The "T" in T+0 and T+1 refers to the number of trading days for a trade to "settle".
To settle means to exchange either the shares (stocks) or bonds for the cash and vice versa.
For example, in a standard T+3 stock trade, when you sell a stock, you have three (3) business/trading days to deliver the stock. When you buy a stock under the T+3 system, you also have three (3) business/trading days to deliver the funds to your broker to pay for the purchase.
So, in most cases, when you sell a stock, you will not see the cash enter your account for three trading days. And, when you buy, although you will receive the shares (and can usually sell at any time after your purchase), the shares will not actually arrive into your account for three days.
In most cases, standard stocks have a T+3 settlement (3 days) and bonds have a T+1 (1 day) settlement. Other trading forms can have other time limits such as commodities, etc.
Best wishes!
---FalconStocks
http://falconstocks.com
You factor it. 81t3 - 81t = 0 Divide by 81: t3 - t = 0 Take out the common factor t: t (t2 - 1) = 0 Here we have a difference of two squares, therefore: t (t + 1) ( t - 1) = 0 If the product is zero, that means that one of the factors must be zero, so you have to solve the three equations: t = 0 OR t + 1 = 0 OR t - 1 = 0. In summary, t can be 0, -1, or 1.
2t^2+5t-3=0 (2t-1)(t+3)=0 2t-1=0 and t+3=0 t=.5 and t=-3
Vrms=sqrt[1/T * integral(v^2(t)dt, 0,t] Irms=sqrt[1/T * integral(i^2(t)dt, 0,t]
This following loop creates a unit ramp input : for t= 0:0.1:5 if(t<=1) y(count) = t; end if(t>1) y(count) = 1; end Rz(count) = tf(y(count)*T,[1 -2 1],T); Rs(count) = d2c(Rz(count)); end % T = 0.005; this is for a discrete system
r=0,Tr-r = 0 = r(T-1), since T != 1, then T-1 is non zero so r must be zero.
t t
The unit step function at t=0 is defined to have a value of 1.
Which one is correct dG(x-0)-dG(x+0)=-1/p(t) or 1/p(t)
Which one is correct dG(x-0)-dG(x+0)=-1/p(t) or 1/p(t)
I will assume that you mean to ask, "What is the arc length of curve C from t=0 to t=1 if curve C is defined parametrically by x=1+2e^t and y=e^t?" I can answer this question. dx/dt=2e^t and dy/dt=e^t. Arc length = a∫b √[(dx/dt)2+(dy/dt)^2] = 0∫1 √[4e^(2t)+e^(2t)]dt = 0∫1 √[5e^(2t)]dt. = 0∫1 [(√5)(e^t)]dt = √5 x (e^1-e^0) = √5 x (e-1) = e√5-√5. Difficult? Maybe. Fun? Hopefully. Accurate? Definitely!
import java.util.Scanner; public class ColEvenOdd{ public static void main (String[]args){ Scanner kb = new Scanner (System.in); System.out.println("Enter the limiting number"); int no = kb.nextInt();//Here put the 50 System.out.println(); int i,j; if(no==0){ System.out.println("ODD"+'\t'+"EVEN"); System.out.println(" "+'\t'+" 0 "); System.exit(0); } if(no==1){ System.out.println("ODD"+'\t'+"EVEN"); System.out.println("1"); System.exit(0); } if(no>1){ System.out.println("ODD"+'\t'+"EVEN"); if(no%2==1){ for(i=1; i<=(no-2); i+=2){ System.out.println((i)+" "+'\t'+(i+1)); }System.out.println(no); System.exit(0); } if(no%2==0){ for(j=2; j<=no; j+=2){ System.out.println((j-1)+" "+'\t'+(j)); }System.exit(0); } } } }
Assume the first term is cos-squared(theta) rather than cos(2*theta). 6cos2(t) + 5cos(t) - 4 = 0 [6cos2(t) + 8cos(t) - 3 cos(t) - 4] = 0 [3cos(t) + 4]*[2cos(t) - 1] = 0 which gives 3cos(t) = -4 so that cos(t) = -4/3 or 2cos(t) = 1 so that cos(t) = 1/2 The first of these is clearly not a possible solution, whereas the second can hve one or more solutions, depending on the domain - which is not specified in the question. If the initial assumption is incorrect and the first term WAS 6cos(2*theta) then using the formula for double angles: cos(2(t) = cos2(t) - sin2(t) = cos2(t) - [1 - cos2(t)] = 2cos2(t) - 1 and so, the equation becomes 6*[2cos2(t) - 1] + 5cos(t) - 4 = 0 or 12cos2(t) + 5cos(t) - 10 = 0 Solve this quadratic equation for cos(t) and then find the values of t (or theta) within the domain specified.