For the first year of junior high school math questions, please see the detailed questions for the q

First of all, EBC = 1/2 B (I don't think so...) The outer angle of one corner of the triangle is equal to the other two angles that are not adjacent to each other (i.e. acd= a+ b), if you don't know, you can calculate, it's quite simple. Then ECD = 1/2 ( A + B), BCE = 180° - ECD = 180° - 1/2 ( A + B).

Then e=180°- ebc- bce=180°-1/2 b-1/2 ( A + b)=180° - 1/2 b - 1/2 A + 1/2 b = 1/2 a.

Call...

(1) 12 (4 20) = 60 (pieces).

2) 60 6 20 = 18 (pieces).

Group IV has the highest number of works, 18.

60 1 20 = 3 (pieces).

The sixth group has a high award rate.

Hope it helps.

I wish you all the best.

(1) The third group has a frequency of 12, and because the third group accounts for 4 parts, so 12 4 = 3, so the first group is 2 * 3 = 6, the second group is 3 * 3 = 9, the fourth group is 6 * 3 = 18, the fifth group is 4 * 3 = 12, and the sixth group is 1 * 3 = 3

So a total of 6 + 9 + 12 + 18 + 12 + 3 = 60

2) From the first question, the fourth group knew the most, 18 copies.

3) The winning rate of the fourth group = 10 18 = 5 9 is approximately equal to the winning rate of the sixth group = 2 3 is approximately equal to.

Therefore, the sixth group has a high award rate.

1) According to the title, the third group accounts for 4 20 = 20%, so the total number is 12).

2) As can be seen from the figure, the fourth group has the highest proportion, that is, the third group has the highest number of works, and the fourth group accounts for 6 20=

60 * pieces) 3) The number of the sixth group is 60 * (1 20) = 3 pieces.

The percentage of winners is 2 3

The proportion of winners in the fourth group is 10 18 = 5 9

Apparently 2 3> 5 9

That is, the sixth group has a higher award rate.

(1) (2+3+4+6+4+1)*(12 4)=60(2) Group 4 (12 4)*6=18

3) Group IV 10:18=5 9

The sixth group has a number of pieces of 3 2:3=4 9

5 9> 4 9 So the fourth group is high.

Solution: Let the number of basketballs purchased be x, volleyball (100-x) (1) 130x + 100 (100-x) < = 11815130x + 10000-100x < = 1181530x < = 1815

x<=So up to 60 pieces.

2）160x+120(100-x)-130x-100(100-x)>=2580

x>=58

Therefore, the buyer must buy at least 58 basketballs, so that the profit of the mall is not less than 2580 yuan, because x>=58

Therefore, the maximum profit w = 160x60 + 120x40-130x60-100x40 = 2600 yuan.

So the mall can make a profit of up to 2,600 yuan.

Solution: (1) If the basketball can be scored at most x, then the volleyball can score at most (100-x) and the equation is obtained: 130x+100(100-x)<=11815, and the solution of x<=60, so the maximum number of basketballs is 60.

2) Equations are listed according to the meaning of the question.

160x+120(100-x)-[130x+100(100-x)]>=2580,130x+100(100-x)<=11815 solution: 58<=x<=60, so the purchaser should purchase at least 58 basketballs.

From the above, when x=60, the maximum profit is (30*60+20*40)=2600 yuan.

(1) If you buy x basketballs, then you can purchase 100-x volleyball and only solve x<= according to the inequality of the column 130x+100(100-x)<=11815

The buyer can buy up to 60 basketballs.

2) The profit for each sale of a basketball is 30 yuan, and the profit for selling a volleyball is 20 yuan, according to the inequality: 30x+20(100-x)>=2580, and x>=58 is solved

The buyer wants to buy at least 58 basketballs.

When buying 58 basketballs, the profit is up to $2,580.

1.Set basketball x, volleyball y

x+y =100

130*x + 100*y <= 11815 then: x< So the maximum number of basketballs is 60.

2.Set basketball x, volleyball y

x+y=100

160-130)*x + 120-100) *y>=2580x>=58 So the minimum basketball is 58.

The maximum profit is to buy 60 basketballs: the profit is 60*30+40*20 = 2600

（1）90

2) The first question is 59, and the second question is that the whole basketball can be sold to make a profit of up to 3000

Solution: The first step is to draw this rectangle and establish the coordinates.

