i dont get this math problem.?
927660
Hi, \n\nThey way to do this problem is to be careful in the way you set up your equations. \n\nD= dimes\nQ= Quarters\nN = Nickels\n\nSince there are twice as many dimes as quarters, you'd have to multiply the number of quarters by 2 to get it to equal the number of dimes. \n\n2Q = D \n\nSimilarly, if there are two more nickels than quarters, you'd have to add 2 to the number of quarters to get the number of Nickels\n\nQ + 2 = N\n\nNow, The total value (the number of each coin times the value of the coin.) is $7.60\n\n(0.25)Q + (0.1)D + (0.05)N = $7.60\n\nNow you have three equations with three unknowns. Now all you have to do is isolate variables, and substitute. \n\nHope that helps, \nMatt
1102373
i wont answer.... log onto www.mathguru.com and discover solutions with explanations.
843943
4, 0, -2, -2, 0\n\nFigure it out! Don't be lazy!
810076
11^(0.35+7) = 3 this is a contradiction\n\nx=3 is a solution but the rest is just a mess.\n\nDo you have any idea of what you are doing?
801562
I dont understand the question
567547
I don't get this riddle.
156773
when x=3\na) 9\nb) 38\nc) 23\nd) 9\ne) 4 and 2/3\n\nwhen x=2\na) 7\nb) 30\nc) 17\nd) 7\ne) 6\n\nwhen x=1\na) 5\nb) 22\nc) 11\nd) 5\ne) 10\n\nwhen x=0\n\na) 3\nb) 14\nc) 5\nd) 3\ne) ERROR, can not divide by zero
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
137681
Ok...\nI see what you are doing...\nYou merely substitute the number in for x...\nso the problem becomes 6-2*(-1)\n6+2\n8
821450
go to a math website off yahoo. it should help.
757452
x=-54\nthis one is incorrect.\ny=-2/3\nx=1\ny=54/2.7\ny=-2 ~
794793
x = 3. Try entering it on your calculator.
912865
bring the a's to one side... u then get 9a = 27 ... divide and answer is a = 3
611639
do the math moron!!!!!!
617282
are u sure its x2? shouldn't it be 2x+10x=(a number) usually there is a number at the end of the equation and u have 2 solve for "x". add details to ur question and then i may be able 2 help u.
1146358
I’m not seeing how this is solvable.  I’m getting two possible answers.\n\nBut here’s how you should work the problem.  \n\nFirst of all, I think you pretty much have to assume that the x,y < 10.\n\nSo we know that Bill can’t tell what his two numbers are just by looking at the product.  As it turns out, there are very few products that allow this to happen: they are 12,16,18,24,36.  So you factor out all the number pairs that would make those products, and the sums are 7,8 (twice), 9, 10(twice), 11(twice), 12, and 13.\n\nSo we know that Albert is seeing one of those sums.  If he sees 7, then he knows that the only way for his numbers to add up to 7, and multiply up to either 12, 16, 18, 24, or 36 is for the numbers to be 3 and 4.  But he says nothing.  Therefore, he isn’t seeing 7.  Neither is he seeing 9, 12 or 13, for the same reason.  So, now we know the sum must be either 8,10, or 11.  And now Bill knows that.\n\nSo, if Bill were seeing 12 as the product, he would know that the numbers were 2 and 6, since they add up to 8.  But he says nothing, so he’s not seeing 12.  Likewise, if he were seeing 18, he would know that the numbers have to be 2 and 9.  But he doesn’t, so he’s not seeing 18, and he can’t be seeing 36 either.  So that means he must be seeing either 16 or 24.\n\nAnd now Albert knows this, so either Albert is seeing a sum of 8, and knows that the numbers are 4 and 4, or he’s seeing 11, and the numbers are 3 and 8.  I just don’t know how to eliminate either of those options.
920569
It looks like you are trying to solve a literal equation, which is an equation with more than one variable, and you have to solve for a specified variable.  You can use the SAME rules to solve this kind of equation as you do with regular equations (those with numbers).  For example, if you had 8=2(b)(9), you would divide by 2 and then again by 9 (which is like dividing by 18).  \n\nIf your problem is a=(1/2)bc^2, then you would divide by 1/2 and then again by c^2.  Remember that to divide by a fraction, you multiply by the reciprocal.