I believe it's 37CM, I had that question on my finals last year.
To answer your question, this could be the possible answer and i hope you understand and interpret it correctly:
<span>[Integrate [0, 1/2] xcos(pi*x
let u=x so that du=dx
and v=intgral cos (xpi)dx
v=(1/pi)sin(pi*x)
integration by parts
uv-itgral[0,1/2]vdu just plug ins
(1/pi)sinpi*x]-(1/pi)itgrlsin(pi*x)dx from 0 to 1/2
(1/pi)x sinpi*x - (1/pi)[-(1/pi) cos pi*x] from 0 to 1/2
=(1/2pi)+(1/pi^2)[cos pi*x/2-cos 0]
=1/2pi - 1/2pi^2
=(pi-2)/2pi^2 ans</span>
Answer:
On occasions you will come across two or more unknown quantities, and two or more equations
relating them. These are called simultaneous equations and when asked to solve them you
must find values of the unknowns which satisfy all the given equations at the same time.
Step-by-step explanation:
1. The solution of a pair of simultaneous equations
The solution of the pair of simultaneous equations
3x + 2y = 36, and 5x + 4y = 64
is x = 8 and y = 6. This is easily verified by substituting these values into the left-hand sides
to obtain the values on the right. So x = 8, y = 6 satisfy the simultaneous equations.
2. Solving a pair of simultaneous equations
There are many ways of solving simultaneous equations. Perhaps the simplest way is elimination. This is a process which involves removing or eliminating one of the unknowns to leave a
single equation which involves the other unknown. The method is best illustrated by example.
Example
Solve the simultaneous equations 3x + 2y = 36 (1)
5x + 4y = 64 (2) .
Solution
Notice that if we multiply both sides of the first equation by 2 we obtain an equivalent equation
6x + 4y = 72 (3)
Now, if equation (2) is subtracted from equation (3) the terms involving y will be eliminated:
6x + 4y = 72 − (3)
5x + 4y = 64 (2)
x + 0y = 8
Answer:
The equation for charge is = current x times
