Can you solve this math problem?
481530
OK -I'll call seating spots  1 2 3 \n                             4 5 6  Joe I put at 5 with  V ball at 4 and C Pres @ 6. that makes 1 the Preacher's kid.  Since Sally likes Vball and aint Preacher's kid, she sits at 3 or 6. Now Frank sits opposite Valdictorian Not Joe-that means  Frank sits at 6 (pres) and Sally at 3 is Valdic-torian .By now we know 2 is cheerleader since Joe cant be-FRank likes cheerleader and this crowd is hetero. Since 2 is cheer.. then Gina sits 1 and Betty is cheer..and that leaves Ron at 4 and Joe the B ball\nSO-preachers kid is Gina @ 1spot-\nBetty @2is cheerleader-\nSally is Valdictorian @ 3 \n-Ron is V ball @4-\n-Joe is B ball @5-\n-and Frank is \nclass president @ 6\nEasier if you start with a seat grid to see it
937830
T=m1*-9.8+m1*a\nT=m2*a--m2*-9.8\n\nset them equal and solve for a\n\ntrust me, i had a test on this today
858530
it is possible when X=33.02163079..... try it
984972
WHAT is the QUESTION ?\n\nThis is NOT an equation.\n\nYou can simplify the expression to:  4x+3\n\n(divide each CONSTANT in the BINOMIAL by 2)
774792
Let the length of side "b" be x ft.\nTherefore side a=2x and side c=2x+3\nAccording to the problem,\nx+2x+2x+3=93\n=>5x+3=93\n=>5x=93-3=90\n=>x=90/5=18\nTherefore.side a=2X18=36 and side c=36+3=39\nAnswer a=36ft. b=18 ft and c=39 ft
1118957
The first number has to be (x+6) and the second number is (x).  There are four possible combinations: 6,0 (60); 7,1 (71); 8,2 (82); 9,3 (93).  60 plus 8 (68) does not equal 06, or 6.  71 plus 8 does not equal 17.  82 plus 8 does not equal 28.  93 plus 8 does not equal 39.\nUnfortunately, this problem is impossible.\nSorry I couldn't be of more help.
1135408
OH NOES THIS IS LIKE A MATH PROBLEM!!!! i can't handle any more variables :( sry.\n\n\nhahahaha.
164104
maybe :\n\nKa = 1.3*10^-2
1020883
i am sure you can find a math tutor that will be happy to help you when you explain what the problem is.\n\nso what IS the problem, anyway?
522619
1 1 1 2\n1 1 2 1\n1 2 1 1\n2 1 1 1\n\nNo other case is possible, So n(A)=4\n\n\n1 1 1 3\n1 1 3 1\n1 3 1 1\n3 1 1 1\n\n1 1 2 2\n1 2 1 2\n1 2 2 1\n\n2 1 1 2\n2 1 2 1\n2 2 1 1\n\nNo other possibility\nn(B)=10
903533
ignore the decimals when u originally do the problem, then however many deciams there are in both of the decimals you're multiplying is how many there will b in the answer. i.e. if you multiply 6.78x4.234 together the answer will have 5 numbers behidnthe dceimal point
924405
Is it possible to have 2 equal signs in a problem?\nI am only in algebra but I haven't seen one like that before.\nSorry I couldn't help.
89158
r u crazy? i dont know maths
720161
Yes, I can:\n4! + 4 \n=\n(4 x 3 x 2 x 1) + 4 = 28\n\n(BTW, 4squared + 4squared = 32)
856711
because this has 2 unknowns it cannot be solved.\n\nbut I can write  x in terms of b\n\n2x = 13-b\nor x =(13-b)/2
70409
Go back to math class and figure it out!
884616
Possible .X....Y\noutcome\nHHHH......3....3\nHHHT......3....2\nHHTH......2....2\nHTHH......2....2\nHHTT......2....1\nHTHT......2....1\nHTTH......1....1\nHTTT......1....0\nTHHH......2....3\nTHHT......2....2\nTHTH......1....2\nTTHH......1....2\nTHTT......1....1\nTTHT......1....1\nTTTH......0....1\nTTTT......0....0\n\nHence,\nP(X=3,Y=3)=1/16\nP(X=3,Y=2)=1/16\nP(X=3,Y=1)=0\nP(X=3,Y=0)=0\nP(X=2,Y=3)=1/16\nP(X=2,Y=2)=3/16\nP(X=2,Y=1)=2/16\nP(X=2,Y=0)=0\nP(X=1,Y=3)=0\nP(X=1,Y=2)=2/16\nP(X=1,Y=1)=3/16\nP(X=1,Y=0)=1/16\nP(X=0,Y=3)=0\nP(X=0,Y=2)=0\nP(X=0,Y=1)=1/16\nP(X=0,Y=0)=1/16\n\n.................JOINT PROBABILITY DISTRIBUTION OF x & y\n\n........0.1....2.3.....|P(X=x)   \n...&#92;Y|__________|\n..X&#92;.|..........................|\n..__|..........................|\n....0|1/16.1/16.0.0.....|2/16\n......|..........................|\n....1|1/16.3/16.2/16.0|2/16\n......|..........................|\n....2|0.2/16.3/16.1/16|6/16\n......|..........................|\n....3|0....0..1/16..1/16|2/16\n......|..........................|\nP(Y=y)|2/16.6/16.6/16.2/16|1\n\nfOR 2nd part we have to find \nP(X=3|Y=1)=P(X=3,Y=1)/P(Y=1)=0\nP(X=2|Y=1)=P(X=2,Y=1)/P(Y=1)=(2/16)/(6/16)=1/3\nP(X=1|Y=1)=P(X=1,Y=1)/P(Y=1)=(3/16)/(6/16)=1/2\nP(X=0|Y=1)=P(X=0,Y=1)/P(Y=1)=(1/16)/(6/16)=1/6\nHence\n      conditonal probability distritribution of X given Y=1\n......X...|0.1.2.3..|TOTAL\n.....___|_______|_____\nP(X=x)|0.1/3.1/2.1/6.|1\n\nForget about those (.........)s. They are provided so that the tables do not collapse.