tan(a+b) = (tan(a) + tan(b))/(1 - tan(a)tan(b))
tan(a-b) = (tan(a) - tan(b))/(1 + tan(a)tan(b))
tan(pi/4) = 1
tan(pi/4 + theta)*tan(pi/4 - theta) = (1+tan(theta))/(1 - tan(theta)) * (1 - tan(theta))/(1 + tan(theta)) → 1
Simply replace t by 4 in your formula:
N(4) = 10*3^4 → 10*81 → 810
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I.e. You start with two 1's and each subsequent term is the sum of the previous two terms.
Almost right EP, but 15/10 is 1.5.
No. The integral is with respect to r not F.
Here's a way of doing the first part of 4):
The second integral can be done in similar fashion (though there are higher powers to deal with). You should find the result is \(\frac{5\pi}{256}a^8\)
You are to be commended for trying owlface.
You are not the only person to make this mistake when averaging speeds. Even the Guiness Book of Records gets it wrong when calculating land speed record attempts!! They take the speeds, calculated seperately, over two distances (there and back) and average them in just the way you did to get an overall average speed. What they get by doing this is not the true average (unless the speeds happen to be the same in both directions)!
Use the fact that a^x/a^y = a^(x-y) and a^x*a^y = a^(x+y)
So 13^-14/13^-7 = 13^(-14 - -7) → 13^(-14 + 7) → 13^-7
8^9*8^5 = 8^(9+5) → 8^14
I'll leave you to try your third one!
Let the distance between where he starts from and Austin be L miles
Time taken to get to Austin = L/40 hours
Time taken to return from Austin = L/60 hours
Average speed = total distance/total time = 2L/(L/40 + L/60) → 2/(1/40 + 1/60) → 48 mph
|z| is the magnitude, as in my original reply.
+33665 
