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the square of the sum of a number and 11

what is the square of the sum of a number and 11

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Square number - Wikipedia


... a square number or perfect square is an integer that is the square of an ... 11 2 = 121 12 2 = 144 13 2 ... The sum of two consecutive square numbers is a ...
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The sum of a number and its square is 110. Find the numbers ...


The sum of a number and its square is 110. ... the sum of -11 and its square, ... Enter your number and we’ll text you a download link.
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square of sum | planetmath.org


square of sum. Primary tabs. View ... The square of a sum is equal to the sum of the squares of all the summands plus the sum of ... This is true for any number of ...
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The sum of a number and its square is 72. What ... - jiskha.com


In a number square, the sum of the numbers in each row, column, and main diagonal is the same. ... CCA Unit 4 Lesson 11 3. If a number is not a rational number, ...
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The square of the sum of a number and 13 - Answers.com


The square of the sum of the number and 13 is (N+13)2 N2 + 26 N + 169. ... The sum of a number and its square is 6 Find the number? There are two solution: ...
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Sum of the Squares of the first n Natural Numbers


... Sum of the Squares of the First n Natural Numbers The sum of the squares ... [3.11 ] Simplifying: [3.12] ... coefficients of the square and the cubic Ken Ward's ...
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The sum of a number squared and three less than ... - Socratic


Let's call the number x "A number squared" =x^2 "Three less than twice the number" =2x-3 "The sum ... If we take out the square, ... SOCRATIC ...
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Translating Verbal Expressions – Terms - Del Mar College


Translating Verbal Expressions – Terms . ADDITION: ... square of the square of a number x2. ... 11. The sum of two consecutive odd ...
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Mathematical Phrase Translation - Cerritos College


Mathematical Phrase Translation The sum of x and 5 x + 5 ... The sum of the product of 8 and a number, and 3. 11. ... The square of the differen ce of the opposite of ...
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Find the numbers such that the square of the sum of the ...


Find the numbers such that the square of the sum of the number and 6 is 169. ... 11/10/2015 ... 3. "the square of the sum of the number and 6 is 169" = (x + 6) ...
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Suggested Questions And Answer :


40 sum from 3 by 3 magic squares

Label the squares A to I: A B C D E F G H I  If each row and column is to sum to 40 then the three rows added together must sum to 120: (A+B+C)+(D+E+F)+(G+H+I)=40+40+40=120, which is the sum of all numbers used. The consecutive numbers 1 to 9 add up to 45, which is 75 short of 120. If we use the numbers 2, 4, 6, ..., 18 the sum would be 90, 30 short of 120. If we bias the numbers by starting from a different number, we can make up the difference. Let the numbers be in order, nx+a where x is between 1 and 9, and a is a bias (a number to be added to each nx) and n is to be an integer, then the sum of all the numbers is n+a+2n+a+3n+a+...+9n+a=45n+9a=120. That is, 5n+a=120/9=40/3. So one solution is n=2 and a=40/3-10=10/3. For numbers 1 to 9, a magic square is typically: 8 1 6 3 5 7 4 9 2 We use the conversion nx+a: 1→16/3, 2→22/3, 3→28/3, 4→34/3, 5→40/3, 6→46/3, 7→52/3, 8→58/3, 9→64/3: 58/3 16/3 46/3 28/3 40/3 52/3 34/3 64/3 22/3  In this square the fractions sum to 40. If we write a=40/3-5n=5(8-3n)/3 or n=(40/3-a)/5=(40-3a)/15 or 8/3-a/5, we can find other values for a and n. Then we can convert the numbers 1 to 9 using the formula x→nx+a. GENERAL 3 x 3 MAGIC SQUARE SOLUTIONS Represent square using letters: A B C D E F G H I A+B+C=S=D+E+F=G+H+I; A+B+C+D+E+F+G+H+I=3S, where S is the common sum. A+E+I=B+E+H=C+E+G=D+E+F=S (A+B+C+D+E+F+G+H+I)+3E=4S; 3S+3E=4S, E=S/3. A+E+I=S, I=S-E-A, I=2S/3-A. H=S-E-B, H=2S/3-B. C=S-(A+B). G=2S/3-C=2S/3-S+(A+B), G=A+B-S/3. D+G=B+C=B+S-(A+B)=S-A; D=S-A-G=S-A-A-B+S/3, D=4S/3-(2A+B). F=2S/3-D=2S/3-4S/3+2A+B, F=2A+B-2S/3. Completed square:           A            B             S-(A+B) 4S/3-(2A+B)    S/3    2A+B-2S/3    A+B-S/3    2S/3-B      2S/3-A So A and B are arbitrary; S must be a multiple of 3 if square is to be whole numbers only. EXAMPLE: A=1, B=5, S=18:   1  5  12 17  6  -5   0  7  11 Since squares can be rotated and reflected, A and B could have been 1 and 17, 12 and 5, 11 and -5, etc., to produce the same square effectively. 
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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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make two magical square with single digit

