Mathematics problems in the third year of junior high school, mathematics problems in the third year

If you want to "solve problems", you must learn masters! And those who are good at learning are now working hard! So I advised LZ to go to the bookstore in person and pay for a difficult practice.

According to the title: the ratio of two polygons is 2, so.

Let one have the number of sides m and the other 2m, then the corresponding internal angles are respectively.

m-2)*180：(2m-2)*180=1:4.(m-2):(2m-2)=1:4

Namely. 4m-8=2m-2

Solution, m=3

2 m = 6, so the number of sides of the two regular polygons is 3 and 6 respectively

Let the number of sides of one be m and the other be n, then the corresponding sum of the internal angles is respectively.

m-2)180.

n-2)180

According to the title: m to n=2

m-2)180.Compare.

n-2)180=4

The solution yields m=6n=3

Let the number of sides of one of the regular polygons be n, then the other is 2n,4 (n-2) 180=(2 n-2) 180,n=3

So the number of sides of these two regular polygons is 3 and 6 respectively

The center of the outer 3 pillars at the top corners can form an equilateral triangle.

Make the triangle high.

Find out that the high is (3 2 under 3 roots).

There is also a radius at the bottom of the equilateral triangle - 1 2 meters and a radius above it - 1 2 meters.

So these pillars are stacked up to (2+3 roots3) 2 meters.

First of all, I would like to ask, is there a bit of a clerical error in the title? 2 Disturbance 9 should be 9 2.

1.If abc is rt, then ao*bo=CO2, and the solution is m=22In ABC, if AC=BC, then the first-time term of the function is 0, that is, M2 -4M +5 2=0, then M2 -4M +9 2=2, Y=X2-4, then A(-2,0), B(2,0), C(0,-4), Sin ACB=4 5

When calculated m=2, the minimum value is prepared, which is 5 4.

In fact, it is noted that no matter what value m takes, when x = -2 y = 0, then when m changes, the quadratic function takes the cluster of functions that pass through the (-2, 0) point (since the coefficient of the quadratic term is unchanged, the shape of the function is unchanged), since the intersection point of the function with the y axis is -2 (m2 -4m + 9 2), when m = 2, the intersection point with the y axis has a maximum value of -1. In this way, the range of the function cluster can be defined, and it is easy to see that the area of abc is the smallest when m is taken as 2.

The answer is 285.

Backwards: Monkey: A1, A2, A3, A4, A5, A6, A7

Before the 7th minute: 320 320 320 320 320 320 2560 (a7=320 6+640).

Before the 6th round: 160 160 160 160 160 2400 1280 (a6=160 5+1280+320).

Before the 5th minute: 80 80 80 80 2320 1200 640 (omitted later...)

Before the fourth minute: 40 40 40 2280 1160 600 320

Before the third minute: 20 20 2260 1140 580 300 160

Before the second split: 10 2250 1130 570 290 150 80

Before the first minute: 2245 1125 565 285 145 75 40

A1 received peanuts from other monkeys for 6 consecutive rounds after giving them to other monkeys, doubling each time to end up at 640, so after giving other monkeys, A1 monkeys have 10 peanuts and can launch a1 a2 a3 a4 a5 a6 a7 10

And so on 32(2a2 10 2a3 2a4 2a5 2a6 2a7) 640, 》a2 a3 a4 a5 a6 a7 15

From this, the following (after consolidation) can be obtained

a3－a4－a5－a6－a7＝20

a4－a5－a6－a7＝25

a5－a6－a7＝30

a6－a7＝35

A7 40.

a1 2245 a2 1125 a3 565a4 285 a5 145 a6 75 a7 40 The sum is 640 7 4480

Since the square OABC is a square, the coordinates of the bright code (x,y) of point b are x=y, and y=1 x are concatenated to obtain x=y=1(x>0). Let the ordinate of the point e be y, then the number coordinate of the horizontal key x=y+1, and y=1 x are bisonic, and the key x, y is the coordinate of the point e. The rest is easy.

From the known conditions, the first row of the odd number series is the square of the odd number, and so on.

The first column of the even row is the square of the even number, and so on.

Since 44*44=1936, i.e. 1st column of line 44 is 1936

It is also known that even numbers act in decreasing sequences, then the first column of line 45 is 1937

The first row of column 45 is 45*45=2025, and the odd number series is also a decreasing number series, so from the first row of column 45 to the next row 17 (17=2025-2008), that is, the 18th line of column 45 is 2008

Solution: Set by the problem, know the bottom area of the water tank s=40 25=1000 (cm 2).

Tank volume v tank = 1000 50 = 50000 (cm 3), iron block volume v iron = 10 10 10 = 1000 (cm 3).

1) If the water depth in the water tank is exactly 50cm after putting in the iron block, 1000A + 1000 = 50000, a = 49 (cm).

So, when 49 a 50, the water depth is 50cm (excess water overflows).

2) If the water depth in the water tank is exactly 10cm after putting in the iron block, 1000A+1000=10000, A=9 (cm) is obtained

So, when 9 a 49, the water depth is (a*40*25+10*10*10) (40*25) = a+1)cm

3) From (2), when 0 a 9, let the water depth be xcm, then.

1000x=1000a+100x, x=10a9 (cm).

Answer: When 0 a 9, the water depth is 10a 9cm;

When 9 A 49, the water depth is (A+1) cm;

When 49 A 50, the water depth is 50cm