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i need to make an equations that equal to 3,0,50 using the numbers 54321

i need to use symbols ( +,- ,x,and divisionsign) ..i cant change anything around

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Graphing using intercepts (old) (video) | Khan Academy


Graphing using intercepts (old) ... by this relationship or this equation. ... equal to when x is equal to 0. So y is equal to-- I want to do it in that pink color. y ...
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Equivalent Equations - Math Dictionary - ICoachMath.com


Equivalent Equations are the equations....Complete information about the equivalent equations, definition of an equivalent equations, examples of an equivalent ...
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Writing Algebraic Equations for word sentences - Math Goodies


Writing Algebraic Equations is presented by Math Goodies. Learn to translate open sentences into algebraic equations. ... Need More Practice?
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Complex or imaginary numbers - A complete course in algebra


OR IMAGINARY NUMBERS. ... we want to be able to say that every polynomial equation has a solution; specifically, ...
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Equation Grapher - mathsisfun.com


If you don't include an equals sign, it will assume you mean "=0" ... If you just want to graph a ... value that is not less than the argument and is equal to a ...
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Imaginary Numbers - Math Is Fun


Imaginary Numbers are not "Imaginary" Imaginary Numbers were once thought to be impossible, and so they were called "Imaginary" (to make fun of them).
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How to Add Parentheses to Make a Statement True | Sciencing


How to Add Parentheses to Make a Statement True ... Our equation will be 1+2x3-4=-3. Make ... Move the parenthesis to go around the last two numbers of the equation; ...
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How can you make the equation 987654321 equal 100 by only ...


Answers.com ® WikiAnswers ® Categories Science Math and Arithmetic How can you make the equation 987654321 equal ... 0+100=100 1+99=100 2+98=100 3+97 ... 100 50+50 ...
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solving for x

I interpret the first number in brackets as the base.You need a calculator to do this because, although the bases 2 and 4 are related to one another because 4 is 2 squared, there's no easy relation between 2 and 3, 3 and 10, or 2 and 10. 10 in base 2 is approximately 3.322; log 2 in base 10 is approximately 0.3010 which is the reciprocal of 3.322. Log 3 in base 2 is approximately 1.585 and log 2 in base 3 is 0.6309 which is the reciprocal of 1.585. Log 4 to base 2 is 2 so log 2 to base 4 is 0.5. To convert log x in one base to another we use the conversion factor of the log of one base of the other base. Let's choose base 2 as the common base. This is what the equation becomes: 0.5logx+logx+logx/log3=logx/log10. Put y=logx (base 2) and we get 0.5y+y+0.6903y=0.3010y. If we were to divide through by y we would have an inequality because 2.1903 is not equal to .3010. Therefore y=0 and logx=0, so x=1. That was the longwinded way of proving the solution. We can see that x=1 makes all the terms in the original equation zero and 0=0 is always true! Why go to all the trouble of using a calculator when the only answer is x=1? It was just possible that the sum of the numbers on the left side of the equation came to the result on the right side of the equation, in which case we would have had an identity instead of an equation, which would have been true for all values of x (except x=0).
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Shawn took in $69.15 in two hours work on Saturday selling Belts for $8.05 and earrings for $4.50. How many of each did Shawn sell?

