Junior high school math homework Don t write the answers directly, you can t understand the answers

You can't draw a function image, you can see for yourself.

First, set the profit to y, and set the house price to increase by x 5 yuan.

y=[(60+5x)*(30-x)]-20*(30-x)] simplify this equation and then use x= -b 2a to find out what is the maximum profit (y) when x is what is.

That's it. If you don't understand it yet, let's ask.

Idea: Suppose the free room is x Rough calculation when x is 1,2,3 The profit is increased, i.e., as the free room increases, the profit increases. Once x increases to a certain value, there may be a flip, i.e., the profit starts to fall, and we design an inequality as follows.

30-x)*(40+5x)>(30-y)*(40+5y)y=x+1x>

x is the number of free rooms before the profit decreases, and y is the number of free rooms after the profit decreases, 60 + 5 * 11 = 115

The equation looks like this.

Solution, set up free room x profit y because.

If the price of each room does not increase by 5 yuan per day, one additional room will be added.

So in fact, an increase of x 5 yuan in the house price is equal to a vacant room.

y=(60+5x)*(30-x)-20(30-x)simplified-5(x*x+14x-240).

Then use x=[-b (b -4ac)] 2a to solve x with two values, x is greater than 0, so x=10

Solution: Let the profit be $w.

If x is increased by 5 yuan, then the number of rooms is (30-x), and the profit of each room is (40+5x) yuan.

w=（40+5x）（30-x）

1200-40x+150x-5x^2

5x^2+110x+1200

5(x-11)^2+1805

So: when x = 11, that is, when the house price is priced at 115 yuan per day, the maximum profit is 1805 yuan.

First of all, list the ergonomic ratios for each group to make tables and stools:

A: 8:10 = 4:5

B: 9:12=3:4

C: 7:11

D: 6:7 in descending order: D, A, B, C. Obviously, a large ratio indicates that it is more efficient to make a table, while a smaller ratio indicates that it is more efficient to make a stool.

Therefore, you can first consider having D and A make a table, and C and B make a stool. In 21 days, you can do:

D and A make a table: 6 21 + 8 21 = 126 + 168 = 294

C and B make stools: 11 21 + 12 21 = 231 + 252 = 483

In this way, the difference between the table and the stool is 483-294=189

For every 1 day of production change, 12 stools can be reduced and 9 tables can be added, that is, 9 + 12 = 21 products can be changed, 189 21 = 9,. Then B produces 9-day tables and 21-9=12-day stools. At this time, :

Total production table: 6 21 + 8 21 + 9 9 = 126 + 168 + 81 = 375

Total production of stools: 11 21 + 12 12 = 231 + 144 = 375

In this way: A and D make tables, B makes tables for 9 days, makes stools for 12 days, and C makes stools, a total of 375 sets can be produced.

If D and A make a stool, it can be done in 21 days: 7 21 + 10 21 = 147 + 210 = 357

C and B make a table, which can be done in 21 days: 7 21 + 9 21 = 147 + 189 = 336

The difference is 357-336=21

In this way, even the production of stools is not as many as the previous plan. It is not advisable to see.

According to the data ratio of the four groups to make tables and chairs, groups D and A were more suitable for making tables, and group C was suitable for making stools.

In 21 days, D and A can make tables (8 + 6) * 21 = 294 sheets, and group C can make 11 * 21 = 231 stools, and there are 294-231 more tables than stools.

Group B used 9 days to make a table 9*9=81, and 12 days to make a stool 12*12=144,144-81=63, which was just right.

Then, according to the above scheme, a maximum of 294 + 81 = 375 sets of tables and chairs can be manufactured.

First of all, we know that the table and stool should be matched and C produces the most stools per day, so let C make stools for 21 days, so if there are still people who produce stools, then there are too many stools, and the four groups should make the most products, and they should each give full play to their strengths Because the table is slow to manufacture, A, B, and D are needed together.

If A produces x day tables, then the remaining 21-x days produce stools, and in the same way, B produces y day tables, then 21-y days of stools then A and B produce a total of 8x+9y tables, and produce 10 (21-x) + 12 (21-y) chairs.

It is the above condition again, the same number of tables and stools are produced, so, 8x+9y=10(21-x)+12(21-y), solve the relationship between y and x, and then write the total number of sets produced with an analytical formula containing x, because the production is 21 days, so the value of c must be less than 21, and two-sevenths is greater than zero, so when the value of x is the largest, the number of sets produced is the largest, which is 375 sets.

If you don't understand, you can continue to ask.

8 + 9 + 7 + 6 = 30 tables.

10 + 12 + 11 + 7 = 40 chairs.

21 x 30 = 630 sets of tables and chairs.

Because the maximum capacity of the four combined is only 30 tables per day, this is a bottleneck. Chairs can do more, but they don't work.

8+9+7+6=30

A 3:4 ratio of competencies and a 4:3 allotted time ratio are sufficient.

