Solve a math problem in a clear and detailed way

Because the period is , so f(365 3) = f(- 3) =sin[(- 3) 2] =sin(- 6) = - sin 30(du) =-1 2

f(x) = sin [(x+2π)/2] x∈[-5π/2, -3π/2]

sin [(x+π)/2] x∈[-3π/2, -/2]

sin [x/2] x∈[-3π/2, -/2]

1) When x [n, n + 2], f(x) = ! f(x) ！Absolutely, can't be beaten)

At x [0, 2], 3f(x) = 1 2 sin(x 2) = 1 6 x = 2 arcsin(1 6).

At x [n , n + 2], x = n + 2arcsin(1 6) (n is an integer).

2) When x [n + 2, (n+1) ], -f(x) = ! f(x) ！Absolutely, can't be beaten)

At x [- 2, 0], -f(x) = 1 2 sin(x 2) = -1 2 x = 2 arcsin(-1 2) = - 3

At x [n, n + 2], x = n - 3 (n is an integer).

But your question is not clear.

According to the meaning of the title, it should be that the four sides of the square are equal, and the rectangle ignores the case of the square.

Let x be the length of the square side, and a and b are the length of the rectangular side.

x^2=ab

4x = 4 root number ab

2(a+b)>=2*2 root number: ab=4x

So, choose A

Of course, you must insist that the square is a special rectangle, and whether the circumference of the square should be filled in is less than or equal to the circumference of the rectangle, which is correct from a theoretical point of view, and the topic may not be very strict.

The correct answer is that a square is smaller.

It can be assumed that the area is 1

Then the side length of the square is 1 and the circumference is 4

Let the length of the rectangle be a, and the width will be 1 a

The circumference is 2 (a+1 a).

From the nature of the inequality, it can be seen that

a+b is greater than or equal to 2ab under the root number

Therefore, the circumference of the rectangle is greater than or equal to 2*(2*a*(1 a))=4, so the circumference of the rectangle is longer than the circumference of the square.

2(a^2 + b^2 ) a+b)^2

If you don't understand, you can prove it

Left - Right = A 2 - 2AB + B 2 = (a-b) 2 > Brigade sum = 0, when a = b, an equal sign is obtained.

c 2 = a 2 + b 2 >=a+b) 2 2 so a+b root number 2 times c

a=b, so the triangle is an isosceles right angle.

By definition, a*b=2(a+2ab+b)=1154

Finishing 2a + 4ab + 2b = 1154

2a+4ab+2b+1=1154+1,2a+1)(2b+1)=1155

Because 2a+1,2b+1 are odd numbers, and because of the symmetry of a,b, there are 32 kinds of ordinal number pairs, specifically 1)2a+1=1, 2b+1=1155, a=0, b=577, so (0,577).

2)2a+1=3, 2b+1=385,a=1,b=192

So (1, 192).

3) 2a+1=5, 2b+1=231, a=2, b=115, so (2,115).

4) 2a+1=7, 2b+1=165, a=3, b=82, so (3,82).

5) 2a+1=11, 2b+1=155, a=5, b=77, so (5,77).

6)2a+1=15, 2b+1=77

a=7, b=38, so (7,38).

7) 2a+1=21, 2b+1=55, a=11, b=27, so (11,27).

8) 2a+1=33, 2b+1=35, a=15, b=16, so (15,16).

By symmetry, get.

a, b can be taken as negative numbers.

So there are 32 kinds in total.

2(a+2ab+b)=2a+4ab+2b

And because 2a+4ab+2b+1=(2a+1)(2b+1)(2a+1)(2b+1)=1155

And because 1155=1*3*5*7*11 or 1155=1*1155, when (2a+1)=3*1 (2b+1)=7*11*5(2a+1)=3*5 (2b+1)=7*11*1(2a+1)=3*7 (2b+1)=5*11*1(2a+1)=3*11 (2b+1)=5*7*1, so there are 4*5=20 pairs; Swap A and B and there are 20 pairs, when 1155=1* At 1155, (2a+1)=1, then (2b+1)=1155, and similarly, (2a+1)=1155, (2b+1)=1, there are 2 pairs.

So there are 42 teams in total! That's how it should be.

2 pairs of shifts to get b = - to make b an integer, then it must be an integer plus form, but when divided by any integer other than 1 that is smaller than 1, it will not have an integer plus form, so a must be 0 at this time, b is 577, and when divided by an integer larger than itself, his value must be less than 1, so let - be an integer, only make b 0, and a is 577

In summary, there are two pairs (577,0) (0,577).

Solution: [ 100(1+10% *8 5 x)][100(1-10%x) 10260

Simplify 8x 2 - 30x +13 <=0 to get 1 2 <=x <=13 4

Because you can't lose money, 100(1-10%x) 80x 2 In summary, 1 2 <=x<=2

100（1+10%x8/5x）」「100（1-10%x）」≥10260

Solution x solution set 1 2, 13 4

Because you can't lose money, 100 (1-10%x) 80x 2 in summary: 1 2,2

x and 1 x must be the same sign, and neither of them is 0

So 2x+1 x =2|x| +1/|x|> 2 root number (2|x|×1/|x|> 2 root number (2|

To everything x r constant established.

So 2a < 2 root number 2

So a < root number 2

If a b and 0

then, a and b are both positive.

The answer is still positive. So AABB, in reverse order with ABBA, the answer is the same.

Then aabb=abba

Wow......abababab fainted.

...Don't be fooled by the title.

ABBA is actually ABB, but the order has changed.

Two are equal.

The commutative law of multiplication aabb=abba is therefore equal.

Set to win x games, because there are a total of 14 games and 5 losses, so it is a draw (9-x) games.

From a total of 19 points, we get the equation 3*x+1*(9-x)=19

Solution x=5

Set to win x games. Flat 14-5-x = 9-x

3x+(9-x)=19

x = 5 won 5 games.