Only those with an IQ higher than 1000 will come to Ha!

2 x = 1 , y = 1 2.

f(3/4)=f(1)sina+(1-sina)f(1/2)=sina+sina-(sina)^2=2sina-(sina)^2

x=3 4 y=1 4.

f(1/2)=f(3/4)sina+(1-sina)f(1/4)sina=[2sina-(sina)^2]sina+(1-sina)(sina)^2

sina = 3 (sina) 2-2 (sina) 3 sina is not equal to 0 so 2 (sina) 2-3 sina + sina = 0 solution ina = 1 2 or 1 (rounded).

Therefore sina = 1 2 a = 30

You should be able to do it below.

。。。Can I say that I can prove that there are a few sick dogs that will go off on the first few days?

In fact, the premise is that everyone has strong reasoning skills.

Proof: Fact1: Only 1 sick dog.

The owner of a sick dog sees that other people's dogs are healthy, indicating that his dog must be a sick dog. , Day 1.

Fact2: There are 2.

The owners of sick dogs all behave exactly the same, so I'll just mention one of them).

The owner of the sick dog (Wang 1) smiled and said nothing when he saw the sick dog of (Wang 2). Went back.

The next day, it was found that the other sick dog was still there. He was immediately depressed.

Is my dog sick?

If not, then there is only 1 dog, and according to fact1, the gun should have gone off on the first day. Contradiction! So my dog is a sick dog. Syllable!

Day 2... Okay, now I'm going to release the mathematical induction. (Used to do this topic.) Khan).

Suppose fact1, fact2, fact3... fact n is true, this is, (the "theorem" that a few sick dogs will fire a gun on the first day is true for positive integers 1 to n), then this fact is also true for positive integers n+1.

Proof: Suppose there are n+1 dogs, and the owner of the sick dog sees only n dogs, he waits for n days, what! The gun hasn't been fired yet? It seems that there are more than n sick dogs, and my dog is also a sick dog. So on the n+1 day he shot his own dog.

Well, now since this "theorem" is true for 1 to 2, it is also true for 3 and 4... Valid from 1 to 50.

Hey, when the landlord asks a question, can you tell me where you went to school? So I know what to say.

Low IQ is a weakness, a vital point. It's hard to make up.

Hello, the answer is: 441

3 7 9 lands are just finished, indicating that it is a multiple of 3 and 7 and 9, that is, an integer multiple of 63; 5 out of 5 take 1 left, indicating that the unit is 1 or 6, and 2 2 2 take the ground and have one left, indicating that the mantissa can only be 1, and all answers are 63x7=441

Hope it helps!

It's so simple, the answer is 81.

Coachman 2 carriage 4 cart 6*2 big cat 6*6*6*4 kitten 6*6*6*4*6*4 mouse 6*6*6*4*6*4*6*4*6*4

Add 4+2+12+864+20736+497664=519282 (strips).

6 taels + 10 taels = 1 catty, in ancient times, half a catty and eight taels, 1 catty = 16 taels. In the imperial unit of weight, one pound is equal to 16 ounces 6 (ounces) + 10 (ounces) = 1 (pound).

4 (hour) + 9 (hour) = 1 (hour) because in time, 4 o'clock + 9 o'clock is equal to 1 o'clock in the afternoon.

4 (weeks) + 9 (weeks) = 1 (quarter) because 1 quarter = 3 months, = about 90 to 93 days = about 13 weeks.

4 weeks + 9 weeks = 1 season. There are 52 weeks in a year, divided into four seasons, and a season of 13 weeks.

4 (hour) + 9 (hour) = 1 (point), is this the answer? If yes, give a good review!

4+9 1 o'clock (13 o'clock is 1 o'clock in the afternoon, 1 o'clock can also be called 1 o'clock.) ）

4 (weeks) + 9 (weeks) = 1 (quarter).

Because there are 52 weeks in a year, there are 13 weeks in a quarter.

Equal to 3 because the addition of the first two digits is equal to the multiplication of the last two digits e.g. 7+5=3x4 9+7=2x8 6+8=7x2 4+5=3x?

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