Maximillian

Nicely done, you nailed it

Dont know if it used to be like this, but having a preview for the latex code that you type out has definitely got to be the best feature here

\(i=i \quad i^2=-1\quad i^3=-i\quad i^4=1\\i^5=i....\)

Strategy that I came up with:

Starting with the exponent at 5 onwards, if you subtract it by 1 then divide by 4 and get a whole number then it equals i. Otherwise if not a whole number then do the process again, subtract the original exponent by 2 then divide by 4 and see if you get a whole number, if you do then it equals -1 ...... if you have to subtract it by 3 it equals -i ...... if you have to subtract it by 4 then it equals 1.

Using this strategy: the exponent of i^10 is 10 soooo:

(10-1)/4 =/= whole so reject i

(10-2)/4 = 2 which is a whole number therefore i^10 = -1.

Now tell me what i^2015 equals =)

Using one of the many trig identities:

\(cos(u)cos(v) = \frac{1}{2}[cos(u-v)+cos(u+v)]\\cos(2x)cos(4x) = \frac{1}{2}[cos(-2x)+cos(6x)]\)

If you take E and R to be variables:

\(\ln{E}=1.2+1.81\ln{R} \\ E = e^{1.2+1.81\ln{R}}\\E = e^{1.2}e^{\ln{R^{1.81}}}\\E=R^{1.81}e^{1.2}\)

Which is an equivalent equation without the natural log

You can only simplify this, this is the equation of a line and doesnt have 1 single answer

Look at the gradients(\(M\)) for both lines

Line 1 gradient = -4 = \(M_1\)

Line 2 gradient = 1 / 4 = \(M_2\)

Definition for a perpendicular line is when the gradient(\(M\)) satisfies:

\(\frac{-1}{M_1}\) is perpendicular to \(M_2\)

remember that \(e^{ln(x)}=x\)

Therefore just raise e by each side of the equation to get rid of the log terms

Edit:

Also a rule that will help you simplify -> \(e^{a+b}=e^ae^b\)

I drew a quick representation of what I thought "above the y - z plane " means. I interpreted it as "above the y plane" means all values of y > 0 and "above the z plane" means all values of z above zero. I completely agree with the rest of your working but this is the only source of confusion that I still have left. Your way seems correct as it returns a lower bound for x whilst my interpretation lets x go down to negative infinity.

Edit:

Did a google search and saw that the x-y plane is where x = x , y = y , z = 0. So in this case the y-z plane would be where x = 0 , y = y , z = z  and "above" means it's on the positive side. My confusion is cleared and I agree with your answer Alan, many thanks =)

Just to make what I'm asking as clear as possible:

The entire surface S is bounded by:   y=0  ,   z =0   ,  4x + y + 2z = 4    above the y-z plane

Part of this surface, \(S_1\), is the surface along y=0.

Find \(\int\int_{S_1} y^2-x dA\)

What are the bounds of integration I should be using here? What is dA meant to be (Is it dx dy or is it dx dz or something else) ?