I don t know how to solve a few math problems, so please help!!

1.The proportional function crosses the origin, so 2-m=0, m=2.

2.Substituting the dot in, 3=k*1, k=3.

3.Substituting two points into it, 3=k+b, -1=-k+b, the solution is k=2, b=1.

4.This problem solves the inequality x+2>3x-2, and the solution is x<2.

5.Substituting this point into two equations, 8=-m+a, 8=m+b, and adding the two equations, a+b=16.

6.Because the function intersects the negative half axis of y, when x=0, y=b<0. Because y decreases as x increases, k <0 (this is conclusive, to remember).

1) When proportional, 2-m=0 is required, so m=2y=2x2)x=1,y=3 into the equation. k=3y=3x3) substituting the coordinates of a and b into the functional equation respectively to solve this system of binary quadratic equations. y=2x+1

4) Draw an image of two equations, when x=2, two straight lines intersect. When x<2 is compliant.

5) Add the equations of the two functions to get 2y=a+b, and substitute y=8 in the intersection coordinates to get a+b=16

6) The negative half axis of the y-axis indicates that when x=0, y=k 0+b<0, that is, b<0; If the value of y decreases with the increase of x, the image is a straight line sloping downward, so k <0

1 m=2 y=2x

2 y=3x

3 y=2x+1

4 x<2

6 k<0 b<0

In order to prevent copying homework, only the answer is given, not the process.

1、 2-m=0 m=2 y=2x

2、 1k=3 k=3 y=3x

3. Combine k+b=3 -k+b=-1 k=2 b=1 y=2x+14, x is less than 2

5. -m+a=8 m+b=8 a+b=166, k less than 0 b less than 0

1、m=2, y=2x

2、k=3.

3、k=2,b=-1.

4. When x is less than 2, the point on the line y=x+2 is above the corresponding point on the line y=3x-2.

5、a+b=16-2m.

6. b is less than 0, less than 0 (is it a writing error, it is a function y=kx+b..))

x²+y²-xy=6

x²+y²=6+xy>=2xy,[x²+y²>=2xy]6+xy>=2xy,xy<=6,x²+y²-xy=6

x+y)²-3xy=6

x+y)²=6+3xy

x y belongs to r+, xy<=6, when xy=6, (x+y) has a maximum, (x+y) =6+3xy=24, x+y maximum: x+y= 24=2 6;

Answer, there are x number of children.

5x>3x+9>=5(x-1) +2

2x>=12

When x=6, apples = 3*6+9=27.

x=5, apples = 3 * 5 + 9 = 24 After verification, there are 6 children who do not meet the conditions, and 27 apples.

There are x number of children.

3x+9=5x-3

2x = 12x = 6 apples = 5 * 6-3 = 27 pcs.

There are 6 children and 27 apples.

No more than 2, then either 1 or 2.

The equation can be listed as 3x+9=5x-3 and 3x+9=5x-43x+9=5x-4

2x=11x represents children. Children can't be half.

So it can only be 3x+9=5x-3

There are x children in total.

Then 3x+9=5x-3

x=6 then the apple has a total of 3*6+9 27

There were 6 children and 27 apples.

The complement of a intersects b={2}, so 2 belongs to b, and substituting x 2-5x+q=0 gives q=6;

Then substituting q=6 into x 2-5x+6=0 to get x=2 or 3, so 3 belongs to a, substituting x 2+px+12=0, and getting p=-7

b=｛x|x 2-5x+q=0}, the complement of a crosses b={2}2 2-5*2+q=0

q=6x^2-5x+6=0

x = 2 or x = 3

So b={2,3}

The complement of a is crossed b = {2}

So 3 is a subset of a.

3^2+p*3+12=0

p=-7

2 times the angle formula, change the shape is OK

The maximum square can be understood as: the square has the largest area of the four small quadrilaterals and the square is the largest of the area it can form. The first understanding of the old ascension is the correct answer, the second understanding:

Because the side length is a natural number, the combination of the above two rectangles is a large rectangle with an area of 13 square centimeters, which can only be 1x13, so its side length can only be 1, and the other side length is 13.

The lower rectangle is a square, one of which coincides and the other adds up to the top 13 (because it can't be 1). The area of the square is greater than the area of the rectangle, let the side length of the square be a, and the side length of the rectangle is a, b

s rectangle = axb, s square = axa, s rectangle then there is b such a combination of a and b has:

There is only one square, so b cannot be 1, otherwise there is a square with a side length of 1, and the last one (12,1) is removed

And the large rectangle is 13 on one side and 1+a on the other side (you can draw it yourself), so the corresponding area is 13x(1+a), that is, all possible values:

All lengths are in centimeters and the area is in square centimeters.

The top two rectangles have a common edge, so the sum of the areas of the top two rectangles is equal to the sum of the common edge multiplied by the other two sides. And the side length of each graph is a natural number. So the length of this common edge must be an divisor of 13, so it can only be 1.

The sum of the other 2 sides is 13.

The largest area of the square means that the maximum side length of the square is 11 (because if it is 12, both sides of one of the above rectangles are 1, and it becomes a square).

So the two sides of the great rectangle are 13 and 12

The area is 156 square centimeters.

As shown in the figure, a, b, and c are natural numbers, and the area of the large rectangle is s=(a+b)(b+c), ac+bc=13, that is, (a+b)c=13, and 13 is a prime number, so c=1, a+b=13, b>a>1

The following is discussed in a sub-situation:

a=2, b=11, large rectangular area s=13 12=156;

a=3, b=10, large rectangular area s=13 11=143;

a=4, b=9, large rectangular area s=13 10=130;

a=5, b=8, large rectangular area s=13 9=117;

a=6, b=7, large rectangular area s=13 8=104

(1-1 4) (1 20 + 1 30) = 9 days.

The two teams completed the rest of the work in 9 days.

A and B cooperate, need to complete 1-1 4=3 4

A completes 1 20 in 1 day, and B 1 day is 1 30

Cooperation completed in 1 day 1 20 + 1 30 = 1 12

A: The two teams worked together in 9 days to complete the rest of the work.