Shawn took in $69.15 in two hours work on Saturday selling Belts for $8.05 and earrings for $4.50. How many of each did Shawn sell? I'm not too sure how ​you are expected to solve this problem. You will end up with a Diophantine Equation (one involving integer coefficients and integer solutions only) If you have never heard of Diophantus, then the precise mathematical solution later on should be ignored. Instead ...   The (Diophatine) equation is 161B + 90E = 1363 You should be able to show that Shawn can only sell a maximum of 8 belts (i.e. if he sells no earrings), and a maximum of 15 earrings (i.e. if he sells no belts). So start with the smaller number (8). Shawn sells up to 8 belts. So put B = 1 in the (Diophantine ) equation and see if B is an integer. If not, then B=1 cannot be a solution, So now try again with B = 2, and so on ..     My precise mathematical solution now follows. Let B be the number of belts sold @ $8.05 per belt. Let E be the number of earrings sold @ $4.50 per pair. Total money made is P = $69.15 Total money made is: B*$8.05 + E*$4.50 Equating the two expressions, 8.05B + 4.50E = 69.15 Making the coefficients as integers we end up with a Diophantine equation. 161B + 90E = 1383 We now manipulate the coefficients 161 and 90. We should end up with a relation between the two that should produce the value 1. This manipulation is in two parts. In the first line, we are writing the larger coefft, 161, as a multiple of the smaller coefft plus a remainder. 1st Part 161 = 1x90 + 71  90 = 1x71 + 19     now we write the 1st multiple as a multiple of the 1st remainder + remainder 71 = 3x19 + 14    we write the 2nd multiple as a multiple of the 2nd remainder + remainder 19 = 1x14 + 5 14 = 2x5 + 4 5 = 1x4 + 1 ============ 2nd Part We rewrite that last equation so that 1 is on the left hand side 1 = 5 – 1x4 We now use the remainder from the next equation up. 1 = 5 – 1x(14 – 2x5) 1 = 3x5 – 1x14 And again we use the remainder in the next equation up 1 = 3(19 – 1x14) – 1x14 1 = 3x19 – 4x14 1 = 3x19 – 4(71 – 3x19) 1 = 15x19 – 4x71 1 = 15(90 – 1x71) – 4x71 1 = 15x90 – 19x71 1 = 15x90 – 19(161 – 1x90) 1 = 34x90 – 19x161 ================ Therefore, 1383 = (-19x1383).161 + (34x1383).90 1383 = -26,277*161 + 57,022*90 Which implies B = -26,277 E = 47,022 which is obviously incorrect! But all is not lost. These two values for B and E are simply two values that happen to satisfy the original Diophantine equation. What we need is the general solution, which now follows. In general, B = -26,277 + 90k    (using the coefft of E) E = 47,022 – 161k    (using the coefft of B) If you substitute these expressions into the original Diophantine equation, the k-terms will cancel out, leaving you with the original eqn. We know that B and E must be smallish numbers and that neither can be negative. For example, if no earrings were sold, Shawn would need to sell over 8 belts to clear $69, and if no belts were sold, then Shawn would need to sell over 15 earrings. So Shawn needs to sell between 0 and 8 belts and between 0 and 15 earrings. Setting k = 292, B = -26,277 + 90*292 = 3 E = 47,022 – 161*292 = 10 B = 3, E = 10 If k is greater than or less than 292 by 1, or more, then either B will be negative or E will be negative. So this is the answer. Check 3*8.05 + 10*4.50 = 25.15 + 45.00 = 69.15 – Correct! Answer: Shawn sells 3 belts and 10 earrings
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magic square 16 boxes each box has to give a number up to16 but when added any 4 boxes equal 34

