The process and answer to a primary school math problem

Length: 5+6 11 (cm).

Width: 2 cm.

Height: 4 cm.

Volume: 11*4*2=88 (cubic centimeters).

Surface area: 11*2+11*4+4*2)*2 148 (square centimeters) A: The volume is 88 cubic centimeters. The surface area is 148 cubic centimeters.

The volume is 48 + 40 = 88 cm3

Type out word by word.

It's hard work.

Volume: the sum of the two bodies 6 * 2 * 4 5 * 2 * 4 = 88 cubic centimeters.

Surface area: the sum of the total surfaces minus the coincident surfaces (6*2 2*4 4*6 5*2 2*4 4*5)*2-(2*4)*2=148 square centimeters.

The 12 balls are divided into 4+4+4 first. Let's take the first two parts and weigh them. 1.

If it is the same, it means that the light ball is inside the other 4 balls. Then divide the four balls into 2+2. Continue to say, divide the light part into 1+1, and you can make it clear 2

If it's different, divide the different part into 2+2. Continue to say, divide the light part into 1+1, and it will be clear.

First divide the 12 balls into three parts, four parts each, take two of them to weigh at will, if the weight of these two parts is the same, the ball with different quality is in the third part (or if the weight of the two points is different, then the distress call with different quality is in the different part), this is the first time; The second time is to divide the found part into three parts, one and two parts are one ball, and the third part is divided into two balls, if the first and two parts are weighed, if they are equal, then the ball that is different is in the third part, and then the two balls of the third part are weighed again, and it can be obtained (or if the first and two parts are not equal, the ball with different quality is in the one or two parts of this weighing).

That's OK, o(o haha

12 divided into 3 equal portions, 4 pieces each.

4 points (2,1,1).

2 is divided into (1,1).

Serial numbering, 3 groups -12), take the first 2 sets of scales, there are two possibilities:

Situation 1: If the results are the same weight, then the defective product is in group 3 (9-12), take any 2 balls in group 1 or group 2, and take the number ball in group 3 and put it on the balance scale, there are two possibilities:

Situation 1-1: If the same weight, the defective product is in the number, and then take 1 10 in the 1 ball, and the 11th scale, if the same weight, the defective product is 12, otherwise it is 11;

Situation 1-2: If it is not the same weight, the defective product is in the number, and then take 1 8 in the 1 ball, and the 9th scale, if the same weight, the defective product is 10, otherwise it is 9;

Situation 2: the result is not the same weight, then the defective product is in the group (1-8), take the first group and the third group scale, there are two possibilities:

Situation 2-1: If it is the same weight, the defective product is in the second group (5-8), take any 2 balls in the first group and the second group in the scale, there are two cases:

Situation 2-1-1: If the same weight, the defective product is in the number, and then take 1 and 4 in the 1 ball, and the 7th scale, if the same weight, the defective product is 8, otherwise it is 7;

Situation 2-1-2: If it is not the same weight, the defective product is in the number, and then take 1 and 4 in the 1 ball, and the 5th scale, if the same weight, the defective product is 6, otherwise it is 5;

Situation 2-2: If it is not the same weight, the defective product is in the first group (1-4), take any 2 balls in the second group and the middle scale in the first group, there are two situations:

Situation 2-2-1: If the same weight, the defective product is in the number, and then take 5 and 8 in the 1 ball, and the 3 scale, if the same weight, the defective product is 4, otherwise it is 3;

Situation 2-2-2: If it is not the same weight, the defective product is in the number, and then take 5 and 8 in the 1 ball, and the 1 scale, if the same weight, the defective product is 2, otherwise it is 1;

Put six balls for the first time, look at the weight, and then put the six balls separately and divide them.

- Step 1.

Divide the 12 balls into 3 portions of 4 each.

Take out 2 servings. If both sides weigh the same, take the remaining portion; If one side is heavier, take the heavier portion.

- Step 1. Divide the selected portion into 2 portions, each serving of 2.

Will be 2 servings. Take a heavier portion.

- Step 1. In the end, only one copy remained (1, 1).

Therefore, just place the two balls in the last part (i.e. divided into two parts) on each side of the scale.

Weigh out which weight is which.

It's a lot of hard work to organize the language.

The first time each weighs six, the second light weight of the six arbitrarily take four and then weigh, if the weight is the same, weigh the remaining two, as the third weighing, you can weigh out, if the weight is not the same, then weigh the light weight of the group, as the third weighing, a total of three times can be weighed.

I hope to divide the paragraphs and sort out the levels.

Divide 12 into 3 parts, 3 (4, 4, 4) 4 and 4 weigh, if it is not the same

The weight of the banana eaten: 5400-2800=2600 The total weight of the banana: 2x2600=5200

Box weight: 5400-5200=200

A: The banana weighs 5200 grams, and the box weighs 200 grams.

Banana net weight 5400-200=5200 grams.

Box weight 2800*2-5400=200g.

5400-2800=2600g 2600*2=5200g (this is the net weight of the banana) 5400-5200=200g (this is the weight of the box) I !!

Box weight 2800*2-5400=200g.

Banana net weight 5400-200=5200 grams.

Fill in the appropriate number in , so that the equation holds.

(15) =(30) =(20) Solution: = + = + because + = + = + so = +

So =15

Fill in the appropriate number in , so that the equation holds.

Numbering: Shadow triangles are 1, 2 for triangles of 2 square cm, 3 for non-shaded triangles without numbers, and 4 for triangles of 3 square cm

s is the area. 2 and 3 are similar triangles, the area ratio is equal to the square of the corresponding side ratio, from the area ratio of 2,4 to 2 3, the ratio of the corresponding side of 2,3 is also 2 3, so s2 s3 = 4 9, so s3 =

s3+s4=s1＋s2

Hence s1=

Rectangle area = 20

Shadow area = 1 2x20-2 = 8

I can't find it because the length of the diagonal is uncertain.

The first one to look at the picture is 60km

At 240km, we rested for an hour.

It traveled 360 km/h in 4 hours and the speed was 90 km/h.

There is no picture, I can't answer it, please send a picture.