Mathematics summer homework for the first year of junior high school should be accurate and step by

Set up a win in x field, a draw in y field, and a series of equations.

3*x+5*y+5*0=19

x+y+5=14

Solve the system of equations.

x=5y=4

So five wins.

Set: Won x games.

The number of draws is (14-5-x).

3x+（14-5-x）=19

The solution is x=5

A: Won 5 games.

Explain that only 9 games have been scored.

Simple method, which can be used for multiple-choice type questions:

If there are 9 draws, there are only 9 points, and the extra 10 points are definitely won by the win, and each win is 2 points more than the draw, then the win is 10 2 = 5

The equation method is set to win the x field.

3x+（9-x）×1=19

x=5 of the solution

Of the 14 games, five losses are equivalent to only nine games and 19 points.

Then there are 3*5=15 that is, 5 wins and 4 draws.

That is, the team record is 5 wins, 4 draws, 5 losses.

Solution: Suppose this team wins x field ball, then the game is tied (14-5-x) that is, (9-x) field ball, according to the title:

3x+1×（9-x）+0×5=19

Solve this equation to get:

x=5A: This team has won 5 games.

Win 5 games. A total of 19 points should be divided into 9 parts. There are only two forms of 9 servings. There are only 4 possible cases (8+1), (7+2), (6+3), (5+4).Substitution of the method of elimination will do.

Looking at the screenshot, the first half of the fifth question is estimated to have been copied incorrectly.

1、（2n-1）/（n^2)

2. The selling price is RMB, and the profit is 20% A RMB. -m^2

x^2+2y^2=2(x^2+y^2)=2[(x+y)^2-2xy]=2(9^2-2*14)=106

I'm just doing the task, please ignore me.

Solution: Let event p be the probability of unlocking, then the basic event space of p is "0,1,2,3,4,5,6,7,8,9".

Since only one number can open the lock, then p=1 10

Hope to be adopted, thank you! ）

The possible values of p are "0,1,2,3,4,5,6,7,8,9" which is a total of 10.

Only one digit is his last password number, so p=1 10

There are a total of 10 possibilities for the last bit, 0-9

There is only one correct number, so the probability is 1 10

The total probability is 10

Correct is 1, so the probability is 1 10

10% since you forgot the last digit on the left, that's one from 0 to 9, one tenth.

1) When panning, s is 3 cm; At this time, the overlap is rectangular, horizontal, 2 wide, and 3

2) When 2 t 4, the area of the figure swept diagonally by a small square is 4 cm; At this time, the overlapping part is the small square, and the small square happens to be in the whole big square, so the area of the overlapping part at this time is the area of the small square is 4

3) When s=2cm, the distance of the small square translation is 1 or 5 cmAt this time, the area is less than three in question 1, indicating that the distance of movement is less than that, then the distance moved is 2 divided by 2 = 1, and there is also a case that when the small square moves out of the large square, the overlapping part is 2, and the length of the overlapping part can be calculated to be 1, so the distance moved at this time is 2+(4-1)=5

I hope it can help you, you can't ask questions, satisfied, this kind of problem has to learn to draw by hand, and it is easy to solve it by drawing the right image according to different exercise times.

(1) translation, then the bottom of the overlapping part is, the height is 2cm, 2) 2 t 4 this range, the distance of the small square is between 2cm and 4cm, and you can get a diagonal line just swept through a small square + a triangle, so the area is equal to 2 2 + 2 2 2 = 6 (cm2).

3) When s=2cm2, the distance = 2 2=1 (cm),

s=2t (0=6) does not overlap the large square.

1) s=2*

2)s=43) 2=2t,t=1 or 12-2t=2,t=5

Solution: Set up large X cans of beverages and small Y cans of drinks.

Column equations: 3x+4y=120 2x+3y=84 2- 3 get:

y=12 substitute y=12

Get: x=24

A: 24 cans of beverages in large pieces and 12 cans of beverages in small pieces.

3 large pieces, 4 small pieces, a total of 120 cans, 6 large pieces, 8 small pieces, a total of 240 cans, two large pieces, 3 small pieces, a total of 84 cans, 6 large pieces, 9 small pieces, a total of 84 * 3 cans, two subtraction.

1 small piece 84*3-240=12 cans.

1 large piece (120-12*4) 3=24 cans.

120-84=36 is 1 large and 1 small.

120-36*3=12 is 1 small.

36-12=24 is 1 large.

Set up x cans for each large piece, and Y cans for small pieces.

3x+4y=120, 2x+3y=84 3 2 get y=12Substituting it to get x=24

The solution is x=24 and y=12

Solution: Set up x cans for each large piece and y cans for each small piece.

3x+4y=120 （1）

2x+3y=84 （2）

Get x 24, y 12

A: 24 cans per large piece and 12 cans per small piece.