The magic square consists of 4X4 boxes. We don't yet know whether all the numbers from 1 to 16 are there. Each row adds up to 34, so since there are 4 rows the sum of the numbers must be 4*34=136. When we add the numbers from 1 to 16 we get 136, so we now know that the square consists of all the numbers between 1 and 16. We can arrange the numbers 1 to 16 into four groups of four such that within the group there are 2 pairs of numbers {x y 17-x 17-y}. These add up to 34. We need to find 8 numbers represented by A, B, ..., H, so that all the rows, columns and two diagonals add up to 34. A+B+17-A+17-B=34, C+D+17-C+17-D=34, ... (rows) A+C+17-A+17-C=34, ... (columns) Now there's a problem, because the complement of A, for example, appears in the first row and the first column, which would imply duplication and mean that we would not be able to use up all the numbers between 1 and 16. To avoid this problem we need to consider other ways of making up the sum 34 in the columns. Let's use an example. The complement of 1 is 16 anewd its accompanying pair in the row is 2+15; but if we have the sum 1+14 we need another pair that adds up to 19 so that the sum 34 is preserved. We would perhaps need 3+16. We can't use 16 and 15 because they're being used in a row, and we can't use 14, because it's being used in a column, so we would have to use 6+13 as the next available pair. So the row pairs would be 1+16 and 2+15; the column pairs 1+14 and 6+13; and the diagonal pairs 1+12 and 10+11. Note that we've used up 10 of the 16 numbers so far. This type of logic applies to every box, apart from the middle two boxes of each side of the square (these are part of a row and column only), because they appear in a row, a column and diagonal. There are only 10 equations but 16 numbers. In the following sets, in which each of the 16 boxes is represented by a letter of the alphabet between A and P, the sum of the members of each set is 34: {A B C D} {A E I M} {A F K P} {B F J N} {C G K O} {D H L P} {D G J M} {E F G H} {I J K L} {M N O P} The sums of the numbers in the following sets in rows satisfy the magic square requirement of equalling 34. This is a list of all possibilities. However, there are also 24 ways of arranging the numbers in order. We've also seen that we need 10 out of the 16 numbers to satisfy the 34 requirement for numbers that are part of a row, column and diagonal; but the remaining numbers (the central pair of numbers on each side of the square) only use 7 out of 16. The next problem is to find out how to combine the arrangements. The vertical line divides pairs of numbers that could replace the second pair of the set. So 1 16 paired with 2 15 can also be paired with 3 14, 4 13, etc. The number of alternative pairs decreases as we move down the list. 17 X 17: 1 16 2 15 | 3 14 | 4 13 | 5 12 | 6 11 | 7 10 | 8 9 2 15 3 14 | 1 16 | 4 13 | 5 12 | 6 11 | 7 10 | 8 9 3 14 4 13 | 1 16 | 2 15 | 5 12 | 6 11 | 7 10 | 8 9 4 13 5 12 | 1 16 | 2 15 | 3 14 | 6 11 | 7 10 | 8 9 5 12 6 11 | 1 16 | 2 15 | 3 14 | 4 13 | 7 10 | 8 9 6 11 7 10 | 1 16 | 2 15 | 3 14 | 4 13 | 5 12 | 8 9 7 10 8 9 | 1 16 | 2 15 | 3 14 | 4 13 | 5 12 | 6 11 16 X 18: 1 15 2 16 | 4 14 | 5 13 | 6 12 | 7 11 | 8 10 2 14 3 15 | 5 13 | 6 12 | 7 11 | 8 10 3 13 4 14 | 2 16 | 6 12 | 7 11 | 8 10 4 12 5 13 | 2 16 | 3 15 | 7 11 | 8 10 5 11 6 12 | 2 16 | 3 15 | 4 14 | 8 10 6 10 7 11 | 2 16 | 3 15 | 4 14 | 5 13 7 9 8 10 | 2 16 | 3 15 | 4 14 | 5 13 | 6 12 15 X 19: 1 14 3 16 | 4 15 | 6 13 | 7 12 | 8 11 2 13 4 15 | 3 16 | 5 14 | 7 12 | 8 11 3 12 5 14 | 4 15 | 6 13 | 8 11 4 11 6 13 | 3 16 | 5 14 | 7 12 5 10 7 12 | 3 16 | 4 15 | 6 13 | 8 11 6 9 8 11 | 3 16 | 4 15 | 5 14 | 7 12  7 8 9 10 | 3 16 | 4 15 | 5 14 | 6 13 14 X 20: 1 13 4 16 | 5 15 | 6 14 | 8 12 | 9 11 2 12 5 15 | 4 16 | 6 14 | 7 13 | 9 11 3 11 6 14 | 4 16 | 5 15 | 7 13 | 8 12 4 10 7 13 | 5 15 | 6 14 | 8 12 | 9 11 5 9 8 12 | 4 16 | 6 14 | 7 13 | 9 11 6 8 9 11 | 4 16 | 5 15 | 7 13 13 X 21: 1 12 5 16 | 6 15 | 7 14 | 8 13 | 10 11 2 11 6 15 | 5 16 | 7 14 | 8 13 | 9 12 3 10 7 14 | 5 16 | 6 15 | 8 13 | 9 12 4 9 8 13 | 5 16 | 6 15 | 7 14 | 10 11 5 8 9 12 | 6 15 | 7 14 | 10 11 6 7 10 11 | 5 16 | 8 13 | 9 12 12 X 22: 1 11 6 16 | 7 15 | 8 14 | 9 13 | 10 12 2 10 7 15 | 6 16 | 8 14 | 9 13 | 10 12 3 9 8 14 | 6 16  4 8 9 13 | 6 16 | 7 15  5 7 10 12 | 6 16 11 X 23: 1 10 7 16 | 8 15 | 9 14 2 9 8 15 | 7 16  3 8 9 14 | 7 16  10 X 24: 1 9 8 16 To give an example of how the list could be used, let's take row 3 in 17 X 17 {3 14 4 13}. If 3 is the number in the row column and diagonal, then we need to inspect the list to find another row in a different group containing 3 that doesn't duplicate any other numbers. So we move on to 15 X 19 and we find {3 12 8 11} and another row in 13 X 21 with {3 10 5 16}. So we've used the numbers 3, 4, 5, 8, 10, 11, 12, 13, 14, 16. That leaves 1, 2, 6, 7, 9, 15. In fact, one answer is: 07 12 01 14 (see 15x19) 02 13 08 11 16 03 10 05 09 06 15 04
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i need the answer for these questions

