Can you cross multiply inequalities?
538748
Yes, but you need to account for two cases. The full solution:\n\n(x-1) / (x-10) < 0\n\nx-1 < 0 [if x > 10] OR\nx-1 > 0 [if x < 10] (by "flip the sign" rule; x cannot be 10)\n\nFirst inequality can never be satisfied.\nSecond inequality means 10 > x > 1.\n\nYou do not use foil for that.\n\nClassical (and Kemmy) probably interpreted it as a number/phase line, which can be faster in a simple case like this. For more difficult inequalities it's best to write out steps.
17334
Geometric -- multiplying the previous term by 4
543381
Yes. the RHS of the first is equivalent to xor(P,Q). This becomes an inclusive or when the or(Q,...) is added. Not(P) as suggested above doesn't give the correct result when Q and P both = true.
408554
yes. take out an x \nyou answer should be x(x+3)
720161
Yes, I can:\n4! + 4 \n=\n(4 x 3 x 2 x 1) + 4 = 28\n\n(BTW, 4squared + 4squared = 32)
217638
simple answer, no.\ncomplex answer, yes.\n\nbut it is not simple, too many hoops, and you dont know if you are over-priting.\n\nbest to find another solution besides user tagging.
754121
Given:\n7<=3-2p<11\n\nYou want to isolate p, and as long, we only add and subtract its OK as long we do it in every term in the inequality.\n\nOf cause, we want to add -3 in the term in the middle, and therefore in rest of the expression, we get:\n7-3<=3-3-2p<11-3 <=>\n4<-2p<8\nNow we want to divide with 2 and we can do so, as long, we do it in all term, but we also want to multiply with -1, but it's also OK if we remember to reverse the inequality sign. \nThis use to be the point of confusion in calculation with inequalities\nHowever, it is easy to se that it is true\nEx 7<8 multiply with -1 gives\n-7<-8 but that not true therefore we have to reverse the inequality sign, if we multiply with a negative number.\n\nBack on track:\n\n4<=-2p<8<=>\n-4/2=> p >-8/2\n.......NB!!.NB!!!\n -2=>p>-4 \nOr if you better like the result in adequate form\n\nx={ x in R l (-4<p<= -2)}\n================
780452
Solve it just like a regular equality:\n\nx-6 ≤ 7\n\nAdd 6 to both sides, to get:\n\nx ≤ 13  ← this is your answer\n\nWhen working with inequalities, you can add and subtract from both sides just as you would in a regular equality.  You can also multiply and divide both sides, but the trick to keep in mind is that if you multiply or divide by a negative number, you need to change the sign!  For example, if you had:\n\n-2x ≤ 10\n\nDivide both sides by -2 AND REVERSE THE INEQUALITY to get:\n\nx ≥ -5\n\nHope that helped!\n\n~ ♥ ~\n\nEDITED TO ADD:  A "regular equality" would be where you have an equal sign rather than a < or > sign.  So x-6 = 7 is an equality.\n\nGood luck kid ... sounds like you will need it!
1148067
You elevate the number to the power 2 like this:\n\n                                    x^2\n\nYou should still do the normal product but it is easier to thing you et the square of a certain number and most calculators already have the elevate to the second power function integrated\n\nCynthia
912333
Well you can, in a sense.\nI don't know a SIMPLE way to do it, in the sense that there is a SIMPLE way to do matrix multiplication. But maybe the following explanation will be helpful:\n\nIf [A] x [B] = [C] (i.e., if the product of matrices A and B equals matrix C), then we can say that [C] / [B] = [A]. So in that sense, we are dividing C by B.\n\nBut what if we don't know matrix A. What if we know only [B] and [C]? How can we divide [C] by [B] to find [A]?\n\nThe answer is that we find the "inverse" of matrix B, and then MULTIPLY matrix C by this inverse matrix (which is shown as [B]^-1). We define [B]^-1 as the matrix for which [B] x [B]^-1 = the identity matrix. (The identity matrix is a square matrix that has 1's on its main diagonal and 0's everywhere else.)\n\nThe problem, then, becomes how to find the inverse of [B]. Fortunately, there are procedures for this, and you can probably find them on line using Google. Not all matrices can be inverted. But if B can be inverted, then you have:\n\n[C] / [B] = [C] x [B]^-1 = [A]\n\nNote that if we substitute [A] x [B] for [C], then this becomes;\n([A] x [B]) / [B] = [A] x [B] x [B]^-1 = [A] x identity matrix = [A]
910422
no \n3x-1>1 \n3x>2 \nx>2/3
765264
It's actually just simple mathematics.  You tally up the unique results of your cross and divide by the total number of the results in the cross.  For example, let's say that you cross AB x AB.  With this cross, your results are 1 AA:2AB:1BB 1/4 is AA, 1/2 is AB, and 1/4 is AA.  4 is just the total number of results.
44629
You can't, the mountain climber is a SCALAR (the cross product requires two VECTORS)
937572
No because 2+ 4 = 6 < 8\n\nsum of 2sides must be greater than 3rd side
368963
Hello i dnt think its possible but u can try with AIEEE this is equilent to B.E
751318
x^2-x-6>0\nor, x^2-3x+2x-6>0\nor, x(x-3)+2(x-3)>0\nor, (x-3)(x+2)>0\nor x>3 and x<2\n\nhope this helps :)
1144280
i'm not exactly sure what you want to multiply, but the only thing related to sets and multiplication is cartesian product, which would be an array (ie ArrayList) of TreeSets - you would need a new class for cartesian product (as it's not a TreeSet). There you could easily add another TreeSet in array (as it needs to be ordered and TreeSet is not).\nother than that, cartesian product of two sets is a set with elements (a,b) where a is from first and b from second set. so your product would return TreeSet with elements of class ie Pair which has Object first and Object second as its properties. combine each of first set with each of second in class Pair and put them in your new TreeSet something like this:\n\nclass Pair {\nObject first;\nObject second;\n}\n\npublic static TreeSet product(TreeSet first, TreeSet second) {\nTreeSet rval = new TreeSet();\nwhile (i = first.next()) {\nwhile (j = second.next()) {\nPair tmp = new Pair();\ntmp.first = i;\ntmp.second = j;\nrval.put(tmp);\n}\n}\nreturn rval;\n}