When Paul crossed the finish line of a 60 meter race he was ahead of Robert by 10 meters and ahead of Sam by 20 meters. Suppose Robert and Sam continue to race to the finish line without changing their rates of speed will Robert beat Sam?
2007-12-29 15:49:20 · 9 answers · asked by GF 1 in Science & Mathematics ➔ Mathematics
Sorry I forgot to mention....how many meters will Robert beat Sam by?
2007-12-30 05:54:11 · update #1
Yes.
2007-12-29 15:54:06 · answer #1 · answered by NARAYAN RAO 5 · 0⤊ 0⤋
observe that if x + a million/x is an integer, so would be (a million + a million/x)^n for any beneficial integer n. we can attempt for some small n's first. to illustrate, n = a million, 2, 3, 4. If n =a million, needless to say. If n = 2, then x^2 + a million/x^2 = (x + a million/x)^2 - 2x(a million/x). considering (x + a million/x) is an integer, so would be (x + a million/x)^2 and 2x(a million/x) = 2. subsequently, x^2 + a million/x^2 could desire to be an integer using fact the RHS of the identity is the version of two integers. If n = 3, then x^3 + a million/x^3 = (x + a million/x)^3 - 3 x^2 (a million/x) - 3 x (a million/x)^2 = (x + a million/x)^3 - 3x - 3(a million/x) = (x + a million/x)^3 - 3(x + a million/x) is an integer! If n = 4, then x^4 + a million/x^4 = (x + a million/x)^4 - 4 x^3 (a million/x) - 6 x^2 (a million/x)^2 - 4 x (a million/x) ^3 = (x + a million/x)^4 - 6 - 4(x^2 + (a million/x)^2). We see that the 1st 2 words are integers and the x^2 + (a million/x)^2 interior the third time era is likewise integer that's proved previously for n = 2. of course, we gained't verify the fact for each n, so the excellent element is do is to apply induction, certainly the reliable induction rather of the vulnerable induction. the version between vulnerable induction (or the induction we use oftentimes) and robust induction is: vulnerable Induction: assume the fact is authentic for n = ok, coach n = ok + a million is authentic. reliable Induction: assume the fact is authentic for all n <= ok, coach n = ok + a million is authentic. right here, we could desire to constantly use reliable Induction, that's, we assume the fact x^n + a million/x^n is authentic for all n <= ok, we could desire to coach x^n + a million/x^n is authentic for all n = ok + a million: x^(ok+a million) + a million/x^(ok+a million) = (x^ok + a million/x^ok)(x + a million/x) - (x^(ok-a million) + a million/x^(ok-a million)) is an integer considering by utilising our assumption, every time n <= ok, x^n + a million/x^n is an integer, so (x^ok + a million/x^ok), (x + a million/x), and (x^(ok-a million) + a million/x^(ok-a million)) are each and every an integer the place n = ok, a million, ok - a million respectively. QED
2016-10-20 09:02:02 · answer #2 · answered by ? 4 · 0⤊ 0⤋
If Rob's and Sam's speeds were constant all along, then Robert will win.
However it is possible that Sam had a bad start and was catching up. In that case, given more distance he would pass Robert.
2007-12-29 15:57:35 · answer #3 · answered by Dr D 7 · 1⤊ 0⤋
Of course he will. Robert runs faster than Sam (by 25%), is 10 m ahead of Sam, and has half as far to go.
2007-12-29 15:57:53 · answer #4 · answered by Philo 7 · 0⤊ 0⤋
Knowing how math books word problems, I think I can safely say that part of this problem is missing. Somewhere in there it should tell you what Sam and Roberts current speeds are.
2007-12-29 16:01:49 · answer #5 · answered by zenock 4 · 0⤊ 0⤋
Yessir.
2007-12-29 15:59:20 · answer #6 · answered by Anonymous · 0⤊ 0⤋
Yes!
2007-12-29 15:57:19 · answer #7 · answered by Anonymous · 0⤊ 0⤋
yes.
i feel so bad that you don't have any stars
im going to give you a star because this question is interesting because the name paul is cool. well not really, but whatever
2007-12-30 00:42:59 · answer #8 · answered by Whatupdawg 3 · 0⤊ 0⤋
Of couse, he will
2007-12-29 16:53:03 · answer #9 · answered by someone else 7 · 0⤊ 0⤋