Part 1 Newton’s Method for Vector-Valued Functions Our system of equations is, f1(x,y,z) = 0 f2(x,y,z) = 0 f3(x,y,z) = 0 with, f1(x,y,z) = xyz – x^2 + y^2 – 1.34 f2(x,y,z) = xy –z^2 – 0.09 f3(x,y,z) = e^x + e^y + z – 0.41 we can think of (x,y,z) as a vector x and (f1,f2,f3) as a vector-valued function f. With this notation, we can write the system of equations as, f(x) = 0 i.e. we wish to find a vector x that makes the vector function f equal to the zero vector. Linear Approximation for Vector Functions In the single variable case, Newton’s method was derived by considering the linear approximation of the function f at the initial guess x0. From Calculus, the following is the linear approximation of f at x0, for vectors and vector-valued functions: f(x) ≈ f(x0) + Df(x0)(x − x0). Here Df(x0) is a 3 × 3 matrix whose entries are the various partial derivative of the components of f. Specifically,     ∂f1/ ∂x (x0) ∂f1/ ∂y (x0) ∂f1/ ∂z (x0) Df(x0) = ∂f2/ ∂x (x0) ∂f2/ ∂y (x0) ∂f2/ ∂z (x0)     ∂f3/ ∂x (x0) ∂f3/ ∂y (x0) ∂f3/ ∂z (x0)   Newton’s Method We wish to find x that makes f equal to the zero vector, so let’s choose x = x1 so that f(x0) + Df(x0)(x1 − x0) = f(x) =  0. Since Df(x0)) is a square matrix, we can solve this equation by x1 = x0 − (Df(x0))^(−1)f(x0), provided that the inverse exists. The formula is the vector equivalent of the Newton’s method formula for single variable functions. However, in practice we never use the inverse of a matrix for computations, so we cannot use this formula directly. Rather, we can do the following. First solve the equation Df(x0)∆x = −f(x0) Since Df(x0) is a known matrix and −f(x0) is a known vector, this equation is just a system of linear equations, which can be solved efficiently and accurately. Once we have the solution vector ∆x, we can obtain our improved estimate x1 by x1 = x0 + ∆x. For subsequent steps, we have the following process: • Solve Df(xi)∆x = −f(xi). • Let xi+1 = xi + ∆x
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Interpolate the data set (1, 150), (3, 175), (4, 185), (6, 200), (8, 300) to estimate the amount of money Gracie may earn if she displays her items for 7 hours

Since 7 is halfway between 6 and 8, Gracie should earn an amount about halfway between 200 and 300, that is, 250 (triangular interpolation). This is the simplest interpolation, but see later. Interpolating for 2 and 5 we get 162.5 and 192.5, that is, respectively, a difference of 12.5 (162.5-150 or 175-162.5) and 7.5 (192.5-185 or 200-192.5), while 250 is a difference of 50 from 200 and 300. So the interpolated figures fluctuate. For a more sophisticated approach, we need to take the whole dataset and look for a formula that best fits. One way to do this is to fit a polynomial F(x)=ax^4+bx^3+cx^2+dx+e into the five given points. This polynomial has 5 unknown coefficients, so with 5 simultaneous equations we should be able to find them. The process can be simplified slightly by taking the lowest "x" coord and using that as the zero starting point. In this case the lowest coord is 1 (hour) so we subtract 1 from the first coord of each pair to get: (0,150), (2,175), etc. F(0)=e=150. So we have the constant 150. The next step is to subtract 150 from each of the other "y" coords so we arrive at the following set of equations: (1) F(2)=16a+8b+4c+2d=25 (2) F(3)=81a+27b+9c+3d=35 (3) F(5)=625a+125b+25c+5d=50 (4) F(7)=2401a+343b+49c+7d=150 and we already have F(0)=150=e. We can now eliminate d from (1) and (2): 2F(3)-3F(2): (162-48)a+(54-24)b+(18-12)c=70-75=-5; (5) 114a+30b+6c=-5. and we can eliminate d from (3) and (4): 5F(7)-7F(5): (12005-4375)a+(1715-875)b+(245-175)c=750-350; 7630a+840b+70c=400 which simplifies to (6) 763a+84b+7c=40 or 109a+12b+c=40/7 We can eliminate c between (5) and (6): 6(6)-(5): (654-114)a+(72-30)b=240/7+5=275/7; (7) 540a+42b=275/7 or 90a+7b=275/42. So b=(275/42-90a)/7. From (6) we have: 109a+12b+c=109a+12(275/42-90a)/7+c=40/7, so c=40/7-109a-12(275/42-90a)/7; c=40/7-550/7+(1080/7-109)a=-510/7+317a/7=(317a-510)/7. We now have b and c in terms of a. We can continue to find d in terms of a. From (1) d=(25-16a-8b-4c)/2=25-16a-8(275/42-90a)/7-4(317a-510)/7; d=25-1100/147+2040/7+(-16+720/7-1268/7)a= (3675-1100+42840)/147+(-112+720-1268)a/7; d=45415/147-660a/7. We have b, c and d in terms of a, so we can find a by substituting into an equation containing all four coefficients (but not (1), because we used it to find d). Let's pick (2) and hope we get a sensible result! 81a+27(275/42-90a)/7+9(317a-510)/7+3(45415/147-660a/7)=35. From this a=5143/2772=1.855. Therefore b=-22.919, c=78.510, d=-67.687, e=150. And F(x)=1.855x^4-22.919x^3+78.510x^2-67.687x+150. This results need to be checked before we use F to find an interpolated value. Unfortunately, this polynomial approach produces inconsistent results, and needs to be discarded. Lagrange's method seems the obvious choice, even if it is tedious to do. We have 5 x values which we'll symbolise as x0, x1, x2, x3, x4 and 5 function values f0, f1, f2, f3, f4. If the function we're looking for is f(x) then: f(x)=(x-x1)(x-x2)(x-x3)(x-x4)f0/((x0-x1)(x0-x2)(x0-x3)(x0-x4))+         (x-x0)(x-x2)(x-x3)(x-x4)f1/((x1-x0)(x1-x2)(x1-x3)(x1-x4))+         (x-x0)(x-x1)(x-x3)(x-x4)f2/((x2-x0)(x2-x1)(x2-x3)(x2-x4))+... x0=1, x1=3, x2=4, x3=6, x4=8; f0=150, f1=175, f2=185, f3=200, f4=300. We want x=7, so f(7) is given by: 4.3.1.-1.150/(-2.-3.-5.-7)+6.3.1.-1.175/(2.-1.-3.-5)+ 6.4.1.-1.185/(3.1.-2.-4)+6.4.3.-1.200/(5.3.2.-2)+ 6.4.3.1.300/(7.5.4.2) This comes to: -60/7+105-185+240+540/7=1600/7=228.57 (229 to the nearest whole number) compared with 250 from the simple interpolation. 
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what is each part of the equation y=mx+b mean and who do you find them using math vocabulary

