Finding a math solution!!! Thank you!!!!

1. 0, 10, -1 3, y=6 7 (2x+1), 1 2, 2, and the last, respectively, calculate the x when y=30, the smallest first, or just look at the slope directly.

3. (1) y=,y=(x-3000)*

4. y=200*(x-1) 2A 1760 B 2000 so choose A.

3.Let A and B be equal to x=5, < 5 for B, > 5 for A.

1. Coordinate origin.

y=6/7(2x+1)

Is there a mistake in the symbol? If not, then it is not your scope of study, and the answer is plural: i 2

2. y=4x+3, because y=30, we get x=29 2,10,33 433 4 is the smallest, so its corresponding function arrives first.

3. (1) y= ; x<=3000

y=;x>3000

2) y=y=3) because y=

1540>1500.So x>3000

Substituting y=, there is 1540=

The solution is x=3050

4. 1. A: y=160x

B: y=200(x-1).

2. A: y=160x11=1760

B: y=200x10=2000>1760

Pick your nails. 3. Order 160x>200 (x-1).

Solution: x<5

1-5 people, choose B.

5 people A and B can be.

More than 5 people will be selected for the first grade.

The topics are the same, except for the change in numbers.

Area of the ring = area of the outer circle - area of the inner circle = (square of the radius of the outer circle - square of the radius of the inner circle).

Dear Students:

Be sure to pay attention to the conditions of absorption and induction skills

Problems 1, 2, 3, 4, and 5 can all be regarded as finding the area of the ring = the area of the great circle - the area of the small circle:

1, multiplied by 2 squared - multiplied by 1 squared = 3

2, multiply 15 squared - multiplied by 3 squared = 216 3, multiplied by 12 squared - multiplied by 2 squared = 140 4, multiplied by 3 squared - multiplied by 2 squared = 5

5, multiplied by 10 squared - multiplied by 5 squared = 75

Why don't you enter all the assignments or the entire exam paper?

22:30 degrees, 2 roots number 3

n+2)!/6

24: Look at the picture.

I should be given that point, right? ）

Question 22: Rotate 30 degrees The height of the trapezoid is 3 times the root number 3 minus 223 Question: 79833600

n+2)!/6

This is the answer I came up with on my own, and it's not necessarily correct! For your reference: (22) 30 degrees (23) 1:1 3*10*11*12=440 2:1 3(n3+3n2+2n) 3:1260

24)1: Because the points p, w, and q are the midpoints of the three sides of the triangle FMN, respectively, the angle pwq=angle wqn=the angle FMN; Angular wqp = angular mpq = angular mfn;

Angle qpw = angle fwp = angle fnm. i.e.: angle pwq = angle fmn;

Angular wqp = angular mfn;

Angle qpw = angle fnm. So the triangle FMN is similar to the triangle qwp 2: because 0 is less than or equal to x less than or equal to 4, the length of the moving point m,n is less than or equal to 4, and the triangle FMN is similar to the triangle qwp, df=2, i.e., fc=4.

So when bn=dm=4, the triangle fmn is a right triangle, that is, the triangle pwq is a right triangle! So when 0 is less than or equal to x less than 4, the triangle FMN is not a right-angled triangle, that is, the triangle pqw is not a right-angled triangle!

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75 hours, that is, 2 and three-quarters hours.

3. The volume should be cubed, not squared, wrong?

8. Five sixths remaining; 299 out of sixths remained; Ate three tenths.

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75 hours. 5/49/dm, 5/6/6kg, 3/10

The circumference of the bottom surface is 2, and the intersection point of the iron wire and the bus bar is set to ABCDE, the cylindrical AB segment, obviously the line segment that connects the two points of AB is the shortest; In the same way, the distance between the line segments connecting BC, CD, and DE is the shortest in the cylindrical BC, CD, and DE segments; When the line segments AB, BC, CD, and DE are in a straight line, AE is the shortest.

At this point, ae= ((8) 2+500 2)=

Untie; Since the side view of the cylinder is rectangular, the shortest length is the diagonal length of the rectangle, the corresponding height is the height of the cylinder 500cm, and the length is the circumference of the bottom surface of the cylinder 2, so the shortest length of the wire is [root number (500 2 + 4 2)], calculate it with a calculator.

Cut the cylindrical iron pipe and pave it into a rectangle, then its length is the circumference of the bottom surface of the cylinder is 2 *, and its width is the length of the cylinder, that is, 5m, so its shortest length is the diagonal length of the rectangle, that is, (the square of the length + the square of the width) and then open the root number ......

Proof : To prove the original proposition, it is only necessary to prove that up 2 lower 0 f(sinx)dx = upper 2 lower 0 f(cosx)dx

The proof is as follows: Up 2 Down 0 F(Sinx)DX =-Up 0 Down (- 2) F(Sin 2-T)DT

Up 2 Down 0 F(Cost)DT = Up 2 Down 0 F(Cosx)DX Completed.

From the verification content, it is obtained that there are 2 down 0 e sinx (e sinx+e cosx) dx

Up 2 Down 0 e cosx (e sinx+e cosx) dx

1 2 ( Up 2 Down 0 E Sinx (e Sinx+E Cosx) DX+ Up 2 Down 0 E Cosx (E Sinx+E Cosx) Dx).

1 2 up 2 down 0 dx = 4

I can't see it clearly! Trouble shines on the point.

The equation for y 2=2px alignment is x=-p 2 on the negative semi-axis of the x-axis and perpendicular to the x-axis.

Circular equation: (x-3) 2+y 2=16

The tangent equation for the circle on the negative semi-axis of the x-axis and perpendicular to the x-axis is: x=-1 so p 2=1 bridge ant 2p = 4

The equation of the throwing pulse is the basis elimination: y 2 = 4x

2.Connecting MF2, ON is the median line of the triangle F1F2M, On=1 2MF2

a=5 2a=10

mf1+mf2=2a=10

mf1=2 mf2=8

So on=4