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what is the sum of 36,37,35,39,35,39,31

i need to find the mean of thoes  numbers

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Romans 8:35-39 NIV - Who shall separate us from the love of ...

Romans 8:35-39. Romans 7 Romans 9 ... Romans 8:35: ver 37-39; Romans 8:35: 1Co 4:11; 2Co 11:26, 27; Romans 8:36: Ps 44:22; 1Co 4:9; 15:30, 31; 2Co 4:11; 6:9; 11:23 ...

Determine The Following Sum: 36 + 37 + 38 + 39 ... - Chegg

Question: Determine the following sum: 36 + 37 + 38 + 39 + .... Show transcribed image text Determine the following sum: 36 + 37 + 38 + 39 + ... + 120 + 121 + 122 ...

38 (number) - Wikipedia

38 (number) This article ... ← 30 31 32 33 34 35 36 37 38 39 ... (which is trivial) and order 3. The sum of each row of an order 3 magic hexagon is 38.

Rasmus - Math, Prime numbers and divisibility, Lesson 1. - 7 ...

Lesson 1. A prime number is a ... If the sum of the digits of the number ... 31: 32: 33: 34: 35: 36: 37: 38: 39: 40: 41: 42: 43: 44: 45: 46: 47: 48: 49: 50: 51: 52 ...

Solved: Find the sum of 36 + 37 + 38 + 39 +......... + 146 ...

Answer to Find the sum of 36 + 37 + 38 + 39 + ... Find the sum of 36 + 37 + 38 + 39 + ... Consider natural numbers 36, 37, 38 ...

Romans 8:35-39 Commentary - Home - Sermon Writer

romans 8:35-39. commentary: romans 8:26-39: ... (vv. 35-39). romans 8:35-36. ... 37-39. we are more than conquerors.

Finding Two Numbers - hoffman

Finding Two Numbers ... Product = 98 35. Sum = 16, Product = 48 36. Sum = 14, Product = 48 37. Sum = 12, Product = 36 38. Sum = 12, Product = 27 39. Sum = 14, Product ...

37 (number) - Wikipedia

← 30 31 32 33 34 35 36 37 38 39 ... (which is the 21st prime number). 37 is the fifth lucky prime, ... Every positive integer is the sum of at most 37 fifth powers ...

What is the sum of the first 30 odd numbers - Answers.com

The answer for the first 30 odd numbers (1,3,5,7,9,11,13,15,17,19,21,23,25,27,29,31,33,35,37,39,41,43 ... the sum of the first 30 odd numbers is 900.1+3+5+7+9+11 ...

use the integers 0, -2, -3, -4, -5, -6, -7, -8, and -10 to fill in a 3x3 magic square

First, find out what the sum is. Consider 3 rows of 3 numbers: each row has sum S. The rows contain all the numbers, so since the sum of the numbers is -45, 3S=-45 and S=-15. Take each number in turn and work out what other numbers can go with it to make -15: 0: (-10,-5),(-8,-7) -2: (-10,-3),(-8,-5),(-7,-6) -3: (-10,-2),(-8,-4),(-7,-5) -4: (-9,-2),(-8,-3),(-6,-5) -5: (-10,0),(-8,-2),(-7,-3),(-6,-4) -6: (-7,-2),(-5,-4) -7: (-8,0),(-6,-2),(-5,-3) -8: (-7,0),(-5,-2),(-4,-3) -10: (-5,0),(-3,-2) Now, consider the square: C1 C2 C3 C4 C5 C6 C7 C8 C9 Each cell participates in various sums. C1, for example, is in the first row and column and on a diagonal, so it is involved in 3 sums. C3, C7 and C9 are also similarly involved. Here's a complete list: C2,C4,C6,C8: 2 sums C1,C3,C7,C9: 3 sums C5: 4 sums Now we match the number of sums with the numbers that can be involved in at least the same number of sums. C5=-5. 2 sums: {C2 C4 C6 C8}={0 -6 -10} (exact) < {-2 -3 -4 -7 -8} (3 sums) 3 sums: {C1 C3 C7 C9}={-2 -3 -4 -7 -8}, one of these can be used to satisfy 2 sums, leaving 4 to match the cells. So {C1 C3 C7 C9}={-2 -3 -4 -7)|{-2 -3 -4 -8}|{-2 -3 -7 -8}|-2 -4 -7 -8}|{-3 -4 -7 -8} and {C2 C4 C6 C8}={0 -6 -8 -10}|{0 -6 -7 -10}|{0 -4 -6 -10}|{0 -3 -6 -10}|{0 -2 -6 -10} Because of symmetry we can assign 0 to C2. So C8=-10: C1 0 C3 C4 -5 C6 C7 -10 C9 C7+C9=-5, which means only -2 and -3. Because of symmetry we can assign either number to C7: C1 0 C3 C4 -5 C6 -2 -10 -3 C1-8=-15 so C1=-7; C3-7=-15, so C3=-8: -7 0 -8 C4 -5 C6 -2 -10 -3 C4=-6 and C6=-4: -7 0 -8 -6 -5 -4 -2 -10 -3

