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Cool, thanks for the help guys!
I am having trouble finding the radius of convergence for:
I am not sure how to deal with the n!
All help is much appreciated!
P.s the sum is from n=0 to infinity...i didnt know how to do this using code
Okay, that makes things alot clearer, thankyou both.
Thanks Jane, but if say lx-1l<0, why must it be equal to -x+1???
Also how do you determine the considered regions, I can understand graphically but not algebraically.
Solve the following inequality:
lx-1l+lx-2l ≥5
I am struggling how to work this out, I tried case analysis but I don't think i'm doing it correctly, all help is appreciated.
Also how do you write the modulus lines using latex?
Thanks, Adam:)
All primes are odd apart from 2, so p+1 either equals 2+1=3 or an (odd number +1)=even number "n^3".
As 3 is not a cubed integer we know that p≠2. This means "n^3" is even, thus meaning "n" must be even as (even)^3=even and (odd)^3=odd, and as 2 is the only even prime number, n=2
therefore, p+1=2^3=8
Meaning p=7
I agree with Identity, if t=9 then:
The first "f" in effort =9+I+1 ...the "1" being carried over.
The second "f" in effort =9+A+1...the "1" being carried over.
Which implies I=A, making it contradictory. :S
what do you mean by "i"???...imaginary number? :S
Thankyou so much!
I understand everything apart from the first line...how do you know that f'(x)=1/(xlna) if f(x)=loga(x) ???
also the answers tell me that the answer is 1+(1/x) :S
thanks though luca, I hate to be a bother
Okay, i've completed 2a) by substituting (ln) instead of log base 10, but I still can't get the correct answers for the other two questions.
well i'm presuming log with base 10 but i dont actually know for sure, i just copied the questions exactly :S
Hi, I'm struggling with the following questions.
1. Differentiate the following with respect to x: y=log(xe^x)
2. Find the derivatives of the following functions at the indicated points:
a) (cosx).log(3x) , x=pi
b) log((x^3)/(x^0.5)) , x=5/2
Any help is much appreciated as I havent a clue what to do here.
Thanks, Adam
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