The math problem of grade one is very simple 1. It takes 3 hours for a ship to go back and forth between a and B. It takes 30 minutes more to go upstream than to go downstream. If the speed of the ship in still water is 26km / h, calculate the current speed 2. Ships a and B start from the upstream and downstream respectively. The speed of a in still water is 20 km / h, and that of B in still water is 16 km / h. It is known that ship a takes two hours to reach the downstream, and the time for ship B to reach the upstream is twice as long as that of ship a, so we can calculate the current speed 3. Xiaohua and Xiaofeng have a swimming competition. Make an appointment for the same distance. Xiaohua goes upstream and Xiaofeng swims downstream. It is known that in still water, the speed of Xiaofeng is 4.8 km / h and that of Xiaohua is 6 km / h. It is known that Xiaohua takes 10 minutes to finish the specified distance, and the time of Xiaofeng is just half of Xiaohua's 4. A two digit number, one digit number is half of ten digit number, transpose the two digits, the two digit number is 36 less than the original number, find the two digit number (Fang Chengjie) hurry! I'll add 100 points!

The math problem of grade one is very simple 1. It takes 3 hours for a ship to go back and forth between a and B. It takes 30 minutes more to go upstream than to go downstream. If the speed of the ship in still water is 26km / h, calculate the current speed 2. Ships a and B start from the upstream and downstream respectively. The speed of a in still water is 20 km / h, and that of B in still water is 16 km / h. It is known that ship a takes two hours to reach the downstream, and the time for ship B to reach the upstream is twice as long as that of ship a, so we can calculate the current speed 3. Xiaohua and Xiaofeng have a swimming competition. Make an appointment for the same distance. Xiaohua goes upstream and Xiaofeng swims downstream. It is known that in still water, the speed of Xiaofeng is 4.8 km / h and that of Xiaohua is 6 km / h. It is known that Xiaohua takes 10 minutes to finish the specified distance, and the time of Xiaofeng is just half of Xiaohua's 4. A two digit number, one digit number is half of ten digit number, transpose the two digits, the two digit number is 36 less than the original number, find the two digit number (Fang Chengjie) hurry! I'll add 100 points!


1. Let the velocity of water flow be x, the equation is obtained
(26+X)*3=(26-X)*3.5
Solve the equation and get x = 2
The current velocity is 2 km / h
2. Let the velocity of water flow be x, the equation is obtained
(20+X)*2=(16-X)*2*2
Solve the equation and get x = 4
The current velocity is 4 km / h
3. Let the water velocity be x, the equation is obtained
(4.8-X)*1/6=(6+X)*1/6*1/2
Solve the equation and get x = 0.4
At that time, the water speed was 0.4 km / h
4. Let the original single digit be x, and the equation is obtained
(2X*10+X)-(X*10+2X)=36
Solve the equation and get x = 4
It turns out that this double-digit number is 84



Write the following decimals or fractions in the form of negative integer exponential power: (1) - 18; (2) 0.000 & nbsp; 1; (3) 164


(1)-18=-2-3;(2)0.000 1=10-4;(3)164=2-6.



1. The length of the thick candle is the same as that of the thin candle. The thick candle can be lit for 5h, and the thin candle can be lit for 4H. If these two candles are lit at the same time, after a period of time, the remaining thick candle is three times as long as the thin candle. How long have these two candles been lit?
(2) when three children divide a packet of candy, the first one gets half of the total, and the second one gets the remaining one third. The third one finds that his candy is just twice that of the second one. What's the total number of candy?
③ A farm worker has to hoe two pieces of land, and the big one is twice as big as the small one. In the morning, the workers hoe on the big one. In the afternoon, the workers split up half and half. Half of them still hoe on the big one, and they finish weeding in the evening. Another worker goes to the small one, and there is still a part left in the evening, which will be hoed by a worker another day, How many workers are there on this farm?
4. If a increases x% to get B and b decreases y% to get a, then the relationship between X and Y is ()
A.x=y B.x+y=0 C.x=100y/100-y D.y=100-x/100+x
(to have a solution, or even higher points!
I'll give a higher score for the best answer today!
Linear equation of one variable


1. The length of the thick candle is the same as that of the thin candle. The thick candle can be lit for 5h, and the thin candle can be lit for 4H. If the two candles are lit at the same time, after a period of time, the remaining thick candle is three times of the thin candle. How long has the two candles been lit? Suppose that the total length is unit "1", and the time is x1-x / 5 = 3 [1-x / 4] x = 40 / 11, that is, lighting



Simplification: 6 (0.9y-x) = 9 (y-30-x)


6(0.9y-x)=9(y-30-x)
5.4y-6x=9y-270-9x
3x-3.6y=-270
5x-6y=-450



Given the square of a + 3A + 1 = 0, find the fourth power of 2A + the cube of 6a-6a-2


a²+3a=-1
2a^4 +6a³-6a-2
=2a²(a²+3a)-6a-2
=-2a²-6a-2
=-2﹙a²+3a﹚-2
=2-2
=0



It is known that a, B and C are unit vectors and satisfy 3A + MB + 7C = 0 vector. The angle between a and B is 60
What is the real number m


3A + MB + 7C = 0, 3A + MB = - 7C, 9 + m ^ 2 + 6mcos60 = 49, M = 5 or - 8



There are many examples in life that can explain the effect of force. The effect is related to the three elements of force. Please give one example each


Playing football will rotate in different directions, which is the point of action
When the spring is stretched short, it is the force
When the spring is elongated, it will point in the direction of the force



5x=0.15+x
How to solve this problem, quick


3.5x-x=0.15
2.5x=0.15
x=0.06



Addition and subtraction of fraction 2 / (x-1) ^ 2-2x / (1-x) ^ 2


2/(x-1)^2-2x/(1-x)^2
=2/(1-x)^2-2x/1-x)^2
=(2-2x)/(1-x)^2
=2(1-x)/(1-x)^2
=2/(1-x)



How to calculate the covariance by computer,


Suppose two samples are Xi, Yi, I = 1... N, e (x), e (y) are the arithmetic mean of two samples respectively
Residual VXI = Xi-e (x); vyi = Yi-e (y);
Covariance s (x, y) = ∑ VXI * vyi / (n-1)