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Look, the 20 numbers from 1 to 20:

Only the following four combinations can be given a 0 at the end of the factorial, note that each number can only be used once.

15 an even number.

You might see 20 15 300, or 20 5 100, or 20 10 200, with two zeros, but that's a combination of two of the four cases above.

So 20! There are 4 zeros at the end of .

Then consider that the square of any natural number within 10 will not end in 0 except that the square of 0 is equal to 0.

So 20! ) has 8 zeros at the end.

Verify with a calculator and see at a glance:

The original question was to say "Don't count 20!" value of the case estimate", right! As a math problem, if you don't let the calculation come alive...

5 10 15 20 These 4.

Square again.

4*2=8.

To put it simply, the case of 0 at the end of the result can only be:

Any 1 even number * 5 or 15, or a 1 multiplier ending with 0 at the end of the above result.

You see: what kind of multiplication of two numbers will result in a zero? It's not hard to know:

1.When 2*5The result is followed by a zero.

2.When the single digit of a number is zero. To sum up the above two cases:

20!The number of zeros in the result of *2 can be equivalent to: 20*15*10*5*2*2 gives 60000, so the result is 4 zeros.

(1) Approximate tens of thousands of accurate to -, there are -2-significant figures, which are -7;9——

2) The quadratic of the approximate number to the fourth power is accurate to -10000-, and there are -3-significant figures, which are -5;0；8.—

3) The approximate number is accurate to the point that there are -3 significant figures, which are -8;0；9—

1.Bringing (1,1) into y=3x+b, we get b=-2, so the straight line is y=3x-2

When x=0, bring in a straight line, y=-2

So, b(0,-2).

2.When x=1, bring in the function y=-x+1 and y=m=0

When x=2, bring in the function y=-x+1 and y=n=-1

So m>n

3.Find the intersection point of the straight line y=3x-2 and the straight line y=-2x+1, and the intersection point of the continuous equation system is, (3 5, -1 5).

When x>3 5, the image of the line y=3x-2 is directly above the image of the line y=-2x+1.

4.According to the straight line y=-2x+k, the intersection points of the straight line and the x,y axis are (1 2k,0) and (0,k), respectively

The area of the triangle enclosed by the line y=-2x+k and the coordinate axis is 1 2(1 2k*k)=9

So, k = 6 or -6

The straight line is y=-2x+6 or y=-2x-6

1.Bringing point a into the function formula gives b=-2, then y=3x-2, and substituting x=0 into it gives y=-2, so b(0,-2).

n=-1, so m>n

3.First, find the intersection point of the two straight lines as (3 5,-1 5), and it can be seen from the image that when x>3 5, the image of the line y=3x-2 is directly above the image of the line y=-2x+1.

or -6, so y=-2x+6 or y=-2x-6.

5.As long as k+b=2 and k>0 are satisfied. For example, y=x+1.

The steel pipe is erected, which is equivalent to the volume of a large circle and the volume of the scattered column minus the volume of a cavity rolling small cylinder, so the volume of the remaining part is v = ( 8 - 6 ) 250 = 7000 (cubic centimeters), and the weight of the steel pipe is 7000 kg.

Don't listen to them, I've seen them all, there is no right one, in fact, it is 10 times.

The car went to pick him up, and the car went 10 minutes less than usual, divided by 5 minutes, that is, the car met him 5 minutes in advance, and at this time the person walked 55-5, that is, 50 minutes, and these 50 minutes are the distance that the car usually travels in 5 minutes, that is, the speed of the car is 10 times that of a person.

Why subtract 5, because at the same time as the person is leaving, the car is also going to the station, but the car picks up the person 5 minutes earlier than usual, and the person walks for 50 minutes, and sees the car 5 minutes earlier than usual.

There is a problem with this topic! He foresaw the car halfway, the speed of the car is greater than the person, and the car must arrive at the factory earlier than the person.

If he rides in the car, he and the car arrive at the factory on time!

If he hadn't taken the car, he would have arrived after the car arrived at the factory on time and he would have been late. I can't be ten minutes early!

Arithmetic Algorithm:

Because the engineer walked for 55 minutes, the car picked up the engineer 10 2 = 5 (minutes) in advance, so the distance that the engineer walked in 55 minutes is equivalent to the distance that the car walked in 5 minutes.

55 5 = 11 (times).

Equation algorithm: Let the engineer walk at a speed, the car at a speed b, and the distance the engineer travels in 55 minutes is 55a

Since the car is in accordance with the usual rules, the engineer is 10 minutes early, which can be regarded as the car driving 10 minutes less.

Since cars meet engineers on the way to factories and stations, the distance from the point of meeting is twice as long as the distance from the meeting point to the station.

The distance the car drives less = 2 55a

Less car travel = 10 minutes car travel.

2×55a=10b

b/a=110/10

11 cars are 11 times faster than engineers.

The information on the topic is incomplete. You're looking at the title.

Set at the beginning of the year** x yuan per kilogram.

2x-36=108

2x=108+36

2x=144

x = 72 A: 72 yuan per kilogram at the beginning of the year.

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At the beginning of 2010, x yuan per kilogram.

2x - 36 = 108

2x = 144

x = 72 yuan kg.

From the perspective of raw materials: x is less than (80000 + 2000) state Lu Shichang 5 x is less than 16400 units.

From the perspective of the return of the personnel sail: x is less than 80 * 2400 12 x is less than 16000

So: x is less than 16000

The first team builds 3 4 = km hours.

The second team repaired 4 5 = km hours.

The third team repaired 5 8 = km hours.

So the second team is the most efficient.

One team, 3 4 = kilometers per hour.

Second team, 4 5 = km h.

Three teams, 5 8 = kilometers per hour.

Because of the second team high.

First team road repair efficiency: 3 4=

2nd team road repair efficiency: 4 5=

The third team's road repair efficiency: 5 8=

So the second team is the most efficient.

The set b is a circle, the center of the circle is (1,1), the radius is r, and from the relation of the necessary and insufficient conditions, we can know that b is a true subset of a, that is, all the points in the circle b satisfy the three conditions in the set a, <=> (1,1) to the distance of the three lines in a is greater than or equal to r, and (1,1) satisfies the three conditions in the set a.

Straight line x 3 = > r 2

x+y-4 0 => r root number 2

y x+2r = > r root number (2 * r) and 1 1 + 2r

So 0 < r < root number 2