nthis has 2 possible solutions:\nt=-1\nor\nt=4
848345
I assume that this is s = (a-rl) / (1-r).\n\nFirst, multiply both sides of the equation by 1-r.\ns(1-r) = a-rl\n\nUse the distributive property on s(1-r)\ns-sr = a-rl\n\nGet the r terms on one side of the equation and the others opposite it.\nrl-sr = a-s\n\nUse the distributive property to factor out the r.\nr(l-s) = a-s\n\nDivide both sides by l-s\n\nr = (a-s) / (l-s)