my young son is having problems understanding the concept of DIVISION sums.  Does any one have an easy way?
535880
How old? Teach addition properly. Specially addition of same numbers repeatedly. It would make him understand multiplication. Give him sticks(match box, ear buds, tooth picks) and ask to give equally to friends, parents etc. Let the concept come in his own way. You are just to help not to stress. With thread/rubber band you may make 'tens' bundle with tying 10 sticks together. And show how 1 n 1 makes 11. Just some confusion is to be cleared. I think 1 week is enough. Be patient have patience.
1091850
This is just something the person who said PEMDAS left out. \n\nYou need to solve it like this: Parenthesis, Exponents, Multiplication-Division, Addition-Subtraction. \n\nSay you have this problem(an example):\n\n3x4/5\n\nYou have both multiplication and division in this problem, which are tied for the third step. Once this happens, you have to solve the problem left-to-right. In this case, you would multiply 3 and 4 first to get 12, then divide 12 by 5.  Addition and Subtraction works the same way.
734575
A place where you can get help cheap, or free is at you local public/private university.  Math Majors would love the extra money to help your son with fractions.  You could prolly pay 20 dollars per session.  Also I know some universities offer free tutoring.
136789
That's an extremely good question. \n\nThe answer is simple. Before that, i will explain what your question means to those who don't get. Well, it means that we must get 24 using 1,3,4 and 6 any number of times however involving operations +,-,* and / only once. \n\n  Then, the answer is very simple :\n\n  {[(6*6)/(3-1)]+[6]}  =  24\n\n---------------------------------------------------------\nWOKRING :\n\n {[(6*6)/(3-1)]+[6]}   =  {[(36)/(2)]+[6]}\n                       =  18 + 6\n                       =  24\n\n----------------------------------------------------------\n\nI hope you post more questions like these. They evoke a lot of thought. I hope the solution serves th purpose.
891438
I used to also. I'll try to give an explanation: forgive me if I go over anything you already know. Alright, think of it this way: you know how to multiply fractions, correct? Alright, so when you multiply two fractions, such as 1/2 and 3/4, you multiply the numerators (those are on the top) (1x3) and the denominators (those are on the bottom) (2x4) and you have your answer: 1/2 x 3/4 = (1x3)/(2x4) = 3/8. Dividing is just the opposite of multiplication. So if you were dividing 1/2 by 3/4, you just flip over the second fraction and multiply: 1/2 / 3/4 = 1/2 x 4/3 = 4/6. \nHere's an answer that's a bit more advanced but might help you understand this better (at least it did for me): if you have some number, a, over some number, b, it can be written this way: a/b. Now, if you divide it by some other fraction, c/d, it can be written this way: \na/b x 1/(c/d). This is the same as saying that when you divide 6/1 by, say, 2 (which equals the fraction 2/1), it is the same as if you are multiplying 6/1 (6/1, or six divided by 1, equals 6) by 1/2. See for yourself: 6 / 2=3, 6 x 1/2= 6/1 x 1/2 = 6/2 which when you reduce (divide top and bottom by common factor, in this case 2) it equals 3/1, or 3. \n1/(c/d) is the reciprocal of (c/d). The reciprocal basically means you flip the top and bottom: for instance, the reciprocal of 2/3 is 3/2, the reciprocal of 1/2 is 2/1, or 2, the reciprocal of 4/1, or 4, is 1/4, etc. So if you flip over c/d, you get: a/b x d/c, which if a=1, b=2, c=3, and d=4, is exactly the same as what we had above.
783134
You were close. Instead of looking at the remainders after division by 2 (1 if odd, 0 if even), you should have considered division by 3.\n\nAll integers can be written as 3k, 3k + 1 or 3k + 2. (If you're wondering, "Why not 3k - 1 or 3k + 5?", the answer is that they are equivalent to 3k + 2, just for a different k.) Their squares are 9k^2, 9k^2 + 6k + 1, and 9k^2 + 12k + 4. These then can be reduced to 3k, 3k + 1, and 3k + 1, for different k's. Thus, squares are always evenly divisible by 3 or exactly one more than a multiple of 3.\n\nSo by the Pigeonhole Principle, there are 3 numbers that give 3k or 3 numbers that give 3k + 1. In the first case, choose those 3. In the second case choose those 3.\n\nYour questions are so much more fun than those "Be my calculator" ones.
123496
You have to use the BODMAS rule.. Brackets OF Division Multiplication Addition & Subtraction (BODMAS)\n\n20 + 5[6+5(4x2-5)]x2+4\n\nLet us resolve the inner most bracket (4x2-5)\nThere is no bracket, so do the Multiplication first 8 - 5 = 3\n\nso the expression becomes: \n20+5[6 + 5(3)]x2+4\n\nNext we do the [6+ 5(3)] bracket where again multiplication comes before addition.. [6+ 15] = [21]\n\n20+5[21]x2+4\n\nThis is the slightly tricky part.. we use the BODMAS rule again to resolve the multiplication first:\n5[21]x2 = 5x21x2 = 5x42 = 210\n\nThe problem is now a simple addition 20 + 210 + 4 = 234
58450
use the distributive property (poor u, u seem to be having a lot of trouble with your math homework tonight!! hope this helps:)\n\nthe property states that a(b+c)=ab+ac, basically you take the "a" and multiply it by the other 2, or distribute it.  say a=2, b=3, and c=1.  your equation would be 2(3+1) and, using the distributive property, make it become 6+2 which equals 8.  so its the same when your using x.  3(-4x+7)= -12x+21!!! yay!! hope this helped!