The normal meaning for this standard linear equation is that x and y are coordinates in a rectangular arrangement of axes. The y axis is North-South while the x axis is East-West. Where they cross is called the origin with coordinates (0,0), that is, x and y are both zero. The equation y=mx+b defines a straight line. It slopes at a value given by m, the slope or gradient, and m is a number which can be a whole number, a fraction, or whatever, as long as it is constant so that the line remains straight. The slope, m, is also known as the tangent, and the tangent of the angle that the line makes with the x axis has a value of m. When the straight line is at an angle of 45 degrees to the x axis, its tangent is 1 so m=1. If the line slopes backwards at 45 degrees to the x axis, it's tangent is -1 and m=-1. Forward sloping lines have a positive m, while backward sloping lines have a negative m. The value of b is also called the y intercept, because it's the point on the y axis where the straight line crosses that axis. It can have a positive or negative value. b is a constant, just like m. mx is m times x. The x axis is divided by equally spaced numbers, 0, 1, 2, 3 etc to the right, and -1, -2, -3 etc to the left of the origin. The y axis is similarly divided, postitive numbers going up and negative numbers going down. By putting numbers in the equation you can work out where points go on the line. m will have a value, like 2, for example, and b a value, say, 3, so we have y=2x+3. If we put x=0 we get y=3 which is the y intercept. So we mark that point (0,3) by going up 3 divisions on the y axis. Now put x=1, then y=5. So we move to the point (1,5), which is right 1 and up 5. We can join that point to (0,3) and continue beyond these points. What we find is that, although we have only plotted two points, other values of x and y actually fit on the line. If we look at where x=3 and go up to meet the line, then go horizontally back to the y axis, we should find it meets the point 9 on the y axis. So the line represents the relationship between x and y as given by the equation for all points including points in between our whole number divisions, like, for example, 1.5 or one and a half, halfway between 1 and 2.
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10 vedic maths rules for class IX