30*12=40*9=360 sets.

There is no capacity to be wasted, so up to 360 sets of tables and chairs can be built.

From xy (x+y) =-2, take the reciprocal (x+y) (xy)=-1 2 which is 1 y+1 x=-1 2....1)

By yz (y+z) = 4 3 , take the reciprocal (y+z) (yz)=3 4 is 1 z+1 y=3 4....2)

From zx (z+x) =- 4 3 , take the reciprocal (z+x) (zx)=-3 4 is 1 x+1 z=-3 4....3)

1)+(2)+(3), 2(1/x+1/y+1/z)=-1/2, 1/x+1/y+1/z=-1/4...4)

4)-(1), 1 z=1 4 , z=4

4)-(2), we get 1 x=-1 , x=-1

4)-(3), 1 y=1 2, y=2

Substituting xyz xy+xz (is this formula written incorrectly).

This problem is equivalent to asking you to solve a ternary equation.

Death can be calculated, and there is no need to dwell on whether there is any other way.

First of all, prove that the triangle abe and the triangle acf are congruent, then be cf, find the perimeter ec cf ef bc ef, as shown in the problem, you can calculate bc 2, if you find the minimum value, you can find the minimum value of ef, ef is equal to the open square of twice the square of ae, ae is the smallest when ae is the high value on the bc side of the triangle abc, and ae 2 2 is obtained, ef is equal to 1, so the minimum circumference is 2 1

The steps are detailed in the figure above.

A polynomial is a quadratic trinomial.

m^2-49)=0

m = 7 or -7

m-7）≠0

m=-7（m+3）=-4

The quadratic function y=ax(x-4) =a(x-2) 2-4a so its vertex is (2,-4a) and its axis of symmetry is x=2

2)(2,-4a)

3) The line y= —2x 3+2 intersects the x-axis at the point n(3,0) to make the quadrilateral pqmn become square, and x=2 is a quadratic function y=ax(x-4).

, so it is also the axis of symmetry of the square pqmn, so the m(1,0) square side length is 2

So p(1,y) q(3,y).

Substituting p(1,y) into the straight line y= —2x 3+2 gives p(1,4 3) Obviously, pm is not equal to 2

So there is no such real number a such that the quadrilateral pqnm is square.

1.Let y=0,ax(x-4)=0,x=0; x=4, so the line segment is 4.

3.If it exists, then it is a series of equations.

y=ax(x-4) vs. y=-2x 3+2;

After solving the coordinates, one is p and the other is q, so that the ordinate of p is equal to the abscissa of q - the abscissa of p, if a has a solution, there is, if there is none, there is none.

It's very, really rare, but if you're bored, just do it.

Question 1: From the coordinates of point a, od=4, ad=2....Let fo be x and fd be 4-x, use the Pythagorean theorem to get fo = five in two, fd = three in half, and do eh at point E perpendicular to DC at point H, because the angle fae = 90 degrees, we can know that the triangle DAF is similar to the triangle AHE, and then the length of AE can be found to be 5....

In summary, the f coordinate is (0,, and the point e coordinate is (5,0), and from this, we find ef:y=

The second question: this kind of problem classification is very easy to do1, ap=pe, through the p point as pn perpendicular to ae, you can find the value of pe is one half of fe, pythagorean to find fe is five two-fifths of the number five, then pe is five quarters of the root number five, calculate the number of seconds, and use the similarity to find the p point coordinates. 2. ae=pe, this is the simplest, ae=5 is known, and the method is the same as above.

3. It seems that there is no more, if there is still don't blame me.

Question 3: If you want to make two angles equal then apc and ahe are congruent, then ap=ah=3, then the coordinates of point p are (a,1), substitute y=1 into the analytic formula of ef, and find the coordinates of p (3,1), then you can find that pe=root number five, then t=1....Most of these problems should need to be classified, you do the circumscribed circle of the right triangle APC, see if it has an intersection with EF, and then ask for it.

1 （a+b）^2+|2b-4|=0 Because the square of a+b is greater than or equal to 0, and the absolute value of 2b-4 is greater than or equal to 0, if the medicine wants to make the sum of these two items equal to 0, a+b=0 and 2b-4=0So b=2, a=-2 ab-[2ab-3(ab-1)]+1=-2*2-【2*（-2）*2-3*（-2*2-1）】+1=10

The second question is ambiguous 5x flat (5x) 2 or 5 times x square If possible, send a sheet** or deduct you to me I'll help you take a look, but you still have to listen to the class It's better to write your homework yourself.

(1) 1 2x can be interpreted as half of the unknown number x (2) (a+b) multiplied by (a-b) can be interpreted as ham, which is eaten by two people; For more answers, go to the "Middle School Math Knows" post bar half x.

Grass: Just 5 points, I want to know the answer, stupid, stupid!