Recent content by tara123

sorry i meant sin(1)=PI/2 but just one other thing, I've bin thinking and y=sin(1/1-x) o<x<1 0 isn't even contained in the set...so how is it a boundary point if 0 is strictly less than x? Help!

yah so for x=0 y=sin(1)=pi and for x=1 i just look at what the graph does as it approaches 0?

yah if u allow for the restriction of n>=8 if u have 8^3=512 and 9^3=729 then there's 23^2=529 and 24^2=576 both of which are between the cubes..

derivative of x^1/2 = 1/2 x^(-1/2)

Induction is the best way to show this proof. the way you should start would be: Let P(n) be the statement 0<x<y then x^n<y^n where n=natural numbers Then your base case is where x and y are the smallest natural numbers such that they apply to your restrictions, then of course you assume...

Im really good at number theory but how to show this statement has me stumped! "Show that among the positive integers greater than or equal to 8, between any two cubes there are at least 2 squares"

Do you understand what mod refers to?

Above ur taking the new function of y to be sin(x) right? .. this seems wrong lol no offence

haha yah don't feel bad I am pretty lost too!

Question on Boundary Points! Determine all boundary points: S={(x,y); 0 < x< 1 and y= sin(1/(1-x)} I'm really confused! I don't understand how I can actually find all the points.. can anyone get me started?? The help is appreciated!