2 Scheme Programming questions I need help with???
403150
you seem to only have trouble figuring out how to take in parameters. if you take some time to read from the link below perhaps it would become clearer to you\n\nyou can work out your logic like this\n\nsomeone is a teenager if:\n  age is greater or equal to 13\n   AND\n  age is less than or equal to 19\n\nsame as:\n  age >= 13 AND age <= 19\n\nhow to write age >= 13 in scheme?\n(>= age 13)\n\nhow to write age <= 19 in scheme?\n(<= age 19)\n\nand how do you AND these results together?\n(and (>= age 13) (<= age 19))\n\nso this tells you age IS a teenager.\n\nAll the operators we've used thus far {and, >=, <=} are binary operators. this means that they each take two parameters. go ahead, look back until this makes sense.\n\nto see if it is NOT a teenager, then we use the not operator. you've tried to use it already, but you passed to it two parameters. this is incorrect because 'not' is a unary operator, not binary. so you only pass it one argument:\n\n(not (and (>= age 13) (<= age 19)) )\n\nand that's your logic.
837082
Hint: Let C=X&#92;O and let f(x)=dist(x,C). The other way is trivial.
782057
look at a few examples and it will make sense\n\nfor example \n\n02 (where the 2 is small)   -> _ O    (the _ is where a number goes)  (2)
703485
1) 9 * (t + 1)\n2) 8 * (v + 7)\n3) 25 * (5 - h)
418804
I'm a bit rusty with BASIC but both of them look ok (assuming the syntax is good).\n\nThe second one is more efficient because of the FOR loop.\n\nIf you always need to calculate the average for 6 students then you don't need to prompt for the number of students, although it does make the program more useful.
1153402
Slope equals rise over run.\nGee, this sounds like fun.\ny = mx + b\nWhat a wonderful sight to see.\nSlope equals rise over run.\nNow my poem is done.
518698
use the inputs i0 - i3 as input and of the 3 selector lines make the last one (s2) 0.\n\nThe possible value will be \n\ns2 s1 s0 out\n0  0  0 - i0\n0  0  1 - i1\n0  1  0 - i2\n0  1  1 - i3\n\nenjoy :)
314789
How to work it? It's not very difficult. You just need to write your code and compile it.\n\nUnfortunately I can't send you anything, but you can check out my first source for more information.
752959
Problem 1: Find an exponential function of the form f(x)=ba^x+c with y-intercept 2, horizontal asymptote y=-2, that passes through the point P(1,4). \n\nYou find the y-intercept by replacing x with zero, which gives f(x) [which is y on the graph] = 2, so b + c = 2 (since a^0 = 1)\n\nThe horizontal asymptote is y = c, so c = -2.  That means b must be [you figure it out]\nNow it is y = 4a^x - 2.  This must be true when x=1 and y=4.\nSo 4 = [you figure it out; answer deleted due to complaint from below]\n\n\nProb. 3.  What is the amount in the account if $500 is invested for 10 years at the annual rate of 5% compounded continuously? \n  Just plug in the values.\nA = 500 e^(.05 x 10)\n\n[answer deleted]
75512
hey champ(Anirudh) !! what happened... so many errors in ur code?\ni hv'nt done much.. just removed the errors.. \nwell ur code was just like an error removing question... that v used to solve in the boards!  i loved solving those easy pickings!!\n\nanyways.. heres the exact code.. \n\n\n#include<iostream.h>\n#include<conio.h>  //for clrscr() & getch()\n\nvoid main()\n{\nclrscr();         //optional\nint array[10],i,least,greatest;\n\n//Input Numbers\ncout<<" Input 10 numbers separated by spaces  : &#92;n";\nfor(i=0;i<10;++i)\n   {\ncin>>array[i];\n   }\n\ngreatest=least=array[0];\n\nfor(i=1;i<10;++i)\n   {\nif(array[i]>greatest)\n     greatest=array[i];\nif(array[i]<least)\n   least=array[i];\n   }\ncout<<"&#92;nLeast in the list is "<<least;\ncout<<"&#92;nGreatest in the list is "<<greatest;\ngetch();//optional\n}
730496
iteration1: a = 3, b = 2\niteration2: a = 13, b = 8\niteration3: a = 51, b = 30\nloop exit:  a = 81, b = 30 // condition a = a + b executed before exit
186035
1) What is the difference between F(n) and f(n)? \n\n2) I cannot comprehend the second condition.\n\nEdit your question, please.
677039
(1) { -2x-y-z=0\n(2) { -x-2y-z=0\n(3) { -x-y-2z=0\n\nLet's look at (2):\n-x-2y-z=0 => x=-2y-z\nNow at (3):\n-x-y-2z=0 => x=-y-2z\nNow we can see that\n-2y-z=-y-2z => *** y=z ***\n\nNow let's put these values into (1):\n-2x-y-z=0 => -2x-y-y=0 => -2x-2y=0 => 2x=-2y => *** x=-y ***\n\nFor now, we know that x=-y=-z\nLet's put all of what we know into (2), and express everything using 'x':\n-x-2y-z=0 => -x-(-2x)-(-x)=0 => -x+2x+x=0 => 2x=0 => *** x=0 ***\n\nThus, x=0 => y=-0=0 => z=0\nSo the only solution is (0,0,0).\n\nGood luck!
839453
1) 21.69\n2) 14.10\n3) 7.59
993026
1. yes.\n\n2. one\n\n3. 2 equal segments (this one's a theorem)
120457
1) 4\n2) 15241578750190521
500368
The links below have good info. The first briefly addresses your first question, and the second is a good brief history of the problem which provides some insight as to the difficulties encountered. I actually played around with this problem a little on an elementary level and it is interesting. Cheers.\n\nhttp://www.sciam.com/print_version.cfm?articleID=000DBA55-B698-1C71-9EB7809EC588F2D7\n\nhttp://www.geocities.com/fermatnow/flt/