anonymous
  • anonymous
Suppose that you take out an unsubsidized Stafford loan on September 1 before your junior year for $4500 and plan to begin paying it back on December 1 after graduation and grace period 27 months later. The interest rate is 6.8%. How much of what you will owe will be interest?
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
i believe its $661.76
myininaya
  • myininaya
cantorset can we use induction on your problem? do you know what that is?.
anonymous
  • anonymous
yes we can

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anonymous
  • anonymous
how did you find me, lol
myininaya
  • myininaya
i went to the homepage and you can see what everyone is doing
anonymous
  • anonymous
the chat program doesnt work for me
anonymous
  • anonymous
ok lets do this one for all n>=5 , n^3 < 3^n
myininaya
  • myininaya
ok
myininaya
  • myininaya
for n=5, we have 5^3=125<243=3^5. Sold the inequality holds for n=5
anonymous
  • anonymous
right, basis case
myininaya
  • myininaya
now we should assume it is true for integers k>=5 Then we show it is true for k+1 the induction step is the hardest so let me think about it
myininaya
  • myininaya
it seems that you are going over logs so i'm pretty sure we may have to involved that somehow
anonymous
  • anonymous
hmmm, they may be unrelated though. here is what i got ,
anonymous
  • anonymous
that was a different question
anonymous
  • anonymous
is there a way to post without having to click my mouse on post?
myininaya
  • myininaya
no i guess not
anonymous
  • anonymous
show that (k+1)^3 < 3^(k+1)
anonymous
  • anonymous
k^3 + 3k^2 + 3k + 1 < 3^(k+1)
myininaya
  • myininaya
omg i got it
myininaya
  • myininaya
i got so excited i clicked out sorry. so we have (k+1)^3
anonymous
  • anonymous
k ^3 + 3k^2 + 3k + 1 < k^3 + 3k^2 + 3k + 3k = k^3 + 3k^2 + 6k < k^3 + k^3 + k^3 = 3k^3 < 3*3^k = 3^(k+1)
anonymous
  • anonymous
(k+1)^3 = k ^3 + 3k^2 + 3k + 1 < k^3 + 3k^2 + 3k + 3k = k^3 + 3k^2 + 6k < k^3 + k^3 + k^3 = 3k^3 < 3*3^k = 3^(k+1)
myininaya
  • myininaya
i got excited ignore what i wrote
anonymous
  • anonymous
(k+1)^3
anonymous
  • anonymous
my proof isnt elegant, but i think it works, might have to do some side induction arguments for the intermediate steps
myininaya
  • myininaya
you are right i got excited
anonymous
  • anonymous
also you better use parenthesses :) more often
anonymous
  • anonymous
the log proof is A LOT easier
myininaya
  • myininaya
yep
anonymous
  • anonymous
what did you get? log_3 n < n if n < 3^n , assuming we raise both sides to 3
anonymous
  • anonymous
or you can start, if we can show that n < 3^n , for n>=1, then taking the log base 3 of both sides we have our result. what was your logic?
myininaya
  • myininaya
ok lets skip the case for n=1 that is easy to do and we know it holds. so assume for integer k, log3(k)
anonymous
  • anonymous
so that proof goes, 1 < 3^n, assume k < 3^k then add 1 to both sides of that inequality we have k + 1 < 3^k + 1 < 3^k + 3 = 3^(k+1)
myininaya
  • myininaya
yea maybe we should base 3 on both sides to get rid of log
anonymous
  • anonymous
you mean exponentiate?
anonymous
  • anonymous
you cant get rid of log with log :)
anonymous
  • anonymous
can you help me with surface area of integral
myininaya
  • myininaya
surface area of a solid?
anonymous
  • anonymous
surface area of a torus
anonymous
  • anonymous
so you cant finish the proof with log, no problemo
myininaya
  • myininaya
one thing at a time
anonymous
  • anonymous
how do you find the question i posted earlier
anonymous
  • anonymous
this website is so confusing
myininaya
  • myininaya
are you given the equation for the torus or is it a parametric curve/
anonymous
  • anonymous
Calculus question: surface area of a torus. it is generated by circle with radius r , off center. Suppose the circle is centered at (x - R)^2 + y^2 = r^2, and R> r. so the distance from (0,0) to the center of the circle is (R,0) which has radius r. Ok the integral i got is int 2pi y sqrt [ 1+ (dy/dx)^2] = int 2pi sqrt ( r^2 - ( x-R)^2) [ 1 + (-2(x-R)/(2sqrt( r^2 - ( x-R)^2) ^2 ]
anonymous
  • anonymous
we are given a circle , and its going to generate the torus by revolving about the y axis
anonymous
  • anonymous
the final answer should be 4pi^2 R*r
myininaya
  • myininaya
so that last thing you have the square cancels the sqrt
myininaya
  • myininaya
ok im just going to do it on paper and scan it and you can open it give a few min
anonymous
  • anonymous
right, good :) ok here is what i got
anonymous
  • anonymous
integral 2pi y dS, remember ds can be sqrt 1 + dx/dy ^2 or sqrt 1 + dy/dx ^2
myininaya
  • myininaya
we need another pi in there somewhere
myininaya
  • myininaya
since are answer has pi^2
myininaya
  • myininaya
this is what i got using the integral you found we can get times another pi it would be correct
1 Attachment
myininaya
  • myininaya
don't look at the name of the file. i know we are not finding volume
anonymous
  • anonymous
right, its surface area, here lets use twiddla
anonymous
  • anonymous
http://www.twiddla.com/521818
anonymous
  • anonymous
if you dont mind, we can draw on whiteboard

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