No that equation is a standard one to get the derivative of any function using the first principle.
Btw how do you guys get those latex or whatever images into your answers?
In case of reflection and refraction of light,its the wavelength of light and not the frequency that changes along with the velocity.May be anyone could look at the problem from that angle.
You get the answer by taking u=a*sec y =>y=arcsec u/a(arcsec-sec inverse)
=>du=a sec y tan y dy
Therefore I=integral[(a sec y tan y dy)/a^2 tan^2 y]
=>(sec y dy)/(a tan y)
=>cosec y dy /a
=>ln(cosec y-cot y)/a
then just substitute value of y to get the answer
There is an infinite square grid with each side of every cell making up the grid of resistance r.
What is the equivalent resistance between two adjacent points on the grid?
To get the antiderivative, you could substitute x=tan y and so dx=(sec y)^2 and thus simplifying
you would end up with (400/sec^2 y) dy
=>(400cos^2 y) dy
=>400(1+cos2y)dy
and integrate this and in the end substitute y=arctan x to get your answer.
If you just want to find a relation between the two pressures, just use both -bernoulli's eqn and continuity eqn and assume a few quantities like r1>r2 .Also in bernoulli's eqn neglect gravitational potential energy.
1/2 d V1^2 +1/2 m V1^2=1/2 d V2^2 + 1/2 m V2^2
v=velocity
d= density
m= mass of...
In ideal conditions, you would continue to oscillate forever.That is because force of gravity increases as you move away from the center and so the force pulling you towards the center would increase with distance from the center.Hence you would get a condition like F=-kx(k is a constant and x...
Answer <if u didnt get it already>
the way 2 do it is this
-3x^2+12x-9=0
divide by -3 lik the other guy said
x^2-4x+3=0
x^2-x-3x+3=0
x(x-1)-3(x-1)=0
(x-1)(x-3)=0
x=3,1
For other harder problems where it is harder 2 factorize use the formula
for ax^2+bx+c=0
x={-b+-sprt(b^2-4ac)}/2a