Could someone please show me how dto do long division - still can't get my head around it...
201140
Try this site:\n\nhttp://www.mathsisfun.com/long_division2.html
21618
Your computer has a calculator built in.  Failing that, try long division.  You're gonna have to learn how to do it anyway.\n\nSheesh.
54748
First get rid of 2x by putting t = 2x\n( Read "eye 1" =  and "eye = " etc... capital letter I doesn't show up well on screen)\nI1 = Integral (sec^3 (t) dt/2)\nFor simplicity I will write Int for integral sign\nI = Int sec^3 t dt = Int sec t sec^2 t dt\nApply product rule\n\nI = sec t * tan t - Int sec t tan^2 t dt\nI = sec t tan t - Int sec t (sec^2 t - 1) dt\nExpand\nI = sec t tan t - Int sec^3 t dt + Int sec t dt\nBut second term is -I hence you have\n2I  = sec t tan t + Int sec t dt\n\nThis is standard form to evaluate using \n\nInt sec t = ln tan (pi/4 + t/2) + c\n I = 1/2 (sec t tan t + ln tan (pi/4 + t/2))+ C\nI1 = 1/2 I
156190
OK, so the annual dividends are:\n\nTime 1   $5\n     2   $5×1.15\n     3   $5×1.15²\n     4   $5×1.15²×1.06\n     5   $5×1.15²×1.06²\n     6   $5×1.15²×1.06³\n     7   $5×1.15²×1.06³\n     etc\n\nDiscounting the first 5 payments:\n\n\n5v + 5×1.15v² + 5×1.15²v³ + 5×1.15²×1.06v^4 + 5×1.15²×1.06²v^5\n\n= 4.6296 + 4.9297 + 5.2492 + 5.1520 + 5.0566\n\n= 25.0172\n\nNow the PV of teh remaining payments is:\n\n 5×1.15²×1.06³( v^6 + v^7 + ...)\n=5×1.15²×1.06³v^6 (1 + v + v² + ...)\n\nInside the brackets is an infinite geometric series with first term a = 1 and common ratio r = v.  Using the sum to infinity formula, a/(1-r), we get:\n\n1/(1 - v) = 13.5\n\nMultiplying this by the  5×1.15²×1.06³v^6 bit is:\n\n13.5 × 4.9630 = 67.000\n\nHence the total PV is:\n\n67.000 + 25.017 = 92.017\n\nSince the share value should be the PV, we get $92.02\n\nHope this helps!
450453
First, you got the divisor & dividend backwards........\n\nFor addition the answer is called the sum.  Not sure about 3, 2\n\nSubtraction:  minuend - subtrahend = difference
361929
I don't know the names of the justifications, but here's the basic idea:\n\nAB=BC  <-midpoint rule, if b is the midpoint it splits into 2 = sections\nBC+CD=BD    (can be seen on the picture, rule of addition??)\nAB+CD=BD   <-transitive/substitution property (look at the first step)
674426
Okay I got the answer 3121 from Scott, but I had a mathematical strategy.\n\nI narrowed down the possibilities to all multiples of 5 times a number with numbers digit of 4 or 9 plus 1.\n\nThis is because to keep the chain of divide by 5 subtract 1 remainder going, we would start with a number x ending in either 4 or 6. We subtract x/5 and 1, so we are basically subtracting the whatever factor times 5 yields the closest muliple. The result from x-(x/5)-1 has to end in either 1 or 6, so that we have a multiple of 5 with remainder 1 and can keep the sequence going. Now if it ends in 1, the factor was even. Ends in 6, factor was odd. If factor was even, we need to subtract the factor plus one, an odd number, to get another number ending in either 1 or 6. If we subtract a number ending in 5, we get a number ending in 6. Therefore, if the factor is even it has to end in 4. And for numbers ending in 6, we have to subtract an even number (odd factor + 1)subtract a number ending in 0, so only factors ending in 9 work.\n\nI was then testing every factor ending in 4 or 9. Take the factor, multiply by 5, add one, subtract (factor + 1), see if the new number minus one divided by 5 yields a factor ending in either 4 or 9.\n\nI was up to 149 when I decided to refresh the page and saw the answer.  Jeez I would've had to go up to 624. -.-;;\n\nAs it is, I only got a chain of 3 which didn't end in a multiple of 5.\n\nSo if I had continued testing throughout the night I would've gotten it. :)
104588
http://en.wikipedia.org/wiki/Time-division_multiplexing
40793
With this given data, it is not possible\n\nalthough the area is totalling to be 70 for both cases (7*10 and two pieces of 8*8 + 1*6) but with one division only, this is not possible\n\nPLEASE DO LET ME KNOW IF THERE IS SOME SOLUTION (i found it very interesting)
425794
