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anonymous
 5 years ago
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?
anonymous
 5 years ago
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?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i believe its $661.76

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0cantorset can we use induction on your problem? do you know what that is?.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how did you find me, lol

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0i went to the homepage and you can see what everyone is doing

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the chat program doesnt work for me

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok lets do this one for all n>=5 , n^3 < 3^n

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0for n=5, we have 5^3=125<243=3^5. Sold the inequality holds for n=5

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0now 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
 5 years ago
Best ResponseYou've already chosen the best response.0it seems that you are going over logs so i'm pretty sure we may have to involved that somehow

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmmm, they may be unrelated though. here is what i got ,

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that was a different question

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is there a way to post without having to click my mouse on post?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0show that (k+1)^3 < 3^(k+1)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0k^3 + 3k^2 + 3k + 1 < 3^(k+1)

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0i got so excited i clicked out sorry. so we have (k+1)^3<K^3+1^3=k^3+1<3^k+1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0k ^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
 5 years ago
Best ResponseYou've already chosen the best response.0(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
 5 years ago
Best ResponseYou've already chosen the best response.0i got excited ignore what i wrote

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(k+1)^3<K^3+1^3=k^3+1<3^k+1 this is false

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0my proof isnt elegant, but i think it works, might have to do some side induction arguments for the intermediate steps

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0you are right i got excited

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0also you better use parenthesses :) more often

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the log proof is A LOT easier

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what did you get? log_3 n < n if n < 3^n , assuming we raise both sides to 3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0or 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
 5 years ago
Best ResponseYou've already chosen the best response.0ok lets skip the case for n=1 that is easy to do and we know it holds. so assume for integer k, log3(k)<k (3is base) now we want to show log3(k+1)<k+1 so we have ...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so 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
 5 years ago
Best ResponseYou've already chosen the best response.0yea maybe we should base 3 on both sides to get rid of log

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you mean exponentiate?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you cant get rid of log with log :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0can you help me with surface area of integral

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0surface area of a solid?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0surface area of a torus

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so you cant finish the proof with log, no problemo

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how do you find the question i posted earlier

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this website is so confusing

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0are you given the equation for the torus or is it a parametric curve/

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Calculus 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  ( xR)^2) [ 1 + (2(xR)/(2sqrt( r^2  ( xR)^2) ^2 ]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we are given a circle , and its going to generate the torus by revolving about the y axis

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the final answer should be 4pi^2 R*r

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0so that last thing you have the square cancels the sqrt

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0ok im just going to do it on paper and scan it and you can open it give a few min

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0right, good :) ok here is what i got

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0integral 2pi y dS, remember ds can be sqrt 1 + dx/dy ^2 or sqrt 1 + dy/dx ^2

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0we need another pi in there somewhere

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0since are answer has pi^2

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0this is what i got using the integral you found we can get times another pi it would be correct

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0don't look at the name of the file. i know we are not finding volume

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0right, its surface area, here lets use twiddla

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if you dont mind, we can draw on whiteboard
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