First, we make all the numbers the same length by prefixing zeroes. The longest number has 8 digits, so we use leading zeroes for numbers with fewer than 8 digits.  0 8 7 6 5 4 3 1  0 6 5 6 6 6 8 7  0 0 8 7 6 5 4 3  7 6 5 4 3 4 3 3  0 0 0 8 7 6 5 5 A gap has been left before each digit to leave space for a marker. Starting on the bottom left we add together the digits in the same column and we insert a marker when this sum exceeds 10, and we keep in mind the residue. We're essentially adding digits a pair at a time. The first column presents no problem so we simply write down 7 (temporarily) as the sum of the first column, but the second column illustrates the method better. Starting at the bottom, we have the digits 0 6 0 6 8. The zeroes we can ignore. So 6+6=12. We place a marker in front of the second 6 and keep in mind the residue 2. We add 2 to 8=10 and we insert another marker to the left of 8 and record the sum as 0. So far we have:  0•8 7 6 5 4 3 1  0•6 5 6 6 6 8 7  0 0 8 7 6 5 4 3  7 6 5 4 3 4 3 3  0 0 0 8 7 6 5 5  7 0 next column: 0 5 8 5 7 we proceed the same. Ignore the zero; 5+8=13; mark the 8 and keep 3 in mind. Add 3 to 5=8, then add 7 to 8=15, and insert another marker before 7. Put 5 in the answer.  0•8•7 6 5 4 3 1  0•6 5 6 6 6 8 7  0 0•8 7 6 5 4 3  7 6 5 4 3 4 3 3  0 0 0 8 7 6 5 5  7 0 5 We continue this way up to the last column:  0•8•7•6 5 4 3 1  0•6 5•6•6•6•8 7  0 0•8 7 6 5•4•3  7 6 5•4•3•4 3 3  0 0 0 8 7 6 5 5  7 0 5 1 7 5 3 9 +2 2 3 2 2 2 1  (sum of markers)   9 2 8 3 9 7 4 9 Now we count the markers. Between the first and second columns we have 2 markers so we add 2 to the result of the first column to give us 9 replacing the 7. Then 0+2=2, 5+3=8, and so on. We do this up to the last column. I've shown the sum of the markers for each column. The Vedic arithmetic only requires you to be able to add only the numbers 1 to 9 together. The faster you can do this the faster you can do Vedic addition (drop tens method).

How many different distributions can the manager make if every employee receives at least one voucher?

There must be 100 vouchers because each is worth RM5 and the total value is 500 ringgits or RM500. (i) Each employee receives at least 1 voucher, so that means there are 95 vouchers left to distribute. We can write each distribution as {A,B,C,D,E}: starting with {0,0,0,0,95}, then {0,0,0,1,94}, {0,0,0,2,93}, ..., {0,0,0,95,0}, {0,0,1,0,94}, ..., {0,0,95,0,0}, ..., ..., {95,0,0,0,0}. So when A=B=C=0, {D,E} range from {0,95}, {1,94}, ..., to {95,0}, 96 ways. When A=B=0 and C=1, {D,E} range from {0,94} to {94,0}, 95 ways. Finally when C=95 so {C,D,E}={95,0,0} we will have covered 96+95+...+1=96*97/2=4656 ways. (The sum of the whole numbers from 1 to n is given by S=n(n+1)/2.) That was for B=0; when B=1, {C,D,E} ranges from {0,0,94} to {94,0,0} to cover 95*96/2=4560 ways. When B=2 it's 94*95/2=4465 ways. So for A=0 we have 4656+4560+4465+...+3+1 = 96^2+94^2+...+2^2 = 4(48^2+47^2+46^2+...+2^2+1^2)=4*48*49*97/6=152096 (the sum of the squares of the whole numbers from 1 to n is given by S=n(n+1)(2n+1)/6. Also, the sum of whole numbers between 1 and n taken in pairs gives us: n(n+1)/2+(n-1)n/2 for each pair. This is n^2/2+n/2+n^2/2-n/2=n^2. For the next pair we get (n-2)^2 and so on.) That was just for A=0! For A=1 we have {1,0,0,0,94} to {1,94,0,0,0}. This will give us 4560+4465+...+10+6+3+1 = 95^2+93^2+...+5^2+3^2+1. There is a formula for this sum. It is S=(n+1)(2n+1)(2n+3)/3, where 2n+1=95, so n=47. So for A=1, the number of ways is 48*95*97/3=147440. For A=2 we have {2,0,0,0,93} to {2,93,0,0,0} which produces 94^2+92^2+...+4^2+2^2 = 4(47^2+46^2+...+1) = 4*47*48*95/6 = 142880. So we alternate between two formulae as A continues to go from 3 to 95.  More to follow...

