What is the answer and process of this math problem?

A positive integer array is known as shown in the figure on the right:

Find a similar, the same solution for you to see!

<> you just go backwards!!

The first n rows are: 1+2+3+...n numbers.

i.e. (1+n)n2

Because 63*64 2=4032=2016

So 2014 is on line 63.

In the odd-numbered row, the number increases from left to right (i.e., from small to large), so 2014 is 61 from left.

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Permutations and combinations. Calculate it yourself. For example, in the vertical column 3, 4 5 6 is from small to large.

Column 3 is taken as 3 3 9 9 2 and the approximation is equal to 5 and the median of the odd number series is obtained. The fourth column that is vertical. 10 9 8 7 from large to small

Column 4 is 4 4 16 16 2 8 The median number of the even number is equal to 8, which is the second smallest number from the right to the left of the even number column. So the extrapolation verification 2014 ...Approximate the range between the maximum and minimum values, and the number of vertical columns [(1) 2] is the maximum value of the maximum value minus the minimum value of the number of vertical columns 1

63 1 2] 63 2016 Maximum Minimum is 2016 63 1 1952 Horizontal total of 63 numbers Maximum is 2016 2016 2014 2 So 63 2 61 should be from right to left. Good luck with your studies! You haven't learned a bit yet, so I'm using my own original idea.

It's not high yet

This problem is an S-shaped arrangement of natural numbers, and then the number of numbers in each row is a series of equal differences, and the first number in the n+1 row can be calculated as d = 1 + (n + 1) * n 2, so that 2014 is substituted into d, and 1 + (n + 1) * n 2 can be obtained< the largest integer n of 2014, we get n=62, that is, 2014 is in line 63, and the first number in line 63 is 1954, and remember that this does not mean that the rightmost number is 1954, because it was emphasized earlier that the numbers are arranged in an S-shape, and now we can judge whether 1954 is on the far right or left side of line 63: we can see that the odd number row is counted from left to right, and the even number row is from right to left, and because 63 is an odd number, so 1954 is on the far left, And 2014-1954=60, so 2014 is the 61st number from left to right in line 63.

1 1 pcs. 3 2 2 pcs.

4 5 6 3 pcs.

Therefore, the number of each row is a series of equal differences 1, 2, 3, 4, 5....According to the formula, (1+63)*62 2=2016>2014, that is, 2014 is in line 63.

And because, the order of the numbers in the odd rows is from left to right, and the order of the numbers in the even rows is from right to left, so the order of the 63rd line is from left to right, and 2016 is in the 63rd, so 2014 is in 63-(2016-2014)=61.

1 in the first row, 2 in the second row, 3 in the third row, and so on, n ...... in the nth rowSo 2014 is the number in line 63, and the order of line 63 and line 3 is the same, and the numbers change from small to large from left to right.

So the arrangement of line 63 is 1954, 1955, 2014, so 2014 is the 61st number from left to right in line 63.

First of all, it is a positive integer arranged in a polylinear arrangement from top to bottom, with knowledge of number sequences, and there are n+1 numbers in the n+1 row, and the beginning is n(n+1) 2, and then the number is enough.

(-3x) 2-(2x+1)(3x-2)-3(x+2)(x-2)=0 Solution: 9x -(6x -4x+3x-2)-3(x -4)=09x -6x +x+2-3x +12=0

x=-14m 2-m+1=0, find the value solution of m 3-3m 2+3m-2011: m -m=-1 [replace it with -1 if you see m -m below] m 3-3m 2+3m-2011

m³-m²-2m²+3m-2011

m（m²-m）-2m²+3m-2011

m-2m²+3m-2011

2m²+2m-2011

2（m²-m）-2011

1.The original formula: 9x -(6x -x-2)-3(x -4)=09x -6x +x+2-3x +12=0

x=-142 is set to: m -m+1=0

m-1)²+m=0

Get m=-(m-1).

m³-3m²+3m-2011

m³-1³-2010-3m(m-1)

m-1)(m²+m+1)-3m(m-1)-2010=(m-1)(m²+m+1-3m)-2010=(m-1)(m-1)²-2010

Substituting m=-(m-1) gives -m(m-1)-2010 to get -m +m-2010

Because there is m -m+1=0 we get -m +m=1, so -m +m-2010

The value of m 3-3m 2+3m-2011 is 2009

The first question is all open x 2 and it is eliminated, and finally x = -14

The second question is that the first equation has no solution, are you typing it wrong?

1。Let the big rope be x

15x+[25-x]×1=221

x=14 substitution, small rope from=11

2。Let the big ship be baix

6x+【10-x】x4=54

x=7 substitution, boat=3

Below the landlord. du, I gave the answer directly. zhi

3 Checkers = 16 Chess = 8

4 Nine of each. dao

5 Large = 15 Small = 16

6 Damaged: 1 pc.

7 11 x 5 jiao 7 x 1 corner.

8 4 triangles 36 rectangles.

You're not going to make a question bank, are you?

Solution: 1. Set up a large rope x and a small rope of 25-x

15x+25-x=221

14x=196

x=14 then 25-x=25-14=11

That is, 14 large ropes and 11 small ropes.

2. Set up x large boats and 10-x small boats.

6x+4（10-x）=54

6x-4x=54-40

2x=14x=7, then 10-x=10-7=3

That is, 7 large boats and 3 small boats.

3. Set chess x vice, checkers 24-x vice.

2x+6（24-x）=112

6x-2x=144-112

4x=32x=8, 24-x=24-8=16

That is, 8 sets of chess and 16 sets of checkers.