3 x 3 MAGIC SQUARE SOLUTIONS Represent square using letters: A B C D E F G H I A+B+C=S=D+E+F=G+H+I; A+B+C+D+E+F+G+H+I=3S. A+E+I=B+E+H=C+E+G=D+E+F=S (A+B+C+D+E+F+G+H+I)+3E=4S; 3S+3E=4S, E=S/3. A+E+I=S, I=S-E-A, I=2S/3-A. H=S-E-B, H=2S/3-B. C=S-(A+B). G=2S/3-C=2S/3-S+(A+B), G=A+B-S/3. D+G=B+C=B+S-(A+B)=S-A; D=S-A-G=S-A-A-B+S/3, D=4S/3-(2A+B). F=2S/3-D=2S/3-4S/3+2A+B, F=2A+B-2S/3. Completed square:           A            B             S-(A+B) 4S/3-(2A+B)    S/3    2A+B-2S/3    A+B-S/3    2S/3-B      2S/3-A So A and B are arbitrary; S must be a multiple of 3 if square is to be whole numbers only. EXAMPLE: A=1, B=5, S=18:   1  5  12 17  6  -5   0  7  11 Single digits can be 1 to 9 (sum=45) or 0 to 8 (sum=36). The common sum is 45/3=15 or 36/3=12. In one case the middle digit is 5 (15/3)  and in the other it's 4 (12/3). In the first case, 5 must be in the middle of the square, and we need to see where 9 fits in. The common sum is 15 so 15-9=6 and the other two numbers must be (1,5) or (2,4). This tells us that 9 can only participate in two sums and therefore it must be in the middle of a side with 2 and 4 on either side of it. So B=9 and A=2. 2 9 4 7 5 3 6 1 8 is a solution. In the case for 0-8 we simply subtract 1 from each square: 1 8 3 6 4 2 5 0 7 and we can reorientate this: 7 2 3 0 4 8 5 6 1 There we have it: two solutions. 
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answer for a magic square where the sum is 35

If the sum is 35 for each row, column and diagonal, the numbers must total 3*35=105. Now we need to find out if the numbers are consecutive. The numbers 1-9, for example, add up to 45. If the numbers form an arithmetic progression (AP): a, a+d, a+2d, ..., a+8d, the sum is 9a+d(0+1+2+...+8)=9a+d*8*9/2=9a+36d=105; a+4d=11+2/3. Since a and d are supposed to be integers, this equation cannot be satisfied because of the fraction 2/3. The numbers 1 4 6 9 12 14 17 20 22 could be used to construct a 3x3 magic square (almost). 20  1 14  6 12 17 all but one sum is 35  9 22   4 The following numbers form a true magic square: 1 11/3 19/3 9 35/3 43/3 17 59/3 67/3: 59/3      1 43/3 19/3 35/3    17      9 67/3 11/3 The numbers 5 20/3 25/3 10 35/3 40/3 15 50/3 55/3 similarly form a true magic square.
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the sum of the consecutive numbers is 35. what is the square of the largest of these numbers?

???????? HOW MANY ??????? sum=35 me assume "number" is INTEGER if yu sum 1 number, biggest number=35 & square=35*35=1225 if 2 numbers, yu hav 17 &18, biggest=18 18*18=324
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the magic square which has both positive and negative numbers and the sum of numbers is same horizontal, vertical, and diagonal

For odd-numbered Magic Squares, we start with the lowest number in the top centre square. If we have a 3X3 square, we start with -4 and end with 4. Let's refer to the square using a grid with rows A, B, C and columns 1, 2, 3. Using the up-1 right-1 rule: Square A2=-4; C3=-3; B1=-2; C1=-1; B2=0; A3=1; B3=2; A1=3; C2=4. 3 -4 1 -2 0 2 -1 4 -3 5X5 square: 4 11 -12 -5 2 10 -8 -6 1 3 -9 -7 0 7 9 -3 -1 6 8 -10 -2 5 12 -11 -4 Both of these squares sum to zero horizontally, vertically and diagonally.. Using the diagonal rule for 4X4: -7 7 6 -4 4 -2 -1 1 0 2 3 -3 5 -5 -6 8 The total is 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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square numbers with the sum of 13

The problem below is from my son's third grade math homework. There is a cube with 9 squares and the instructions are below. Fill in the boxes with the numbers 1,3,5,7,9,11,13,15,and17 such that the sum of the numbers in the horizontal,vertical and diagonal positions are the same. Each number can only be used once.
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I really need help with these optimization problems. Thanks.

Well the difference between them is even though they are the same numbers, you said negative and positive. I see -25 and +25 are different because -25 is below zero and +25 is above zero. That is what is different between them.
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What is the value of SS (sum of squared deviations) for the following sample?

data=2,3,4, 7 num numbers=4, su sum=16, averaej=4, median=3.5, biggest =7 sum av squares=14 varians = (sum av squares) / (num numbers)=3.5 std deviashun=1.87083
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