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I also need some help on this problem:
that satisfy for all nonzero x.
Find all functionsI know I'm supposed to substitute (x-1)/x for x, but I don't seem to be getting anywhere with that. Thanks in advance!
Being a homework problem I can give only hint.
In the previous problem you went from A to B and directly returned to A. Now try going from A to B , B to C and from there to A.
Thanks thickhead ,
What will be the probability if A cannot get through the centre of E but B can ?
Product of the 2 probabilities integrated over the region gives the required probability.0.08618
http://www.wolframalpha.com/widgets/view.jsp?id=8ab70731b1553f17c11a3bbc87e0b605
I am originally from Karnataka, now staying in Thane ,Maharashtra.
Yes, something similar but the pronunciation is clearly "d" but not "dh" associated with it.
Daabaa is roadside eating place but it has no connection.
No reason why.
Some people used to call it "daabaa" like "daabaa y by daabaa x'
Please see bobbym's signature.It says 'In mathematics, you don't understand things. You just get adjusted to them". It is true for percentage.Percentage is defined as (final-initial)/initial *100. If a quantity changes from 100 to 200 it is 100% increase. From 200 if it comes back to 100 it is only 50% down. It is defined this way to reduce heart attacks resulting from crash in stock market.Actually every body is confused about the definition but is not aware of it.
The point is the quickest time calculated on the basis of rowing speed of 2 mi/hr is not valid now since the rowing speed has changed. Now i want to know how you handled the first part so that I can continue from that.It won't help if the 2 parts are disjoint.
Simple. Show that quadrilateral IFBD is cyclic. And use the property of inscribed angle vs.measure of arc.Finished in two lines.
I think my solution is clear 9/sqrt(13) matching with your book answer.I never mentioned 9/sqrt(5).And what you did not understand?My post no.8 or the answer?
Yes. I am getting
continuing with my previous post#8fulfill the tub= fill the tub full ; However that is Indian English.
That is excellent.but my query was for your posting in #14 removing 4 from A and weighing it (3:3) vs.C.
but in any way Salem_Ohio's start of 8,8,1 is to be admired as it is less taxing and shows the strength of binary division to the core(8 vs 8 ;4 vs 4; 2 vs 2 when rotten apples break down) even though it requires 3 weighings.
With 7,7,3 how do you deal with 2:2:1 break of rotten apples?
Hi Salem_ohio,
As I had written earlier I had a wrong start.6:6:5 combination. but you had a better start. 8:8:1
on first weighing if A and B are equal, you conclude either it is all good apples or both lots have 2 rotten apples each.
Go for second weighing 4:4 of any of the lots. Split verdict indicates 1 rotten apple on each side. the third weighing 2:2 is the decider as this apple can not be divided.
My start with 6:6:5 combination was bad as it gave rise to 0:0:5,1:1:3 or 2:2:1 distribution of rotten apples.on second weighing C lot had to be weighed against A or B with one apple removed.It failed against 2:2:1 distribution. But wait, this also looks plausible.Let us say on second weighing we remove one from B and weigh the remaining 5 with C. a 0:0:5 or 1:1:3 distribution will make it unbalanced.If they are balanced it means it is 2:2:1 distribution and the one removed is a rotten apple. Add it to A to make make it 7 apple lot with 3 rotten apples in it.Add 2 apples from C to B. Now B may at the most contain 2 rotten apples and will be unbalanced against A in the third weighing.
In polar coordinates r is always positive.