The two cars run from east to west at the same time. Car a runs 56 kilometers per hour and car B 48 kilometers per hour The two cars meet 32 kilometers away from the midpoint. How many kilometers is the distance between the East and the west? No equation, no equation) just need to solve the problem. The idea should be clear and easy to understand

32 × 2 = 64 km, because we meet at 32 km from the end point, it is twice as fast as the slow one (that is, a is faster than B)
56-48 = 8 km / h a is 8 km faster than B
64 △ 8 = 8 hours, so it was 8 hours driving when we met
8 × (56 + 48) = 832km
A: the distance between the East and the west is 832 kilometers
I don't know which step to follow

If the inequality x ^ 2-2x + 4-m ^ 2x about X is less than or equal to 0, it is consistent on [1,3], and the value range of real number m is obtained
It's m ^ 2 times X

When x = 1, 1 - 2 + 4 - M & # 178; ≤ 0, M & # 178; ≥ 3, m ≤ - √ 3 ∪ m ≥ 3
When x = 3, 9 - 6 + 4 - M & # 178; ≤ 0, M & # 178; ≥ 7, m ≤ - √ 7 ∪ m ≥ 7
Take its intersection to get m ≤ - √ 7 ∪ m ≥ √ 7

The toll standard of a highway toll station for passing vehicles is: bus 30 yuan, minibus 15 yuan, sedan 10 yuan. The ratio of the number of buses and minibuses passing through the toll station is 5:6, the ratio of minibus and sedan is 4:11, and the toll of the minibus is 210 yuan more than that of the bus. How much is the number of the three kinds of vehicles passing through on this day?

Bus: minibus = 5:6, minibus: sedan = 4:11, then bus: minibus: sedan = 10:12:33, bus money: minibus money: sedan money = (10 × 30): (12 × 15): (10 × 33) = 10:6:111, share money = 210 ÷ (11-10) = 210 (yuan) bus money: 210 × 10 △ 30 = 70

If y = cos2x + 2psinx + Q has a maximum value of 9 and a minimum value of 6, find the value of real numbers P and Q

Let SiNx = t, t ∈ [-1, 1,1], y = 1-sin2x + 2psinx + QY = - (sinx-p) 2 + P2 + Q + 1 = - (T-P) 2 + P2 + Q + 1 t = t, t ∈ [- 1,1], y = 1-sin2x + 2 psinxx + QY = - (sinx-p-p) 2 + P2 + Q + 1 = (T-P) 2 + P2 + Q + Q + 1 = - (t-p-p) 2-P) 2 + P2 + P2 + Q + Q + 1 + 1 - (t-p-p) 2 + P2 + 2 + P2 + P2 + Q + Q + 1 = 6, we get P = 34, q = 152, which is the contradiction of P < 1 and P < p < 1; the contradiction of P < p < 1; when p > 1, [[- 1, [[- 1 [[- 1, when p \1] is the increasing interval of function y, ymax = y | t = 1 = 2p +Q = 9, Ymin = y|t = - 1 = - 2p + q = 6, P = 34, q = 152, which is contradictory to P > 1; when - 1 ≤ P ≤ 1, ymax = y|t = P = P2 + Q + 1 = 9, then when p ≥ 0, Ymin = y|t = - 1 = - 2p + q = 6, P = 3-1, q = 4 + 23; when p < 0, Ymin = y|t = 1 = 2p + q = 6, P = - 3 + 1, q = 4 + 23  P = ± (3-1), q = 4 + 23

Party A and Party B took the same money to buy apples, a 6 kg, B 14 kg, B to give a 12 yuan, ask how many apples per kg

Apples per kg:
12*2/(14-6)
=24/8
=3 yuan

If x-yi = 4-3i (x, y ∈ R), find the value of x.y

x-yi=4-3i
x=4 y=3

It is known that 100 tons of cement can be transferred from warehouse A, 80 tons from warehouse B and 70 tons from warehouse a,
It is known that warehouse a can transfer out 100t of cement and warehouse B can transfer out 80t of cement; place a needs 70t of cement and place B needs 110t of cement. The distance and freight from two warehouses to a and B are as follows:
Distance (km) freight (yuan / T)
Warehouse A, warehouse B, warehouse A, warehouse B
A: 20 15 12
B: 25 20 10 8
If the freight is 37100, how many tons of cement should be transported from warehouse A to place a?
(it's urgent. It's due on Sunday. It's a one variable equation.)

Suppose that warehouse A is transported to place a, then warehouse A is transported to place B (100-x), warehouse B is transported to place a (70-x), and warehouse B is transported to place B [110 - (100-x)]
20*12x+15*12(70-x)+25*10(100-x)+20*8[110-(100-x)]=37100
The solution of the equation is x = 70

Cosa = 3 / 5, then sin2a =, cos2a=

cosa=3/5
Sina = - 4 / 5 or Sina = 4 / 5
Sin2a = 2sinacosa = 2 * (4 / 5) * (3 / 5) = 24 / 25 or = - 24 / 25
cos2a=2cos²a-1=18/25-1=-7/25

Car a and car B climbed over Saturn 1 mountain. Car a traveled 30 kilometers in 20 minutes and car B 40 kilometers in 30 minutes. Which car is faster? How much faster? Please help to write the formula. Thank you

A speed per hour = 30 * 3 90 km / h
B speed per hour = 40 * 280km / h

This is the seventh question on page 126 of Volume 2 of grade 9!
The three views of a geometry are as follows, and the expanded view of the solid figure is drawn according to the views
Top view: a circle!
Left view: a rectangle!
Top view: square!
Sorry, the first one is the main view

The stereogram is a cylinder, and the expansion is a rectangle. As for the top view is a square, it means that the height of the cylinder and the diameter of the circle are equal. So after expansion, the length of one side of the rectangle is x, and the length of one side is π times X

If the inequality x ^ 2-2x + 4-m ^ 2x about X is less than or equal to 0, it is consistent on [1,3], and the value range of real number m is obtained It's m ^ 2 times X