Thank you for the math problems in the first year of junior high school

Set a win x game, a draw y game, and a loss (12-x-y) game.

3x+y=19

x, y, and x+y are all in the range of 0-12.

The second question is solved by x=4, y=7 or x=5, y=4 or x=6, y=1. Appearance fee 500 12 = 6000 yuan.

The three results of the first question are:

Therefore, the maximum w is 10900 + 6000 = 16900 yuan.

Team points are mainly related to the number of wins and draws.

A win is worth 3 points, and a draw is 1 point.

Now the integral is 19 points, and the remainder of 19 3=6 is 1

So at least 1 of them is a draw.

Probable. Draw wins.

The rest is not established.

The number of games set to a draw is a

w=500*12+700*a+[(19-a)/3]*15006000+700*a+500*19-500*a15500+200a

A: The maximum amount of 7 is 15500 + 200 * 7 = 16900 yuan.

I'm not good at math, don't blame me for my miscalculations!

There are 3 situations in the team's record.

Wins, 4 draws, 7 losses, 1

Wins, 5 draws, 4 losses, 3

Wins, 6 draws, 1 loss, 5

In the first case, the most is obtained, which is 10900 + 500 * 12 = 16900

Set up x wins, draw y, and lose z.

1, then. 3*x+1*y+0*z=19

x+y+z=12

The solution is: x1=4, y1=1 z1=7

x2=5 y2=4 z2=3

x3=6 y3=1 z3=5

It can be seen from the results of 1.

Bring the above results into the comparison w=(x*(1500+500)+y*(500+500)+z*500)*11

The maximum w = 170,500 yuan.

,16;

2x-3,(x+3)/2,17;

The coefficient of the term of the cube of a and the cube of b is exactly zero.

25；n=-3, m=2, is 3

26；Because the number of hundreds is 2 larger than the single digit, and the number of swaps is small, the difference between the two numbers is 198, so the result is the same.

27；3b*+2b+4.

Question 13: When M<60, should pay RMB, when M>60, should pay (48+ RMB.

(1) The equation relationship between angle 1, angle 2 and angle 3 is: 1+ 2 = 3 Prove that if you pass the point p as a straight line m l1 divides 3 into sum , then m is also parallel to l2 because m l1, according to the theorem 'two straight lines are parallel to each other and the wrong angles are equal'

De: 1= , similarly: 2=

2) The degree of BAC is 85°

1.The power of the BAC is 85°

Proof: According to (1), DBA+ ECA= BAC is known DBA=40°; ∠eca=45°

bac=∠dba+∠eca=40°+45°=85°2.The degree of angle 1 + angle 2 is 180°

Proof is that according to the theorem, 'two straight lines are parallel and their internal angles are complementary';1 and 2 are the same side inner angles between parallel lines

1 + 2 = 180° completed.

1. Solution: Let the water of pool A be x, and the water of pool B will be y

x+y=400

x+4=y-8

Solving the above equation gives x=194 y=206

Answer: Pool A has 194kg of water, and Pool B has 206kg2 of water.

5*18/60+5x=14x

The solution is x=1 6

i.e. 10 minutes.

Let the car arrive and return to the walker at x hours, 5 (1 + 35 60 + x) + 60x = 35;x=5 12 hours 1 question answer is 1 + 35 60 + 5 12 = 2 hours 2:60 * (35 60 + 5 12 + 5 12) = 85 km 4, Assuming that the time of the encounter is t, then.

A time + B time = 100

A + B) Time = 100

time=100

time=100

Let A catch up with B after x seconds, according to the equivalence relationship: the distance traveled by A in x seconds = the distance traveled by B in x seconds + the distance traveled by B in 1 second

The column equation gets: 7x=, solves: x=13, and answers: A can catch up with B after 13 seconds

Set t seconds after A catches up with B.

The speed of B is less than that of A, and B in the same direction is 10 meters in front of A

t=20 (sec).

A: After 20 seconds, A catches up with B.

I wish you progress in your studies!!

1.Solution: Pool A ---400+4-8) 2-4=194 kg Pool B (400+4-8) 2+8=204 kg.

2.Solution: (5*18 60) (14-5)=1 6 hours=10 minutes3Solution: (1) x hours after that. Rule.

5x+60(x-1)=35*2 x=2(2) after arrival: 35+2(35-5x)=85 kilometers.

4.Solution: (1) 100 (7+ seconds.)

2) (100+ seconds.)

3) 10 (seconds.)

The armor setting speed is x

Then B 4 hours is 5 4

So in 2 hours A has 5 more than B 4

So 2(x-5) 5 4

x-5≥10

x 15A: A is riding at least 15 km/h.

B has advanced a total of 5*(4+2)=30 kilometers in the previous 4 hours, and A must ride at least 30 2=15 kilometers in order to catch up.

Solution: 1 2(1-x)=1+k

1-x=2+2k

x=-2k-1

3 4(x-1)-2 5(3x+2)=k 10-3 Shanji2(x-1).

15(x-1)-8(3x+2)=2k-30(x-1)15x-15-24x-16=2k-30x+3021x=61+2k

x=(61+2k)/21

The solutions are inverse numbers of each other.

2k-1+(61+2k)/21=0

42k-21+61+2k=0

40k=40

Brother Qiao k = 1

Xie Fang's hand is repentant, and Cheng Bizheng is taking a trillion.

x=-2k-1

x=(2k+43)/23

2k-1)+(2k+43)/23)=0

46k-23+2k+43=0

20=44k

k=5/11