Concentrate on the ^2, because that will have the most significant impact on the value of the expression being raised to the second power:\n\nFirst I tried just raising the 4 to the second power.  4^2 = 16, so "n" / 16 = 6, therefore n = 96.  If you can group the other numbers to reach 96, you have an answer.  But working that out for a while got me nowhere, so I started working leftwards to bring the next number and operator into the square:\n\n(6/4)^2 = 2.25\n\nThat is not an integer so it probably won't work.\n\nEven bringing the next number and operator in, which is 2 -, I still end up with a non-integer:\n2 - (6/4)^2 = -0.25\n\nBut look -- the next number and operator are "8 x " so lets work that out:\n\n8 x (2 - (6/4)^2) = 8 x (2 - 2.25) = 8 x (-0.25) = -2\n\nso that might work if we can come up with a number which works out for ___ + (- 2) = 6\n\nWhat plus -2 is =6?  It is 8 of course.  Now we just need to group 3 x 5 - 3 + 2 to equal 8:\n\n(3 x (5-3)) + 2\n(3 x (2)) + 2\n(3 x 2) + 2\n6 + 2\n8\n\nAha!  We have it.\n\nSo then, or grouping is:\n(3 x (5-3)) + 2 + (8 x (2 - ((6 / 4) ^2))) = 6\n\nTechnically some of those parenthesis are extras because the order of operations says multiplications and divisions are always done first, you don't need to explicitly show that they are (but it helps keep things clear).  So this would be a simpler form:\n3 x (5 - 3) + 2 + 8 x (2 - (6 / 4) ^2) = 6\n\nBut anyway, putting it back in the more complex form and proving it true works like this:\n(3 x (5-3)) + 2 + (8 x (2 - ((6 / 4) ^2))) = 6\n(3 x (5-3)) + 2 + (8 x (2 - ((6 / 4) ^2))) = 6\n(3 x ( 2  )) + 2 + (8 x (2 - (( 1.5 ) ^2))) = 6\n(3 x   2   ) + 2 + (8 x (2 - (    2.25   ))) = 6\n(3 x   2   ) + 2 + (8 x (      -0.25        )) = 6\n(6) + 2 + ( -2) = 6\n6 + 2 - 2 = 6\n6 = 6 <<<<< TRUE!\n\n\nHere is another grouping that also works out:\n3 x (5-3) + 2 + 8 x ((2 - 6) / 4^2) = 6\n3 x    2    + 2 + 8 x     (-4 / 16) = 6\n3 x    2    + 2 + 8 x (-0.25) = 6\n3 x    2    + 2 - 2 = 6\n6 + 2 - 2 = 6\n\nand another, this one involves only integers in the proof so might be more what you are looking for:\n3 x (5-3) + 2 + (8 x (2 - 6)) / 4^2 = 6\n3 x 2 + 2 + (8 x (-4)) / 16 = 6\n3 x 2 + 2 + (-32) / 16 = 6\n3 x 2 + 2 + (-2) = 6\n3 x 2 + 2 - 2 = 6\n6 + 2 - 2 = 6\n6 = 6
321442
SOrry, I can't help you. If I had the formulas I could do it, but there is no way I remember that off the top of my head! Good luck
183907
Using an integral,\n\nWords = S 3t(3t^2 + 1)^(-1/2) dt\n\nlet u = 3t^2 + 1\nthen du = 6t dt\n\n= (1/2) S u^(-1/2) du\n= (1/2)(2u^(1/2))\n= u^(1/2)\n= (3t^2 + 1)^(1/2)\n\nEvaluating from 1 to 4,\n= (3(4)^2 + 1)^(1/2) - (3(1)^2 + 1)^(1/2)\n= 49^(1/2) - 4^(1/2)\n= 7 - 2\n= 5\n\nAnswer A.
453548
I understand your frustration in receiving no Answer yet to this question, but since I do not have any direct knowledge of how to perform the calculation(s) you need ... I hope the link below serves you well.  It appears to explain the mathematics behind all the concepts you mention.
297938
Above answer is ok except for not using the -- operator:\nNeed to start at 10 and to decrement i use the statement:\n\ni--;\n\nThis changes the problem somewhat.  Another problem in the first answer is that fact is not initialized, so it will always equal zero resulting in a division by zero error.
384935
[(x^3)/(2 - x)]^4/3 \ndy/dx = (4/3)[(x^3)/(2 - x)]^1/3[3x^2(2 - x) + (x^3)]/(2-x)^2\ndy/dx = (4)[(x^3)/(2 - x)]^1/3[6x^2 - 3x^3 + (x^3)]/(3(2-x)^2)\ndy/dx = 8x^3(3 - x)/(3(2-x)^7/3)
481030
The long hand method for dividing polonomials is much like normal division\n\n\nput the denominator to the left of your square root looking symbol and then your numerator under your that symbol....\n\n\n1.take the first term from the denominator and muliply it by whatever you need to in order to get the first term in the numerator in this case that is 3x(keep this in some part[if you use normal division method it will be on top of symbol as it is part of your answer)\n\n2. now multiply the operand(3x in this case) by your entire denominator for this= 3x^4 + 3x^3 - 27x\n\n3. now take the numerator and subtract what you got from it\n(3x^4 - 6x^3 - 14x^2 - 6x + 18) - (3x^4 + 3x^3 - 27x) = -9x^3 -14x^2 + 21X + 18\n\n4. now repeat steps 1-3, with your product from #3(every time) as your new numerator, untill you need a negative exponent to complete step 1\n\nyour overall answer is 3x - 9 - {(5x^2 + 21x - 63)/(x^3 + x^2 -9)}\n\nsorry my steps are so confusing but I rarely help others... so I came here to kind of practice for tutoring
64038
Divide the decimal number by 16.   Treat the division as an integer division.   \nWrite down the remainder (in hexadecimal). \nDivide the result again by 16.  Treat the division as an integer division.   \nRepeat step 2 and 3 until result is 0. \nThe hex value is the digit sequence of the remainders from the last to first. \ncheck here: http://www.permadi.com/tutorial/numDecToHex/