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

- katieb

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- anonymous

i believe its $661.76

- myininaya

cantorset can we use induction on your problem? do you know what that is?.

- anonymous

yes we can

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## More answers

- anonymous

how did you find me, lol

- myininaya

i went to the homepage and you can see what everyone is doing

- anonymous

the chat program doesnt work for me

- anonymous

ok lets do this one
for all n>=5 , n^3 < 3^n

- myininaya

ok

- myininaya

for n=5, we have 5^3=125<243=3^5. Sold the inequality holds for n=5

- anonymous

right, basis case

- 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

it seems that you are going over logs so i'm pretty sure we may have to involved that somehow

- anonymous

hmmm, they may be unrelated though. here is what i got ,

- anonymous

that was a different question

- anonymous

is there a way to post without having to click my mouse on post?

- myininaya

no i guess not

- anonymous

show that (k+1)^3 < 3^(k+1)

- anonymous

k^3 + 3k^2 + 3k + 1 < 3^(k+1)

- myininaya

omg i got it

- myininaya

i got so excited i clicked out sorry.
so we have (k+1)^3

- 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

(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

i got excited ignore what i wrote

- anonymous

(k+1)^3

- anonymous

my proof isnt elegant, but i think it works, might have to do some side induction arguments for the intermediate steps

- myininaya

you are right i got excited

- anonymous

also you better use parenthesses :) more often

- anonymous

the log proof is A LOT easier

- myininaya

yep

- anonymous

what did you get?
log_3 n < n if n < 3^n , assuming we raise both sides to 3

- 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

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

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

yea maybe we should base 3 on both sides to get rid of log

- anonymous

you mean exponentiate?

- anonymous

you cant get rid of log with log :)

- anonymous

can you help me with surface area of integral

- myininaya

surface area of a solid?

- anonymous

surface area of a torus

- anonymous

so you cant finish the proof with log, no problemo

- myininaya

one thing at a time

- anonymous

how do you find the question
i posted earlier

- anonymous

this website is so confusing

- myininaya

are you given the equation for the torus or is it a parametric curve/

- 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

we are given a circle , and its going to generate the torus by revolving about the y axis

- anonymous

the final answer should be 4pi^2 R*r

- myininaya

so that last thing you have the square cancels the sqrt

- myininaya

ok im just going to do it on paper and scan it and you can open it give a few min

- anonymous

right, good :) ok here is what i got

- anonymous

integral 2pi y dS, remember ds can be sqrt 1 + dx/dy ^2 or sqrt 1 + dy/dx ^2

- myininaya

we need another pi in there somewhere

- myininaya

since are answer has pi^2

- myininaya

this is what i got using the integral you found we can get times another pi it would be correct

##### 1 Attachment

- myininaya

don't look at the name of the file. i know we are not finding volume

- anonymous

right, its surface area, here lets use twiddla

- anonymous

http://www.twiddla.com/521818

- anonymous

if you dont mind, we can draw on whiteboard

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