1) b(10,0) c(10,-8) d(0,-8)(2) let p meet x seconds after departure.

x+2x=10+8+10 x=28 3 seconds (3) There may be two situations in this question, you think that the distance between them at the beginning is 10+8+10=28 meters, and then p,q start to move, and after a period of time the distance narrows to 25 meters, but if you continue to move until they meet, their distance will continue to increase until Q reaches point A, you also have to look at the distance between p and q at this time, this situation does not meet in this question (you don't need to write these words when you answer the question). So.

Let the distance between each other be 25m after x seconds, then there is x+2x+25=28 x=1 second, so we meet in 1 second.

4) Let the APQ area be S

The calculated relation is:

s= { 4x （0<=x<5）

x2+9x (5<=x<9)

0 (9<=x<10)

x2+24x-140 (10<=x<=14)

The minimum subsidy is 4x500 + 3x300 = 2900 yuan, 4500-2900 = 1600 yuan, as long as the remaining subsidy is not more than 1600 yuan. （1）2x500+2x300=1600 ,(2)3x500=1500, (3)5x300=1500 ,(4)2x500=1000 ,(5)4x300=1200, (6)3x300=900 (7)1x500+3x300=1400(8)1x500=500 (9)2x300=600 (10)1x300=300

Final plan: (1) 6 Class A and 5 Class B, costing 4500 (2) 7 Class A and 3 Class B, costing 4400

3) 4 Class A and 8 Class B, cost 4400

4) 6 Class A and 3 Class B, cost 3900

5) 4 Class A and 7 Class B, cost 4100

6) 4 Class A and 6 Class B, cost 3800

7) 5 Class A and 6 Class B, cost 4300

8) 5 Class A and 3 Class B cost 3400

9) 4 Class A and 5 Class B, cost 3500

10) 4 Class A 4 Class B cost 3200

11) 4 Class A and 3 Class B cost 2900

A: 11 scenarios.

Solution: Suppose there are x (x 4) students in category A and y (y 3) in category B students in need of funding, according to the meaning of the question, there is an inequality:

500x+300y 4500, simplify:

5x+3y≤45

Scenario 1: When x=4, y=8

Scenario 2: When x=5, y=6

Scenario 3: When x=6, y=5

Scenario 4: When x = 7, y = 3

Solution: Set a class x people, B class Y people.

500x+300y 4500 x 4 y 3 When y is minimum = 3, substituting in , get:

500x+900≤4500 x≤

Combine: 4 x

So there are 4 types: 4 people in category A and 8 people in category B;

5 in Category A and 6 in Category B;

6 in Category A, 5 in Category B;

7 in category A and 3 in category B;

According to the minimum standard, you need to spend 4x500 for class A students, 300x3 for class B, and the remaining 1600 yuan, and the remaining students can be funded as follows:

xa+yb = <1600, where x,y are the natural numbers of 0,1,2,3,4,5. A total of 6 + 4 + 3 = 13 types.

4*500+3*300=2900

Also save 1600

Situation: A 4+0 B, 6 types, 1600 300, rounded +1

A 4+1 b 4 1100 300 rounded +1

A 4+2 b 3 600 300 rounded +1

A 4+3 B 1 100 300 rounded +1

The answer is 6+4+3+1=14

There are x people for class A students and y people for class B students, then.

500x+300y≤4500,x≥4,y≥3,。

1，x=4,y=3/4/5/6/7/8

2，x=5,y=3/4/5/6

3，x=6,y=3/4/5

4，x=7,y=3

You can get y=2x+4 (3) from (2), substitute (3) into (1) to get the unary equation 2x-7(4+2x)=8 to get (x=-3), and then substitute the value into (3) to get (y=-2), so as to get the solution of the original equation as (x=-3 y=-2).

The following system of equations is solved by substitution elimination: 2x-7y=8 (1).

y-2x=4 ( 2)

You can get (y=2x+4) (3) from (2)), substitute (3) into ((1)), get the unary equation (2x-7(2x+4)=8) to get (x=14 3), and then substitute the value into (3) to get (y=40 3), so as to get the solution of the original equation as (you will definitely do this).

Solution: Set a x yuan, B y yuan.

x+y=500, solve this system of equations).

Classmates, I think if you think hard, you will be able to do it independently! o( o solution: Let the original ** of the two commodities A and B be x and y yuan respectively, then:

x+y=500

Solving the system of equations yields: x=410, y=90

Answer: A, has two kinds of goods original** respectively 410 yuan and 90 yuan.

Set the original price of commodity A x and the original price of commodity B y

x+y=500

x=410 y=90