4. Set x chickens and 18-x rabbits.

2x+4（18-x）=54

4x-2x=72-54

2x=18x=9, 18-x=18-9=9

That is, 9 chickens and 9 rabbits.

5. Set up x large barrels and 31-x small barrels.

4x+x=15

then, 31-x = 31-15 = 16

That is, 15 large barrels and 16 small barrels.

6. Set up x damage.

250×2-（20+2）x=478

22x=500-478

22x=22

x=1 means 1 damaged.

7. If there are x pieces in 1 corner, then there are 18-x pieces in 5 corners.

x+5（18-x）=62

5x-x=90-62

4x=28x=7, 18-x=18-7=11

That is, there are 7 pieces in 1 corner and 11 pieces in 5 corners.

8. There are x triangular cards and 40-x rectangular cards, and 3x+4(40-x)=156

4x-3x=160-156

x=4, then 40-x=40-4=36

That is, there are 4 triangular cards and 36 rectangular cards.

1. Let all the big ropes 15 come 25 221 154 source 154 (15-1) 11 25-11 14 Answer: 14 large ropes and 11 small ropes.

2. Set all large ships 6 10-54 6 6 (6-4) 3 10-3 7 Answer: 7 large boats and 3 small boats.

3. Set all checkers 6 24-112 32 people 32 (6-2) 8 pairs 24-8 16 pairs Answer: 8 sets of chess, 16 sets of checkers.

4. Let all rabbits 18 4 54 18 18 (4-2) 9 18-9 9 Answer: 9 chickens and rabbits.

5. Set all to large barrels 4 31-100 21l 21 barrels 31-14 17 barrels Answer: 17 large barrels and 14 small barrels.

6. Set all good 250 2 478 22 yuan 22 2 22 1 Answer: A total of 1 is damaged.

Yuan 2 jiao 62 jiao set all 5 jiao 5 18 62 28 jiao 28 5 1 7 18-7 11 Answer: 11 pieces of 5 jiao. Seven pieces of 1 corner.

8. Let all be rectangles 4 40 156 4 4 4 4 3 4 40-4 36 Answer: 4 triangles and 36 rectangles.

If there is someone behind me who is exactly like me, it must be plagiarized from me.

Hope it helps.

It's too little... There are so many questions, and it is not difficult for young students to do the questions by themselves, and you can't expect others to take the exam.

1.Untie the big rope x

15x+（25-x）=221

14x=196

x = 14 small ropes: 25-14 = 11 pieces.

2.Column equation: Let the big ship have x bars, then the small boat has (10-x) bars.

6x+4(10-x)=54

6x+40-4x=54

2x+40=54

2x=54-40

2x=14x=14/2

x=710-x=10-7=3

2x+6y=112

The solution is x=8 y=16

4.If there are x chickens, then there are (12-x) rabbits, 2x+4(12-x)=36

x=612-x=6 (only) so there are 6 to chickens and 6 rabbits.

5.Let the large barrel x the small barrel 31 -x 4x + x) =100

x =1531-x =16

15 large barrels and 16 small barrels.

6.Broken x.

250-x)*20-100*x=4400

5000-20x-100x=4400

120x=600

x=57.Solution: If there are x 1 jiao coins, then there are (18-x) 5 jiao coins 6 yuan 2 jiao = 62 jiao 1 x + (18-x) 5 = 62

x+90-5x=62

x=718-7=11 (pcs).

8.Solution: If all the cards are rectangular, then there are:

40*4=160 corners.

160-156=4 Now there are fewer 4 corners, that is, the triangle is less than the rectangle.

And a triangle is one corner less than a rectangle, so there are a total of 4 triangles and 36 rectangles.

1. A large rope.

x small rope from y x+y=25 15x+y=221 to solve 14 large ropes and 11 small ropes.

2. 3 small boats, 7 large boats 3, elephant 6, 16 bright chess 4, 9 chickens and rabbits 5, 15 large 16 small 6, damaged 1.

7 corners, 5 corners, 11 8, 4 triangles, 36 rectangles.

1.The small rope has (25 15—

221) copy(15—bai1)=11 large rope has du25—zhi11=14 2Small boats have (6 10-54) (6-4) = 3 Large boats have 10-3 = 7 The following questions are answered in this way, if not, don't do it.

degrees) 6 divided by 9 = 2/3

The balloon is two-thirds of a kilometer high.

4 divided by 2 = 2.

Then the total freight price A to gold A to Shu B to gold B to Shu y = 35x + 30 (100-x) + 40 (90-x) + 45 (x-40) y = 10x + 4800

x is in a closed range between 40 and 90.

It's a function.

So the minimum value x is taken as 40 and y=5200

The plan is that A will send 40 to gold, 60 to Shu, B to send 50 to gold, Shu will not send high school will be almost forgotten math, I don't know if it's right.

Find the center of the circle first, the center of the circle is on the perpendicular line of the straight line AB, the midpoint of AB on the perpendicular line of the straight line BC is (9 2, 3 2), the slope of AB is (3-2) (4-5)=-1, then the perpendicular line of AB is y-3 2=x-9 2, that is, Y=X-3, (1) the midpoint of BC is (3, 1), and the slope of BC is 2 (5-1)=1 2 then the BC perpendicular line is Y-1=-2(X-3), that is, Y==2X+5, (2) (1), (2) The simultaneous solution obtains X=-8, Y=-11, i.e., the radius of the center of the circle (-8, -11) r 2=(-8-1) 2+(-11-0) 2=202, then the circle equation is.

x+8)^2+(y+11)^2=202