Math manipulation problems, how to do math problems?

If you have 100 numbers and half of each cycle, you need to do it 7 times.

Suppose you start with 1, then (k=0,1,2,3,..

The first time to cross 1+2k, the last 99 and then skip 100, the second time to start from 2+4k, 100 is still not divisible and the remaining 100 is skipped.

The third time starts from 4+8k, this time 100 is divisible, the last time is crossed out of 100 the fourth time because 1,2,3,4,5,6,7 is crossed, 8 has to be skipped again, so starting from 16 (k+1), the rest is a period of 16, starting from 8, that is, 8+16k

You may wish to list the values that meet the requirements within 100 and then draw them, (note that 8 is the beginning of the previous round, so 8 cannot be crossed out at this time, you need to skip 8, start with 24) and the last remaining number is 72

If you ask for a number of 45 to start, then you can get it according to the correspondence.

x+100k---45

x should be equal to 74, at which point k takes -1

So starting with 74, the jump is crossed out of 100 numbers, and the last number is 45

Nonsense... Obviously it's 16

I've done the math.

Obviously, the number crossed out each time = half of the original number.

After 2 n<100<2 (n+1).

Get n=6, if you start from 1, then the last remaining number is 2, 6=64, so starting from 100, the last remaining number is 63

From [100-(63-45)]=82, the rest is 45

You can classify the rectangles of numbers.

Since they are all stacked in an orderly manner, you can do it without worrying about the size of the number, and simply classify the number (the land bridge is the first row and the third row, although they are different in size, but the arrangement is the same. This is very different from the first question, the first question is disorderly stacking, a large rectangular early base fierce shape on the left and the 4 on the right are obviously not one category, this kind of should be separated).

This question from Example 6.

a single rectangle, 21 in all; vertical Japanese characters are placed, 14; horizontal Japanese characters are placed, 18; 7 of the main words; horizontal eyes, 15; Tian character, 12 ......

Total 21 + 14 + 18 + 7 + 15 + 12 + 12 + 9 + 6 + 6 + 3 + 5 + 4 + 3 + 2 + 1 + 2 = 140 (pcs).

Same for the last question.

16 + 12 + 12 + 8 + 8 + 4 + 3 + 3 + 2 + 2 + 1 = 75 (pcs).

What to do with math problems? Finch letter.

Count squares. <>

Answer: The general practice is to add a number of one by one, add two or two numbers, add three or three numbers or excited, and add four or four to the number of rounds ,......It can't be omitted, it can't be repeated, and it needs to be careful.

<> Anna 53, Banksy 17, Coty 20

Don't know how to ask me.

Anna initially had $36, Banksy had $18 and Coty had $54.

Triangle area = base height 2

s=ah÷2

Rectangle area = length and width.

s=absquare area = side length side length.

s = a 2 parallelogram area = base height.

s=ah trapezoidal area = (upper bottom + lower bottom) height 2

s=（a+b）×h÷2

Circle area = pi square of the radius.

s=πr^2

Sector area = pi square of radius number of angles of the center of the circle and number of angles of circumference.

s=πr^2×n÷360

Box surface area = (length and width + length and height + width and height) 2s = (ab + ah + bh) 2

Cube surface area = 6 squared of the edge length

s=6a^2

Box volume = length, width, height.

v = abh cube volume = cube of edge length.

v = a 3 cylindrical side area = bottom perimeter height.

s=ch=πdh=2πrh

Cylindrical surface area = side area + base area 2

s=2πrh+2πr^2

Cylindrical volume = base area of the tall eggplant.

v=sh=πr^2h

Cone volume = base area height 3

v=sh÷3=πr^2h÷3

Counts: All are the same times, for example, if the number of times on the left side is 2, then the right side is all 2

Number of items: If there are n numbers in the left parentheses, then there are n+n(n-1) 2 items on the right.

Coefficient: The coefficient of the single letter on the right is 1, and the coefficient of the product of the two letters is 1

There are five groups of 1 brigade limb group: 1 and 92 and 83 and 74 and 6

In this way, any combination of five groups can be feasted. That is: 2 to the 5th power = 32 to list too many to name a few:

The same goes for empty sets.

Square the two sides of the known conditions, and I believe you can make it. Hehe! The result is still 2

You'll know when you learn calculus. All formulas have a fixed solution.

2.Let the equation of 1 yuan and 1 order.

10 minutes = 600 seconds, if Wang Qiang runs x seconds at a speed of 6 meters, then the time he runs at a speed of 4 meters is (600-x) seconds.

6x+4*(600-x)=3000, the solution is x=300, note that what you get here is the number of seconds, which should be converted into meters, so 6*300=1800 meters, get Wang Qiang ran 1800 meters at a speed of 6 meters and seconds.

Set up the purchase of type A TV x units, B type TV 50-x units, 1500 * x + 2100 * (50-x) = 90000

x=25，50-x=50-25=25

Set up the purchase of a type of TV Y station, C type TV 50-Y station, 1500 * Y + 2500 * (50-y) = 90000

y=35，50-y=50-35=15

Set up the purchase of B TV Z station, C TV 50-z station, 2100 * Z + 2500 * (50-Z) = 90000

z=greater than 50, this option is rounded.

Therefore, there are two schemes: 25 units each for A and B; or 35 units of A and 15 units of C.

Therefore, in order to make the most profits, you should not buy B TV, so you should choose 35 units of A and 15 units of C.

Solution: Because d=|ax+by+c|Divide by the root number (a square + b squared) according to the title.

d=|3·2+4·2+（—6）|Divide by the root number (3 + 4) so, d = 8 5

The formula for the distance from the point to the straight line is: ax+by+c, the absolute value of the above root number below a square plus b square, that is, 3*2+4*2-6 than the upper root number 9+16, the result is 8/5

According to the formula for the distance from a point to a straight line d=|3*2+4*2-6|/（3^+4^）^1/2)=8/5

The narrative is that in absolute terms, 3 times 2 plus 4 times 2 minus 6 divided by the square of 3 under the root number plus the square of 4.

The formula for the distance from the point po(xo,yo) to the straight line l:ax+by+c=0 is:|axo+byo+c|Divide by the square of a + the sum of the square of b and open again.

d=|3*2+4*2-6|/（2^2+2^2)^1/2=2*2^1/2