1: What is the ratio of 1:3 cement mortar undercover (volume or mass?)

1: What is the ratio of 1:3 cement mortar undercover (volume or mass?)

Mass ratio, cement: sand --- 1:3
Undercover is the meaning of undercover, such as paste floor tiles generally say: with 20 thick 1:3 cement mortar undercover, cement paste

1: 3 weight mix proportion of cement mortar?

It's the mass ratio
The mixture of cement, sand and water is called cement mortar. Generally speaking, 1:3 cement mortar is mixed with 1 part of cement and 3 parts of sand. In fact, the composition of water is ignored. Generally, the proportion of water is about 0.6, that is, it should be 0.6:1:3, and the density of cement mortar is 20KN / m ^ 3

A and B start from a and B at the same time and go opposite each other. When they start, the speed ratio of a and B is 5:4. After meeting, the speed of a decreases by 20%
The speed of car B is increased by 20%, so that when car a arrives at place B, car B is still 10 kilometers away from place A. find the distance between two places of ab. (formulate and analyze or formulate equations, but do not use the equation of two variables once, but only one unknown number)

The arithmetic solution is as follows: after meeting, the speed ratio of Party A and Party B is 5 × (1-20%): 4 × (1 + 20%) = 5:6. When meeting, Party B walks 4 ÷ (5 + 4) = 4 / 9 of the whole journey, so after meeting, Party A goes 4 / 9 of the whole journey, so after meeting, Party A goes 4 / 9 of the whole journey, and Party B walks 4 / 9 △ 5 × 6 = 8 / 15 of the whole journey, so Party B walks 4 / 9 of the whole journey

Application examples of plane vector
Given that the line L: MX + 2Y + 6 = 0 and the vector (1-m, 1) is parallel to L, what is the value of the real number m?

The normal vector of line L is (m, 2), which is the coefficient of line L equation
Then vector (m, 2) is parallel to vector (1-m, 1)
M (1-m) + 2 = 0, M = - 1 or 2

The unit price of apple is 1 / 2 of that of pear. The price of pear is equivalent to the price of () kg of apple; the price of 6kg of apple
The price is equivalent to () kg Li0; the price of 3kg pear and 6kg apple is equivalent to () kg pear or () kg apple

The unit price of apples is equivalent to 1 / 2 of that of pears. The price of pears is equivalent to the price of (2) kilograms of apples
1÷1/2=2
The price of 6 kg apple is equivalent to that of (3) kg pear
6×1/2=3
The price of 3 kg of pear and 6 kg of apple is equivalent to the price of (6) kg of pear or (12) kg of apple
3+6×1/2=6
3÷1/2+6=12

Let X and y be real numbers, and X (1 + I) + y (1 + 2I) = 5 (1-3i), find the value of 2x + 3Y

x+xi+y+2yi=5-15i
x+y+xi+2yi=5-15i
x+y=5
x+2y=-15
Solution
x=15
y=-20
2x+3y
=2*15+3*(-20)
=30-60
=-30

Party A and Party B race around the lake. A week around the lake is 400 meters. Party B walks 80 meters per minute. The speed of Party A is 1.25 times that of Party B. now Party A is 100 meters ahead of Party B. how many minutes later
To solve the equation
Explain ideas

Let's catch up in X minutes
100x=80x+(400-100)
20x=300
x=15
The idea is to list the equations according to the conditions given in the title, and then solve X
The way to do this kind of problem is to grasp the known conditions given by the problem, then set up the unknowns, list the equations, and then solve X

Taking Sina = 4 / 5 and a as the acute angle, the value of sin2a + cos2a is obtained

cosa=3/5
-->sin2a=2sinacosa=2*3*4/25=24/25
cos2a=cos^2a-sin^2a=16/25-9/25=7/25
And 31 / 25

The speed of car a is 30 km / h, and that of car B is 40 km / h. The two cars start from the same place at the same time and go in the opposite direction. After what time, the distance between the two cars is 140 km
The speed of car a is 30 km / h, and that of car B is 40 km / h. The two cars go in the same direction on the same ground. When does the distance between the two cars reach 140 km?
Solve the equation

The distance is 140 km in X hours
X*(30+40)=140
70X=140
X=140/70
X=2

Here's another math problem
How many can nine numbers from 1 to 9 make up——
(1) Three digits without repetition?
(2) A three digit number that can be divided by 5 without repetition
(3) A three digit number greater than 500 without repetition

(1) Three digits without repetition? 9 * 8 * 7 = 504
(2) The number of three digits that can be divided by 5 without repetition = 8 * 7 = 56
(3) The number of three digits greater than 500 without repetition = 5 * 8 * 7 = 280