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A and B cars run from AB to each other at the same time. A car travels 48 kilometers per hour, B car travels 50 kilometers per hour. When they meet, B car travels 10 kilometers more than a car. How many kilometers is the distance between AB and B
Meeting time: 10 ÷ (50-48) = 5 hours
The distance between the two places is (50 + 48) × 5 = 490km
Car B travels 2 kilometers more per hour than car a
When they met, car B traveled 10 kilometers more than car a.
It took five hours
（48+50）*5=490
Distance between two places of AB: 10 ÷ (50-48) x (48 + 50) = 490 (km)
490, 5 hours
1) 10 / (50-48) = 5 (hours)
Distance: 5 * (48 + 50) = 490 (km)
Seeing the question you sent out, I think junior high school students can figure it out,. Now I have been out of society for more than six years. I can still figure it out when I am studying. Now I can't figure it out. one hundred and fifty-seven million eight hundred and seventy thousand six hundred and eighty-three
Four hundred and ninety
A and B travel from a and B at the same time. It takes 6 hours for a to complete the whole journey, which is 50% faster than B. when they meet, a travels 180 kilometers more than B. how many kilometers per hour can b travel?
(1 + 50%): 1 = 1.5:1 = 3:2180 ^ (33 + 2-23 + 2), = 180 ^ 15, = 900 (km); 900 ^ 6 ^ (1 + 50%), = 150 ^ 1.5, = 100 (km); a: car B travels 100 km per hour
A and B vehicles travel from AB to ab at the same time. A vehicle travels 50 kilometers per hour, B vehicle travels 48 kilometers per hour. They meet at 8 kilometers from the midpoint of ab
How many kilometers is the distance between a and B?
The distance difference between them is 8 + 8 = 16 km
Encounter time = 16 / (50-48) = 8 hours
Distance between the two places = 8 * (50 + 48) = 784 km
The distance between a and B is 784 km
The distance between the two places is x km
When they met, car a walked X / 2 + 8 kilometers, and car B walked X / 2-8 kilometers
X/2+8 : X/2-8 = 50:48
It can be concluded that x = 784km, and the distance between the two places is 784km
8 * 2 / (50-48) * (50 + 48) = 784km
Seven hundred and eighty-four
A and B are walking in the same direction on the same road from A. B, which is 5km away. A's speed is 5km / h, and B's speed is 3km / h. A takes a dog. When a catches up with B, the dog first catches up with B, then returns to meet a, and then returns to catch up with B, and then reverses in turn until a catches up with B. the known speed of the dog is 15km / h
Can you tell me the formula
The speed of a relative to B is 2 km / h, and the speed of dog relative to B is 12 km / h. OK, now it's very good. Forget it
If the equation 2-2x = mx-m is a univariate linear equation about X, then M=
2-2X=MX-M
2+M=MX+2X
2+M=﹙M+2﹚X
So m ≠ - 2
2-2X=MX-M
(M+2)X=M+2
M is not equal to - 2
-2-m is not equal to 0
M is not equal to - 2
With 8 tons of cars to transport goods, each car loaded 4 tons, there are still 20 tons left! If each car loaded 8 tons, then the last car is not satisfied, ask how many cars there are
More than five methods（
Method 1: 8 (x-1) + 1 ≤ 4x + 20 ≤ 8 (x-1) + 7
So 21 / 4 ≤ x ≤ 27 / 4
So 5.25 ≤ x ≤ 6.75
So x = 6
Method 2: the remaining 20 tons can be filled with another 5 cars, so there must be another one that is not filled
That's six
Who can help me solve an equation x * cosx = 126.86,
X is the angle multiplied by the number = number? Is there something wrong with your question? Even if it is converted into a number, it should be multiplied by π / 180
This kind of equation can not be solved directly and must be solved by computer numerical method
Newton iteration method is used to get the formula and the result is obtained step by step
x(n+1)= x(n)-(x(n)cosx(n)-126.86)/(cosx(n) -x(n)sinx(n))
A and B go from a place 25.5km away to B place at the same time. A rides a bike and B walks. A's speed per hour is more than three times of B's speed per hour. A stops for 45 minutes after arriving at B place, and then returns to a place from B place. On the way, they meet B. at this time, the time from their departure is exactly 3 hours?
Suppose B's speed is x km / h, then a's speed is (3x + 1) km / h. A's time from a to B is 25.5 / (3x + 1) h, and its stopover time is 45 / 60 = 0.75 h. When a and B meet, B's distance is 3x km, so a's distance is (25.5-3x) km, and the time of this distance is (25.5-3x)
If the equation (k-1) x ^ 2 + (4K + 3) x + 3k-5 = 0 is a linear equation with one variable, then the value of K is ()
In this case, the equation can only be one-time
With 8 tons of cars, each car loaded 4 tons, there are still 20 tons left! If each car loaded 8 tons, the last car is not empty
How many cars are there
Is there any inequality system? / / / thank you
8(x-1)+1≤4x+20≤8(x-1)+7
So 21 / 4 ≤ x ≤ 27 / 4
So 5.25 ≤ x ≤ 6.75
So x = 6
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