There are a few math problems that need help urgently, and there are a few math problems that I hope

1) Let the original quality be A, and the purchase price be B

Then, the profit is a90%*b110%-ab=-ab%<0, so the loss is. 2) Set the price to increase by x

then a90%*b(1+x)-ab ab35%x 50%, so the minimum increase is 50%.

First, we get x a, x2a+4x

x<-1-2a

Then -1-2a=2

a=-3/2

5/2）x-m=（x/2）-3

2x=m-3

x=(m-3)/2

The solution of the equation is between 1 and 3, which is 1 x 3

So 1 (m-3) 2 3

2≤m-3≤3

5 m 61) sales income is (1-10%) * 1 + 10%) = 99% of the purchase cost, so it is a loss.

Therefore, the minimum price should be increased by 50%.

Let the ten digits be x and the single digit be x+3

10*x+(x+3)<55

mx+3=3x

3-m)x=3

x=3/(3-m)

The solution of the equation is positive, i.e., x>0

So 3 (3-m) >0

3-m>0

m<35/2）x-m=（x/2）-3

2x=m-3

x=(m-3)/2

The solution of the equation is between 1 and 3, which is 1 x 3

So 1 (m-3) 2 3

2≤m-3≤6

5 m 9 so m = 5, 6, 7, 8, 9

(1) 90% 110% = Answer: The supermarket is a loss.

2）135%=90%x

x=135%÷90%

x=150%

A: 50% increase

2、a 3、 3x-1/4>a+2x/2

3x-2x/2>a+1/4

3x-x>a+1/4

2x>a+1/4

x>(a+1/4)/2

x<2(a+1/4)/2<2

a+1/4<4

a<17/4

1. (1) 90% 110% = Answer: Loss.

2）135%=90%x

x=135%÷90%

x=150%

A: 50% increase

2、a 3、 3x-1/4>a+2x/2

3x-2x/2>a+1/4

3x-x>a+1/4

2x>a+1/4

x>(a+1/4)/2

x<2(a+1/4)/2<2

a+1/4<4

a<17/4

Exhausted! Give it a point! Thank you!

2008baby200808 Comrade, how can you copy me!

Solution 3x-4 is greater than or equal to x+7

x is less than or equal to -33 2

2、x＞2+a/a-3

3. Dissolve the inequality; x＞2/5a+1/5

Because x 2 then 2 5a+1 5=2

So a=2 9

1. Let this number be x, that is.

x 3-4 x+7, so 2x 3 -11, so x -33 22, because a 3, then a-3 0, then (a-3) x 2+a, i.e. x (x+a) (a-3).

x-1 2 (a+2x) 4, i.e. 12x-2 a+2x, so 10x a+2, so x(a+2) 10

So (a+2) 10=2, so a=18

1. Let this number be x1 3x-4 is greater than or equal to 7+x, and x is less than or equal to -33 2

3. First, simplify the inequality 3x-1 2 a+2x 4 to get x (2a+1) 5, because its solution set is x 2So we get (2a+1) 5=2 and we get a=9 2, so when a=9 2 the inequality gives the solution set as x 2

2、（a-3）x＜2+a

x＜ （2+a）/（a-3）

x＜（a-3+5）/（a-3）

x 1+5 (a-3) because a 3 so a-3 0 so x 1 point give me a hehe.

4mp=1

Combining similar terms, the coefficient of x 2 is a+b-c, because it is three sides of a triangle, a+b-c>0 is completed.

1.Let the side length of the square be xm, then (x+5)(x+2)=54xsquare+7x+10=54

x square + 7x-44 = 0

x = -11 or 4

Because the side length cannot be negative, the side length is 4m

2.Substituting x, then 1-5 + p squared - 2p + 5 = 0 (p-1) square = 0

So p=13The formula is simplified.

a+b-c)xsquared-bx+2b=0

Because abc is the three sides of a triangle, the sum of the two sides is greater than the third side, so a+b-c>0, the side cannot be minus 2b>0

So the equation is a quadratic equation with respect to x.

Square + 3x-2011 = 0, so ap as the two roots of the equation, the sum is -3, and the product is 2011

A square + p square + 2ap - 2ap + 3 (a + p) = (a + p) square - 2 ap + 3 (a + p).

Substitute the values of A+P and AP.

Get 4022

Solution: 1, (1) substituting the coordinates of the point (-2, 0) into y=3 2x+m can solve m=3, so the analytical formula is y=3 2x+3, the image has been.

One, two, and three quadrants.

2) Look at the process diagram.

2. The image of the primary function y=1 2x-3 passes the first, third, and fourth quadrants, because k=1 2>0, it must pass.

1. The three quadrants intersect with the y-axis at (0,-3) points, so the four quadrants are also passed.

(1) 3 2x-2+m=0 m=3 123 (2) The ratio of the intersection point to the horizontal and vertical coordinates to the origin is 2:3, so it intersects with the x-axis (6,0) and the y-axis (0,-9).

y=3/2x-9

3) One, three, four.

1)0=3 2*(-2)+m, m=3, y=3 2x+3 through one, two, three quadrants.

2) Area s=(2 3)m*m=27, m1= m2=image passes through the first and third. Four. Quadrant m is a negative value m=