Help, it s all math problems I don t understand, and I can only do it if I do it

1. If there are x volleyballs, then basketball is 5 3x and volleyball is 5 2x.

Basketball is 5 3:5 2 = 2:32, set the purchase price x, the original price y

y-x=500

Get y=800 then x=300

The purchase price is 300 yuan.

3, the first two draw numbers don't matter, the third is when there are 8 left, and there is only one on the 8th.

Then that's 1 8 probability.

1.Volleyball is three-fifths of basketball, and football is two-fifths of volleyball, so football is six-fifths of basketball. 3/5 * 2/5 = 6/25

2.Original price (500-300) (800.)

The purchase price is 800-500=300

I. 25:6

Solution: Volleyball: Basketball 3:5, you can think of the number of volleyballs as 3, the number of basketballs as 5, then the number of football is 2 5 of the number of volleyballs, then the number of football is 3 times 2 5 6 5, so basketball: football 5 6 5=6 25

Second, 300 solution: let the purchase price be y, the original price is x, then there is:

1、y=x-500

2、y=75%x-300

Solve x=800, and then bring in y=300 three

1。The number of soccer balls is 3/5 less than the number of volleyballs = > soccer ball is 2/5 of volleyball, so the ratio of soccer balls to volleyballs is 2:5

So the ratio of football, volleyball, and basketball is.

The ratio of football and basketball is 6:25

2。The original price is 200 yuan less than 75% off, so the original price is (500-300) (800, so the purchase price is 800-500=300 yuan.)

3。There are 8 questions left, and the probability of drawing one from them is equal, which is 1 8

1 If the number of volleyballs is 3x and basketball is 5x, then the soccer ball is 3x-3x*(3 5)=(6/5)x

So the ratio of basketball to football is 25:6

2 Let the original price be x, the input price be y, and the system of equations x-y=500, and the solution will be x=800 and y=300

3 Draw No. 8 is entitled Event A

p(a) = 1 in 8.

Even if I don't know much about myself.

1. The area of the rectangle: 160 * 100 = 16000 The area of the small square below: 40 * 40 = 1600

The area of the upper trapezoid: 1 2 * (40 + 160) * (100-40) = 6000 The area of the shadow is: 16000-1600-6000 = 84002, and the area of the large square: 6 * 6 = 36

The area of the small square: 4*4=16

The area of the triangle: 1 2 * (6 + 4) * 6 = 30 The shadow area is: 36 + 16-30 = 22 square centimeters.

Answer: Both of these questions are not very difficult, the idea is the same, just subtract the white area from the total area

1.The total area is 160*100, and the white part is square (40*40) + trapezoidal area ((40+160)*(100-40) 2).

16000-1600-6000=84002.The total area is 6*6+4*4=52, and the white part is triangular (6+4)*6 2=30

The idea is: subtract the white area from the total area

1.The total area is 160*100, and the white part is square (40*40) + trapezoidal area ((40+160)*(100-40) 2).

16000-1600-6000=84002.The total area is 6*6+4*4=52, and the white part is triangular (6+4)*6 2=30

is a simple question in math with the following short answers.

The total area of the first graph is 160 times 100 is 16000 minus the area of the blank space is the area of a trapezoid and the area of a square, the trapezoid is (160+40) multiplied by the height (100-40) divided by 2 to get 6000, and the area of the square is 40 times 40 is 1600, so the area of the shaded part of the first question is 16000-6000-1600=8400

Question 2: Add up the area of two squares, find the sum, and then subtract the area of a teaching triangle to get the result, the area formula of the triangle between is the base multiplied by the height divided by two, because the figure is not clear, so the result is not counted.

1) Because aob=90°, aoc=30°, boc=120°. Because OM bisects BOC, com=12 BOC=60°. Since on bisects aoc, con=1 2 a oc=1 2 30°=15°.

So mon = com con=60° 15°=45°;

2) When aob= and other conditions remain unchanged, imitation (1) can obtain mon= 1 2 (i.e., (a+ aoc) 2- aoc 2= 1 2).

3) Imitation (1) can be found mon= com con

1. Obviously, after the rotation moves, oa oc on coincides, then, dob=80°-40°=40°, then mod=20°, then mon=40°+20block=60°

2. In the same way, it is 60°

Solution: (1) Let the sum of these 11 numbers be a times of 20, according to the title.

1+3+9+x+3+7+y+5+8=20a, i.e. x+y=20a-36

0 x+y 18

0≤20a-36≤18

The solution is an integer. a=2

x+y=20×2-36=4 ．

2) There are a total of 5 pairs, and the probability of playing a pair number is 1 5