2 instead of 5: 34/5 can be calculated by multiplying 34 by 2 instead of dividing by 5. 34*2=68. We just move the decimal point one place to the left: 34/5=6.8. 124/5=24.8 because 124*2=248. Move the decimal point: 248 becomes 24.8. 34*5 is the same as 34/2=17 but we add a zero to make 17 into 170. 73*5 is the same as 73/2=36.5 then move the decimal point one place to the right (or add zero): 36.5 becomes 365=73*5. So we only need to know how to multiply and divide by 2 to divide and multiply by 5. We just move the decimal point. Divisibility by 9 or remainder after dividing by 9. All multiples of 9 contain digits which added together give 9. As we add the digits together, each time the result goes over 9 we add the digits of the result together and use that result and continue in this way up to the last digit. Is 12345 divisible by 9? Add the digits together 1+2+3=6. When we add 4 we get 10, so we add 1 and zero=1 then we add 5 to get 6. The number is not exactly divisible by 9, but the remainder is 6. We can also ignore any 9's in the number. Now try 67959. We can ignore the two 9's. 6+7=13, and 1+3=4; 4+5=9, so 67959 is divisible by 9. Multiplying by 11. Example: 132435*11. We write down the first and last digits 1 ... 5. Now we add the digits in pairs from the left a digit step at a time. So 1+3=4; 3+2=5: 2+4=6; 4+3=7; 3+5=8. Write these new digits between 1 and 5 and we get 1456785=132435*11. But we had no carryovers here. Now try 864753*11. Write down the first and last digits: 8 ... 3. 8+6=14, so we cross out the 8 and replace it with 8+1=9, giving us 94 ... 3. Next pair: 6+4=10. Again we go over 10 so we cross out 4 and make it 5. Now we have 950 ... 3. 4+7=11, so we have 9511 ... 3. 7+5=12, giving us 95122 ... 3; 5+3=8, giving us the final result 9512283.  Divisibility by 11. We add alternate digits and then we add the digits we missed. Subtract one sum from the other and if the result is zero the original number was divisible by 11. Example: 1456785. 1 5 7 5 make up one set of alternate digits and the other set is 4 6 8. 1+5+7=13. We drop the ten and keep 3 in mind to add to 5 to give us 8. Now 4 6 8: 4+6=10, drop the ten and add 0 to 8 to give us 8 (or ignore the zero). 8-8=0 so 11 divides into 1456785. Now 9512283: set 1 is 9 1 2 3 and set 2 is 5 2 8; 9+1=0 (when we drop the ten); 2+3=5; set 1 result is 5; 5+2+8=5 after dropping the ten, and 5-5=0 so 9512283 is divisible by 11. Nines remainder for checking arithmetic. We can check the result of addition, subtraction, multiplication and (carefully) division. Using Method 2 above we can reduce operands to a single digit. Take the following piece of arithmetic: 17*56-19*45+27*84. We'll assume we have carried out this sum and arrived at an answer 2365. We reduce each number to a single digit using Method 2: 8*2-1*9+9*3. 9's have no effect so we can replace 9's by 0's: 8*2 is all that remains. 8*2=16 and 1+6=7. This tells us that the result must reduce to 7 when we apply Method 2: 2+3+6=11; 1+1=2 and 2+5=7. So, although we can't be sure we have the right answer we certainly don't have the wrong answer because we arrived at the number 7 for the operands and the result. For division we simply use the fact that a/b=c+r where c is the quotient and r is the remainder. We can write this as a=b*c+r and then apply Method 2, as long as we have an actual remainder and not a decimal or fraction. Divisibility by 3. This is similar to Method 2. We reduce a number to a single digit. If this digit is 3, 6 or 9 (in other words, divisible by 3) then the whole number is divisible by 3. Divisibility by 6. This is similar to Method 6 but we also need the last digit of the original number to be even (0, 2, 4, 6 or 8). Divisibility by 4. If 4 divides into the last two digits of a number then the whole number is divisible by 4. Using 4 or 2 times 2 instead of 25 for multiplication and division. 469/25=469*4/100=1876/100=18.76. 538*25=538*100/4=134.5*100=13450. We could also double twice: 469*2=938, 938*2=1876, then divide by 100 (shift the decimal point two places to the left). And we can divide by 2 twice: 538/2=269, 269/2=134.5 then multiply by 100 (shift the decimal point two places left or add zeroes). Divisibility by 8. If 8 divides into the last three digits of a number then the whole number is divisible by 8. Using 8 or 2 times 2 times 2 instead of 125 for multiplication and division. Similar to Method 9, using 125=1000/8. Using addition instead of subtraction. 457-178. Complement 178: 821 and add: 457+821=1278, now reduce the thousands digit by 1 and add it to the units: 278+1=279; 457-178=279. Example: 1792-897. First match the length of 897 to 1792 be prefixing a zero: 0897; complement this: 9102. 1792+9102=1894. Reduce the thousands digit by 1 and add to the result: 894+1=895. Example: 14703-2849. 2849 becomes 02849, then complements to 97150. 14703+97150=111853; reduce the ten-thousands digit by 1 and and add to the result: 11854. Squaring numbers ending in 5. Example: 75^2. Start by writing the last two digits, which are always 25. Take the 7 and multiply by 1 more than 7, which is 8, so we get 56. Place this before the 25: 5625 is the square of 75. The square of 25 is ...25, preceded by 2*3=6, so we get 625. All numbers ending in 0 or 5 are exactly divisible by 5 (see also Method 1). All numbers ending in zero are exactly divisible by 10. All numbers ending in 00, 25, 50 or 75 are divisible by 25. Divisibility by 7. Example: is 2401 divisible by 7? Starting from the left with a pair of digits we multiply the first digit by 3 and add the second to it: 24: 3*2+4=10; now we repeat the process because we have 2 digits: 3*1+0=3. We take this single digit and the one following 24, which is a zero: 3*3+0=9. When we get a single digit 7, 8 or 9 we simply subtract 7 from it: in this case we had 9 so 9-7=2 and the single digit is now 2. Finally in this example we bring in the last digit: 3*2+1=7, but 7 is reduced to 0. This tells us the remainder after dividing 2401 by 7 is zero, so 2401 is divisible by 7. Another example: 1378. 3*1+3=6; 3*6=18 before adding the next digit, 7 (we can reduce this to a single digit first): 3*1+8=3*1+1=4; now add the 7: 4+7=4+0=4;  3*4=12; 3*1+2+8=5+1=6, so 6 is the remainder after dividing 1378 by 7.  See also my solution to: http://www.mathhomeworkanswers.org/72132/addition-using-vedic-maths?show=72132#q72132
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find all reconstuctions of the sum: ABC+DEF+GHI=2014 if all letters are single digits

