For the first question, the math master will come and use it urgently

1. Pyramid volume formula: bottom area * height 3

When 3 and 4 are used as the axis, the bottom radius is 4, 3, the height is 3, and when 5 is the axis, the bottom radius is 3*4 5=, and the height is 5, and the rest is calculated by yourself.

2. Set up person x, when it is divided into 6 books, the child still lacks a book, which is enough for 6 books 4x+20=8x-a

4x=20+a

Note that the value range of a is greater than 0 and less than 8

a=4 x=6

3. Set the minimum angle x, the maximum number of angles 4x, and the third angle nx, where n is greater than 1 and less than 4, then 1+4+n is greater than 6 and less than 9

1+4+n)*x=180 degrees, then x is greater than or equal to 20 and less than or equal to 30 integer acute triangles, each angle should be less than 90 degrees, 4x should be less than 90, x should be less than 23x=20,21,22

4. Set monkey n, and each monkey gets peach x

n+1/4)x=164

x=656/(4n+1)

Obtained from 656 = 2 * 2 * 2 * 41.

n=10 5, the freight per ton of the house is 1

Then D is transported to B with a freight per ton of 2 3, C is transported to A per ton of freight, C is transported to B per ton of freight is set to D to A x tons.

x+2/3（60-x）+

The rest is your own.

1.With 5 as the base. Person.

Only. 5.Speechless.

degrees, 20 degrees, 21 degrees, 22 degrees.

5.60 tons of grain from D were transported to A, 30 tons of grain from C were transported to A, and the remaining 70 tons were sent to B.

The volume with 3 as the axis is 4*4*3, the volume with 4 as the axis is 3*3*4, and the volume with 5 as the axis is 12 5*12 5*5, then the answer is 5:3

C is like B transporting 70, transporting 30 D to A and transporting all of them to A.

，x+2，2x

2.Where are the shadows???

3.Half, which is 1 3a squared.

Solution: Let xm be part a, then part b of (6-x)m is 3*(4x)=240*(6-x).

12x=1440-240x

12x+240x=1440

252x=1440

x=350/63

Answer: 350 63 cubic meters of steel should be used as a part, and 28 63 cubic meters of steel should be used as a B part, how many sets of instruments should be matched into this kind of instrument? shu

I feel that the data is wrong, so let's check it and ask back.

It's 40 A's. 4 is too little ......

X steel is used for A parts, and 6-X steel is used for B parts, which is exactly how many sets of such instruments are matched;

1：3=40x:240(6-x);

240(6-x)=120x;

240*6=240x+120x;

36x=24*6;

6x=24;

x=4;b:6-x=2

It happens to be equipped with 160 sets of such instruments.

If 6m is used, the total number is (240+4) 6=1464 to make a x and b(1464-x).

3x=（1464-x）1

3x+x=1464

4x=1464

x=266b 1464-266=1198

Dissolve to make a x and b (1680-x).

3x=（1680-x ）1

4x=1680

x =420

b 1680-420=1260

If xm steel is used as part A, then there is: 40*x:240(6-x)=1:3 solution: x=4(m).

Then use 6-4=2m steel to make the B part.

It can be equipped with 4*40=160 sets of instruments.

If xm steel is used for part A, then (6-x)m steel is used for part B.

3×40x=240×（6-x）

Solution: x=4,6-x=2

Therefore, if 4m steel is used for A part, then 2m steel is used for B part, which is exactly how many sets of this instrument are matched.

Let a be x and by. With 1m, you can make 40 A parts or 240 B parts, so A needs a quarter and B needs 1 240

Column relation: x 40 + y 240 = 6And because there is 1 part A and 3 parts B, x=3y. Solve binary linear equations.

4 square meters do A, 2 square meters do B

The first question is equal to a 2a 2a 2-3a + 1 = 6

Solution: 4a-3a=7

a=7 and the rest of the algebra will do the trick.

Question 2 We haven't learned yet.

Question 3 x+6y

Question 4: t-(-1)=t+1

Question 5 12 and Question 6 n+2

Question 7: 4a+2b

1 From the title:

a-b=2;

c-3=2;

2a-6b=2;

Solution: a=b=c=5

2 Typing the wrong question.

3. The same applies to the first question.

field, because 18+2>16+3

You can list equations 18+x>16+(4-x).

1) Because 1x2 = one-third 1x2x3-0x1x2) 2x3 = one-third (2x3x4-1x2x3) 3x4 = one-third (3x4x5-2x3x4) so 1x2+2x3+3x4+.100x101 =1/3（1x2x3-0x1x2）+1/3（2x3x4-1x2x3）+1/3（3x4x5-2x3x4）+.1 3 (100x101x102-99x100x101) After proposing 1 3, we get 1 3 (1x2x3-0x1x2+2x3x4-1x2x3+3x4x5-2x3x4+....+100x101x102-99x100x101) From the question 1x2+2x3+3x4=one-third x3x4x5=20, it can be seen that the one plus and one subtract in parentheses are reduced Finally, 1 3 (0x1x2+100x101x102)=1 3x100x101x102=343400 (2) is the same as the previous question: 1x2+2x3+3x4+...

n(n+1) =1/3 n(n+1)(n+2) (3) 1x2x3+2x3x4+3x4x5+..n（n+1)（n+2） =1/4(n+1)(n+2)(n+3))

1) The process of adding 3 equations:

1x2+2x3+3x4=1/3（1x2x3-0x1x2）+1/3（2x3x4-1x2x3)+1/3（3x4x5-2x3x4）

1/3（1x2x3-0x1x2+2x3x4-1x2x3+3x4x5-2x3x4)

1/3（3x4x5）=20

So 1x2+2x3+3x4+.100x101=1/3（100x101x102）=343400

2）1x2x3=1x2x3（4-0）x1/4 2x3x4=2x3x4（5-1）x1/4

From the addition of the above questions, we can get 1x2x3+2x3x4+3x4+3x4x5+.n（n+1)（n+2）= 1/4（n+1)（n+2）（n+3）

Each item is a single digit.

1+2+3+..n=1/2*n(n+1)

Each item is a two-digit number.

1*2+2*3+3*4+..n(n+1)=1/3*n(n+1)(n+2)

Each item is a three-digit number.

1*2*3+2*3*4+3*4*5+..n(n+1)(n+2)=1/4*n(n+1)(n+2)(n+3)

80 degrees, extending CE, intersects AB at F, BFE=100 degrees, so 3=80 degrees.

He is not wrong, the angle of incidence is equal to the angle of reflection.