what is this pattern? 2, 5, 9, 19, 40, 77, 137

Compound Arithmetic Sequence a0          a1          a2          a3          a4          a5          a6         b0          b1         b2          b3          b4           b5  ------- 1st differences               c0          c1          c2          c3          c4  -------------- 2nd differences                        d            d            d             d   ------------------- 3rd differences = constant We have here an irregular sequence (a_n) = (a_1, a_2, a_3, ..., a_k, ...). The differences between the elements of (a_n), the 1st differences, are non-constant. So also are the differences between the elements of (b_n), the 2nd differences. However, the differences between the elements of (c_n) are constant. This makes (c_n) a regular arithmetic sequence, and so we can write, cn = c0 + nd,  n = 0,1,2,3,... The sequence (b_n) is non-regular, but we can write, b_(n+1) = b_n + c_n Using the expression for the c-sequence, b_(n+1) = b_n + c_0 + nd Solve the recurrence relation for b_n Develop the terms of the sequence. b1 = b0 + c0 b2 = b1 + c0 + 1.d = b0 + 2.c0 + 1.d b3 = b2 + c0 + 2.d = b0 + 3.c0 + (1 + 2).d b4 = b3 + c0 + 3.d = b0 + 4.c0 + (1 + 2 + 3).d ‘ ‘ ‘ b(n+1) = b_0 + (n+1).c0 + sum[k=1..n](k) * d b_(n+1) = b_0 + (n+1).c0 + (nd/2)(n + 1) b_(n+1) = b_0 + (n+1)(c_0 + nd/2) We could also write c0 = b1 – b0 in the above expression. Now that we have a rule/expression for the general term of the b-sequence, let us analyse the a-sequence. Since the sequence (a_n) is non-regular like (b_n), we can write, a_(n+1) = a_n + b_n Substituting for the general term of the b-sequence, a_(n+1) = a_n + b0 + n(c0 + (n-1)d/2)      (co can be replaced later with c0 = b1 – b0) Solve the recurrence relation for a_n Develop the terms of the sequence. a1 = a0 + b0 a2 = a1 + b0 + 1.c0 + 1.0.d/2 = a0 + 2.b0 + 1.c0 + 1.0.d/2 a3 = a2 + b0 + 2.c0 + 2.1.d/2 = a0 + 3.b0 + (1 + 2).c0 + (1.0 + 2.1).d/2   (N.B. here, 1.0 = 1*0, 2.1 = 2*1) a4 = a3 + b0 + 3.c0 + 3.2.d/2 = a0 + 4.b0 + (1 + 2 + 3).c0 + (1.0 + 2.1 + 3.2).d/2 . . . a_n = a0 + n.b0 + sum[k=1..n-1] k * c0 + sum[k=1..n-1](k(k-1)) * (d/2) a_n = a0 + n.b0 + (c0/2).n(n-1) + sum[k=1..n-1](k^2) * (d/2) – sum[k=1..n-1](k) * (d/2) a_n = a0 + n.b0 + (c0.n/2)(n-1) + (d/2){(n/6)(n-1)(2n-1) – (n/2)(n-1)} a_n = a0 + n.b0 + (c0.n/2)(n-1) + (nd/6)(n^2 – 3n + 2) Substituting for c0 =  b1 – b0, b1 = a2 – a1, b0 = a1 – a0 Then, c0 = a0 – 2a1 + a2. These substitutions give, a_n = (1 + (n/2)(n – 3)).a0 – n(n – 2).a1 – (n/2)(1 – n).a2 + (nd/6)(n^2 – 3n + 2) Now substituting for a0 = 2, a1 = 5, a2 = 9 and d = 5, the rule is, a_n = 2 + n(n – 3) – 5n(n – 2) – (9/2).n.(1 – n) + (5/6).n.(n^2 – 3n + 2)

arrange 0,7,5,1,-1,-4,-5,-2,2 in circular order such that every three digit summation is 0

Let's look at groups of 3 digits that sum to zero (5 groups): (0,5,-5), (0,1,-1), (0,-2,2), (7,-5,-2), (5,-1,-4) 0 appears in 3 groups; 7 appears in 1 group; 5 appears in 2 groups; 1 appears in 1 group; -1 appears in 2 groups; -4 appears in 1 group; -5 appears in 2 groups; -2 appears in 2 groups; 2 appears in 1 group. Three numbers also appear with negative counterparts: (1,-1), (2,-2), (5,-5). We have 5 intersecting regions: Region A contains the numbers 7 -5 -2 Region B contains the numbers 5 -1 -4 Region C includes (overlaps) -1 from Region B, 0 and 1 Region D includes (overlaps) -2 from Region A and 0 from Region C and 2 Region E includes (overlaps) -5 from Region A, 5 from Region B and 0 from Region C Each region contains numbers whose sum is zero. The singletons 7, 1, -4 and 2 lie outside the intersecting parts of their respective regions (A, C, B and D). 0 is part of overlapping Regions C, D and E and occupies the central position in the regions. The picture below shows 7 circular or elliptical shapes. The central circle containing 0 is enclosed by three ellipses but sits in between two other circles, the contents of which sum to zero, Region A being the right-hand circle and Region B the left. The contents of each ellipse also sum to zero. Region C is the ellipse leaning to the right, Region D the ellipse leaning to the left and Region E is the horizontal ellipse.

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.

What is the sum of all even numbers between 0 and 398?

sum even numbers, start=0, end maebee=398 if inklude 0 & 398, hav 200 numbers first=0, last=398 sum=n*averaej, n=200 averaej=(0+398)/2=199 sum=200*199=39,800