There is no solution if A to I each uniquely represent a digit 1 to 9, because the 9's remainder of ABC+DEF+GHI (0) cannot equal the 9's remainder of 2014 (=7). Let me explain. The 9's remainder, or digital root (DR), is obtained by adding the digits of a number, adding the digits of the result, and so on till a single digit results. If the result is 9, the DR is zero and it's the result of dividing the number by 9 and noting the remainder only. E.g., 2014 has a DR of 7. When an arithmetic operation is performed, the DR is preserved in the result. So the DRs of individual numbers in a sum give a result whose DR matches. If we add the numbers 1 to 9 we get 45 with a DR of zero, but 2014 has a DR of 7, so no arrangement can add up to 2014. If 2 is replaced by 0 in the set of available digits, the DR becomes 7 (sum of digits drops to 43, which has a DR of 7). 410+735+869=2014 is just one of many results of applying the following method. Look at the number 2014 and consider its construction. The last digit is the result of adding C, F and I. The result of addition can produce 4, 14 or 24, so a carryover may apply when we add the digits in the tens column, B, E and H. When these are added together, we may have a carryover into the hundreds of 0, 1 or 2. These alternative outcomes can be shown as a tree. The tree: 04 >> 11, >> 19: ............21 >> 18; 14 >> 10, >> 19: ............20 >> 18; 24 >> 09, >> 19: ............19 >> 18. The chevrons separate the units (left), tens (middle) and hundreds (right). The carryover digit is the first digit of a pair. For example, 20 means that 2 is the carryover to the next column. Each pair of digits in the units column is C+F+I; B+E+H in the tens; A+D+G in the hundreds. Accompanying the tree is a table of possible digit summations appearing in the tree. Here's the table: {04 (CFI): 013} {09 (BEH): 018 036 045 135} {10 (BEH): 019 037 046 136 145} {11 (BEH): 038 047 137 056 146} {14 (CFI): 059 149 068 158 167 347 086 176 356} {18 (ADG): 189 369 459 378 468 567} {19 (BEH/ADG): 379 469 478 568} {20 (BEH): 389 479 569 578} {21 (BEH): 489 579 678} {24 (CFI): 789} METHOD: We use trial and error to find suitable digits. Start with units and sum of C, F and I, which can add up to 4, 14 or 24. The table says we can only use 0, 1 and 3 to make 4 with no carryover. The tree says if we go for 04, we must follow with a sum of 11 or 21 in the tens. The table gives all the combinations of digits that sum to 11 or 21. If we go for 11 the tree says we need 19 next so that we get 20 with the carryover to give us the first two digits of 2014. See how it works? Now the fun bit. After picking 013 to start, scan 11 in the table for a trio that doesn't contain 0, 1 or 3. There isn't one, so try 21. We can pick any, because they're all suitable, so try 489. The tree says go for 18 next. Bingo! 567 is there and so we have all the digits: 013489576. We have a result for CFIBEHADG=013489576, so ABCDEFGHI=540781693. There are 27 arrangements of these because we can rotate the units, tens and hundreds independently like the wheels of an arcade jackpot machine. For example: 541+783+690=2014. Every solution leads to 27 arrangements. See how many you can find using the tree and table!
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Factor by grouping

The idea behind factoring is to group as many like terms as possible making the equation simpler and easier to manipulate. 1) Seperate the like terms.  This means you group the terms that have either the same variable or are products of the same number.  I will demonstrate by putting like terms in parentheses. 0 = (x^3 + 4x^2 -3x) - 18 2) Now you will basically 'un-FOIL' the part of the equation inside the parentheses.  Since you know how to FOIL(asumming you do) you just work backwards. You know that after FOILing you have 4 terms, so try separating it into 4 terms.  The only term that can be split in this instance is 4x^2. 4x^2 = 2x^2 + 2x^2 3) un-FOIL x^3 +2x^2 + 2x^2 - 3x = (x-2)(x+3)^2 4) To find the value of 'x', you simply set it equal to each term inside parentheses. x = x-2 x = (x+3)^2 Ans) x = -3          x = 2
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the integral of xtanx dx

