Thank you for the detailed process and ideas for the third question

The cosine theorem is used to solve the problem by using the angle EOG. Suppose that the triangle OEG could be an isosceles triangle, and it could only be go=ge= 4-x (because OE=OG is not possible because OG= 4-x, OE= y, if the two are equal, then it means y=4-x, which contradicts the triangle OEB that the sum of the two sides is greater than the third side X+Y>4), and see if you can find the value of x.

In the triangle EOG to calculate the cosine of the angle EOG, there is cos(angle EOG) = y 2 [ 2y(4-x) ] ] = y [ 8-2x ].

In the triangle EOB to calculate the cosine of the angle EOG, there is cos( angle EOG) = [y 2 + 16-x 2] [ 8y ].

To calculate the cosine of the angle EOG in the triangle COB, there is COS(angle EOG) = [32-x 2] 32

Thus there is cos(angular eog) = y [ 8-2x ] = [y 2 + 16-x 2] [ 8y ] = [32-x 2] 32

Wherein, using the law of fractions, the law that the value does not change after the difference of the last equation [y 2+16-x 2] [ 8y ] = [32-x 2] 32 numerator and the difference of denominator is obtained, cos(angle eog) = y [ 8-2x ] = [y 2+16-x 2] [ 8y ] = [32-x 2] 32 = [y 2-16] [ 8y-32 ] = [y+4] 8

Then for y [ 8-2x ] = [y+4] 8, we get cos(angular eog) = y [ 8-2x ] = [y+4] 8=4 [2x] =2 x by using the law that the fractional numerator is the difference and the denominator is the difference between the values and the denominator is unchanged, so we get x=16 (y+4).

Bringing x=16 (y+4) into cos(angular eog)=[32-x 2] 32=2 x gives x 3 - 32x + 64=0, introducing 4x 2 into x 3-4x 2+4x 2 - 32x + 64=0, i.e. (x-4)(x 2+4x-16)=0The solution is x=4 (this result is not plausible on the image) or x=-2+2*root 5 or x=-2-2*root 5 (this value is negative, rounded).

Therefore, only x=2+2*root number 5= can be taken. This value is also the radius of the circle b, which is the length of the b.

Additional comments: It's a high school question, and it's only 3 points. I feel like I've thought too much about the answer, but in fact, the meaning of the question is that the OEG mentioned in the first paragraph is impossible, so the triangle OEG cannot be an isosceles triangle.

If you have poor grades, go to jingrui to make up for it.

1 **, 6 small plates, 1 remainder.

2**, 4 small plates, 3 remaining.

3**, 3 small plates, just full.

4**, 1 under, 2 remaining.

Answer: 3 **, 3 small plates, just full.

Because the total number is an odd number, the number of small disks must be an odd number, and the remaining 4 ** of 5x7=35 cannot be filled, so it is only possible to have one. There are 34 left after a small plate that cannot be filled **, then it is only possible, and the answer is 3 small plates and 3** after trying separately.

3 pieces of 24**, 15 pieces of 3 small plates.

（1）2kclo3 +2mno2=2kmno4+ cl2↑ +o2↑

2) (3) According to 2kClO3 = 2kCl + 3O2 and 2kClO3 + 2Mno2 = 2kmNO4 + Cl2 +O2

It can be seen that the ratio of kclo3 participating in the reaction to the amount of kclo3 participating in the reaction is: [(1553 - 7) 2 3] :7 2) = 1546 :21

The mass of potassium chlorate reacted according to the above formula accounts for the total mass of potassium chlorate reacted: 21 21 + 1546) 100% =

There are three Q points:

Because C is 3 units upwards and then 8 units to the right are translated to A, the Q1 coordinates are translated by 3 units upwards from B, and then 8 units to the right: Q1 (8,9), in the opposite direction, Q2 (-8,3), B to A is translated down 6 units, then 8 units to the right, C is translated 6 units downward, and then 8 units are translated to the right, and Q3 (8,-9) is obtained

Dengjinling Phoenix Terrace (Li Bai).

Use elementary row transformations.

a=1 6 1 7

4 2 22 3 r5+2r3,r3-2r1,r4-3r1~1 6 1 7

0 4 4 3 r5-r2,r2/4,r5/2,r2-1/4*r5,r3+14r5,r4+22r5

0 0 0 1 r1-6r2,r3+11r2,r4+9r2

0 0 0 1 r1-7r5, swap row order 1 0 -5 0

The maximally linear independent groups a1, a2, and a4 are obtained

And a3 = -5a1 +a2