559152
are u sure its not supposed to be the other way around. Cuz that would make a little more sense.\n\nBut if not, then it would probably be:\n1 + ((4s squared - 11) / (s to the 4 - 3s squared -11))\n1 the quotient. the rest the remainder.
443017
Set it up like a long division problem:\n\nx - 1 ) x^5 + x^4 - 2x^3 + 2x^2 - 3x - 7\n\nNow focus on the x terms.  x is on the outside, and x^5 is on the inside.  Just like long division with numbers, how many times does x go into x^5?  x^4 times.  That goes on top, and then you multiply that by x - 1 and subtract, again like normal long division.\n\nx^5 + x^4 - 2x^3 + 2x^2 - 3x - 7\n-(x^5 - x^4)\n--------------------\n2x^4 - 2x^3 + 2x^2 - 3x - 7\n\n\nRepeat process...x goes into 2x^4 2x^3 times:\n\n2x^4 - 2x^3 + 2x^2 - 3x - 7\n-(2x^4 - 2x^3)\n--------------------\n2x^2 - 3x - 7\n\nx goes into 2x^2 2x times:\n2x^2 - 3x - 7\n-(2x^2 - 2x)\n--------------------\n-x - 7\n\nFinally, x goes into -x -1 time:\n-x - 7\n-(-x + 1)\n--------------------\n-8\n\nSo you have x^4 + 2x^3 + 2x - 1 with a -8 remainder, which you can write as -8/(x - 1).
514419
3x+6/x2-9 / 6x2+12x/4x+12\nIs this all one problem?\n(3x+6)/(x^2-9) / [(6x^2+12x)/(4x+12)]\nWhen dividing by a fraction, flip the second fraction and multiply.\n= (3x+6)/(x^2-9) times (4x+12)/(6x^2+12x)\nFactor everything first.\n3x + 6 = 3(x+2)\nx^2 - 9 = (x+3)(x-3)\n6x^2 + 12x = 6x(x+2)\n4x + 12 = 4(x+3)\n(3x+6)/(x^2-9) times (4x+12)/(6x^2+12x)\n= 3(x+2)/[(x+3)(x-3)] times 4(x+3)/[6x(x+2)]\nNow you can cancel anything that is on top of the fractions with anything that is on the bottom of the fractions.\n(x+2) and (x+3) cancel.  3 times 4 is 12 and 12 divided by 6 is 2.\n= 2/[x(x-3)] or 2/(x^2 - 3x)
352026
I would like to be able to give you the answer but I don't think I would be doing you a favour.  If you are to get practice in division, it is better for you to work it out on your own.  Does your teacher know that you are getting homework answers here on Yahoo?
817481
We can use faux-algebra to solve these. You probably haven't taken algebra yet. With algebra these would be extremely easy to solve. It's why algebra was invented.\n\na. 16 x ___ has to be -42, because if we add 51 to it we get 9. Thus, ____ has to be -42/16, or -21/8.\n\nb. 2 x ____ has to be 22 - 9 x _____, because if we subtract 8 from it we get 14 - 9 x ____. And 22 has to be 11 x ____, because if we subtract 9 x ____ from it we get 2 x ____. Thus, ____ has to be 22/11, or 2.\n\nc. 9 x ___ has to be 3 x ____ + 3, because if we add 2 to it we get 3 x ___ + 5.  And 3 has to be 6 x ____, becuase if we add 3 x ____ to it we get 9 x ____. Thus, ____ has to be 3/6, or 1/2.\n\nWith algebra, no such thinking would be required. You just plug in the formulas and grind away. You'll see when you study algebra.
803007
I don't understand your question.  Do you mean using the associative law?\n\nabc*de = \ncba*de =\nbca*de =\nacb*de =\ncab*de = \nbac*de \nNow all of these are multiplied by the two ways to postiion de\n\nThat's 3! = 6\ntimes  2! = 2\n\nfor a total of 12.  \n\nFor a 5th grader I'm sure that you're just supposed to list all 12.  For convenience, note that I've put each letter in the middle and switched the other two.  For neatness, I'd recommende just putting another column to the right of each of these 6 with de and ed .\nWhen doing counting like this, called combinotorics, it's helpful to arrange things neatly.
1054159
First, get rid of the brackets (multiply the terms inside the brackets by -1):\n\n7-2x^2 +2x -2 -2x -5 +2x^2\n\nCollect like terms:\n\n2x^2 -2x^2 +2x -2x +7-2-5 = 0\n\nIt's easy to see the sum is zero.
808154
To the first part - to prove your division - you multiply back out.\n6 divided by 3 = 2     2 x 3 = 6\n\nbut 0 times ANYTHING is always 0\nand you cannot say 0 div by 0 = 439  although 439 x 0 = 0\n\nthe second part has me stumped - been out too long!
795656
The first thing you should remember is the phrase PEMDAS.  This is shorthand for parenthesis, exponents, multiplication/division, and addition/subtraction.  This is the order of operations in math.  You should solve all equations in this order.  So for the first one, you should first distribute (multiply) -7 with (3-3x).  After that combine like terms, i got 25x -30; you can also factor that to be 5(5x-6).  The same steps apply for the rest of the problems.  I hope it helps.