Consider the function tanx=a0+a1x+a2x^2+...+a(n)x^n where a(n) is the coefficient of x^n. We need to find a(n). We can do this by applying calculus (effectively Taylor's theorem). If we integrate tanxdx we get -ln(cosx). If we integrate the power series we get C+a0x+a1x^2/2+a2x^3/3+...+a(n)x^(n+1)/(n+1), where C is a constant of integration. This is a power series for -ln(cosx) or strictly, -ln|cosx|, because we can only take logs of positive numbers. Also cosx can only assume values between 0 and 1, so the log is negative, and we can write -ln|cosx| as ln|secx|. |secx| is always 1 or more. Going back to the expansion for tanx, we know tan0=0, so a0=0. Therefore ln|secx|=C+a1x^2/2+...+a(n)x^(n+1)/(n+1). We'll see this pop up later when we integrate xtanx by parts. The derivative of tanx is sec^2x=S(na(n)x^(n-1)) where S is the sum of n terms (from n=1), since a0=0. When x=0, sec^2x=1 and when x=(pi)/2, sec^2x=0. The only term for S not containing x is a1, so a1=1. So far the series for tanx is: x+S(na(n)x^(n-1)) for n>2. Not much to go on yet. The next derivative is 2sec^2tanx=S(n(n-1)x^(n-2)) for n>2 (differentiation by substitution: let u=secx; du=secxtanxdx; d/dx=d/du*du/dx=2u*secxtanx=2sec^2xtanx). When x=0, tan0=0 so this derivative is zero, making 2a2=0, so a2=0. The next derivative is 4sec^2xtan^2x+2sec^4x (differentiation by parts: u=2sec^2x, v=tanx; du=4sec^2xtanxdx, dv=sec^2xdx; d(uv)=vdu+udv=(tanx)(4sec^2xtanxdx)+(2sec^2x)(sec^2xdx)). This derivative is 2 when x=0, so 6a3=2 and a3=1/3 (from n(n-1)(n-2)a(n)x^(n-3) where n=3). The 4th derivative is 8tanxsec^3x+8sec^2tan^3+8sec^4xtanx, which is zero when x=0 and a4=0. The 5th derivative is: 8tanx(3sec^3xtanx)+8sec^3x(sec^2x)+8sec^2x(3tan^2sec^x)+8tan^3(2sec^2tanx)+8sec^4x(sec^2x)+8tanx(4sec^4tanx) This derivative is 16 when x=0. The relevant term is 120a5, so a5=16/120=2/15. tanx=x+x^3/3+2x^5/15+... xtanx=x^2+x^4/3+2x^6/15+... integrate: x^3/3+x^5/15+2x^7/105+... Another way of approaching the series is to use the power series for sinx and cosx because tanx=sinx/cosx. Just as we found the coefficients of the power series for tanx, we can do the same for sinx, when we get sinx=x-x^3/3!+x^5/5!-x^7/7!+... And cosx is derivative of sinc, so cosx=1-x^2/2!+x^4/4!-... Also tanx*cosx=sinx, so we can use this identity to derive the coefficients for tanx. (a0+a1x+a2x^2+...)(1-x^2/2!+x^4/4!-...)=x-x^3/3!+x^5/5!-...=a0+a1x+...+a1x-a1x^3/2!+a1x^5/4!-...+a2x^2-a2x^4/2!... By equating the coefficients for a particular power of x we can work out the unknown coefficients a(n). For example, because there are no even powers of x in the expansion of sinx, a0=0 (which we already discovered), a2, a4, etc. are all zero. to find a1, we need all terms involving x. Since a1x is the only one, a1=1; to find a3, we have -a1x^3/2=-x^3/6 so a1=1/3; a1x^5/24-a3x^5/2+a5x^5=x^5/120, 1/24-1/6+a5=1/120, a5=1/120-1/24+1/6=(1-5+20)/120=16/120=2/15, as we discovered earlier. What is a7? To get the coefficient of x^7 we need to combine x with x^6, x^3 with x^4, x^5 with x^2 and x^7. The coefficients are a1, a3, a5 and a7 from tanx; -1/6!, 1/4!, -1/2! from cosx; -1/5040 from sinx. -a1/720+a3/24-a5/2+a7=-1/5040; -1/720+1/72-1/15+a7=-1/5040; a7=1/720-1/72+1/15-1/5040=(7-70+336-1)/5040=272/5040=17/315. Now we return to integral xtanxdx. Let u=x, then du=dx; dv=tanxdx, then v=ln|secx|, as we discovered earlier. d(uv)=vdu/dx+udv/dx=ln|secx|dx+xtanxdx. So integral(xtanxdx)=xln|secx|-integral(ln|secx|dx)=xln|secx|-(x^3/3+x^5/15+2x^7/105+...)+C.      
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