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1. There are three kinds of rectangles: 4 (width) * 3 (height) * 10 (length * 2 * 15

4*3*10: The surface area is: 2*(4*3+3*10+4*10)=164 (the minimum) Volume: 4*3*10=120

3*2*20: The surface area is: 2*(3*2+3*20+2*20)=212

4*2*15: The surface area is: 2*(4*2+4*15+2*15)=196

2. The cube is cut into 27 pieces, then each side has 9 small squares, and the surface area of the small cube is: a, and the surface area of the large cube is: (a 6) * 9 * 6 = 270, then a = 30

3. If you have a question, you can know that the cuboid is 1 (width) * 1 (height) * a (length), after cutting into 3 breaks, 4 1 * 1 surface areas are added to the original surface area, that is, (1 * 1 * 2 + 1 * a * 4) (for the original surface area) 1 * 1 * 2 + 1 * a * 4 + 1 * 1 * 4 = 806 to get a = 200, the volume is: 1 * 1 * a = 200

After a few minutes, the volume of B is: 5*200ml=1000ml, then the height: h=1000 25=40cm

A and B are of the same height, and the volume of A is: 4000ml + 5 * 200ml = 5000ml

Then the base area of A is: 5000 40 = 125cm

Make an extension of DF and CE and intersect at Q

Take advantage of proportional selection. ab bp = ef fq i.e. 2 3 = 4 fq so fq = 6Similarly, ab bp = cd dq i.e., 2 3 = cd (df+fq), and df = 12;fq=

1.There are double seats for X people and four seats for Y people.

So. y=39xx+y=40x<70

So x" because x is an integer greater than 0, so x=1, y=39, a total of 40 students come to the meeting.

2.If the degree of electricity consumption must be an integer, then it can only be established if the degree of electricity used by A is greater than 24 degrees and the degree of electricity consumed by B is less than 24 degrees.

Therefore, if A uses (24+x) kWh, B uses (24-y) kWh.

Rule. 2x+ so. x=

Because 0x=3y=4

So. A used 27 kWh of electricity, and B used 20 kWh.

Set the purchase price of each unit x to get:

x(1+50%)*90%-50=370

x=2800 9 (yuan).

Because sn=na1+(1 2)n(n-1)d

So sn n=a1+(n-1)(d 2).

So it is an equal difference series (the tolerance is half of the tolerance d of the series), and since s10 10 = 10, s100 100 = 1 10, its tolerance is (1 10-10) 90 = -11 100 (i.e. d 2 = -11 100).

sn=a1+a2+..an=a1+a1+t+a1+2t+..a1+(n-1)t=na1+(n-1)nt/2

sn n=a1+(n-1)t 2 -> equal difference series s10 10=100 10=10 s100 100=10 100=10 10 subtract s100 100=(100-10)t+s10 10 t=(1 10-10) 90=-11 100

s110/110=(110-100)t+s100/100=10*(-11/100)+1/10=-1

s110=-110

5 times any number of single digits that is non-zero, i.e. 5

From the result ten digits 0 to know which of the 6 is 4, then the first three digits can only be 9, and the result hundred digit 9 knows that the one above it is 3 (because the following digit is also carried 1), so the first two digits are 45

And because the result is that the number of 10,000 digits is only 1, the first three digits and hundreds can only be 3, so it is 309 times 45 = 13905

According to common sense:

The sum of odd numbers must be odd, and cannot be even, i.e., it cannot be equal to 30. There is no solution to this question.

But since the question shows that 30 does not have to be a conventional understanding to do thirty, but can be variant thinking to find a way to understand him as other numbers, that is, to change other decimal systems to understand, to change 30 into odd numbers, the simplest is to use odd decimal system, you can use the decimal system as an example, 1 is equal to 1 in decimal system, 15 is equal to 11 in decimal system 1 + 5 = 16, 30 is equivalent to 3 11 = 33 in decimal system, then 1 + 16 + 16 = 33, so you can fill in 1 + 15 + 15 = here30, there are many similar methods, and the difficulty lies in jumping out of the conventional thinking of the decimal system. More generally, it can be concluded that as long as two numbers greater than 10 and less than ten are added, the correct answer is the decimal system, and the decimal system is the addition of the single digits of the three numbers.

In decimacimal: 1+15+15=30 3+13+15=30 5+13+13=30 5+11+15=30 7+11+13=30 9+11+11=30

In decimacimal: 3+15+15=30 5+13+15=30 7+13+13=30 7+11+15=30 9+11+13=30

In decimacimal system: 5+15+15=30 7+13+15=30 9+13+13=30

In hepadecimal: 7+15+15=30 9+13+15=30

In hexadecimal system: 9+15+15=30

This problem can also be understood as using these numbers to form decimals, such as etc., which are done in the decimal case.

This question can also be understood as the numbers given are in the form of cards, 6 and 9 are common, there are 6 + 13 + 11 30, of which 6 is the '9' of the original question.

All in all, don't stick too much to conventional thinking, and think of a logic that can satisfy the meaning of the topic, which can be counted as a correct solution.

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