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The speed of the train is 75% of the speed of the bus. The two trains leave from a and B at the same time and meet at a distance of 6 km from the midpoint
How many kilometers are there between a and B?
They meet at the midpoint of 6 km, so the distance difference between the two vehicles is 2 * 6 = 12 km
Divide the distance difference by the speed percentage difference, which is the speed of "unit one" bus. 12 / (1-75%) = 48 km / h is the speed of bus
The train speed is 48 * 75% = 36 km / h
Because of the distance difference of 12 km, I drove for 1 hour
Therefore, the distance between stations a and B is (48 + 36) * 1 = 84km
Solving inequality: 0 ≤ x ^ 2 + 4x + 3 ≤ 8
Find 0 ≤ x ^ 2 + 4x + 3 first
0 ≤ (x + 1) (x + 3): X ≥ - 1 or X ≤ - 3
Find x ^ 2 + 4x + 3 ≤ 8 again
x^2+4x-5≤0
If (x + 5) (x-1) ≤ 0: - 5 ≤ x ≤ 1
By combining the two methods: - 5 ≤ x ≤ - 3 or - 1 ≤ x ≤ 1 can be obtained
The distance between a and B is 450 km. The passenger train starts from a place for 2 hours, and the freight train starts from a place. After 3 hours, the two cars meet. Every hour, they travel 10 km more than the train
fast
When they met, the bus drove for five hours and the truck for three hours
If the passenger car travels 10 kilometers more than the freight car per hour, the speed of the freight car is set as X, and the speed of the passenger car is set as x + 10
Then 3x + 5 × (x + 10) = 450
Find x = 50
Therefore, the speed of passenger cars is 60km / h and that of freight cars is 50km / h
According to your topic, I make this understanding, I don't know, right
In the triangle ABC, a (3,1), B (7, y), C (- 5,7) and G (x, 4). X, y ∈ R are known
(1) Find the value of X, y (2) if the trisection point of line BC is m n in turn, find the coordinates of vector am and vector an
(1) Let the midpoint of AC be D, then d (- 1,4). Connect BD. vector BD = (- 1,4) - (7, y) = (- 8,4-y). Vector BG = (x, 4) - (7, y) = (x-7,4-y). ∵ vector BG = (2 / 3) vector BD = (2 / 3) (- 8,4-y) = (- 16 / 3,8 / 3-2y / 3) ∵ (x-7,4-y) = (- 16 / 3,8 / 3-2y / 3). X-7 = - 16 / 3, x = 7-16 / 3 = 5 / 3; 4-y
It took six days for the engineering team to build a 1290 km long railway. Team a built 15 km more than team B. how many kilometers did team a and team B build?
Let a fix x every day and B fix y every day
（x+y）*6=1290
6x-15=6y
y=x-2.5
Bring in
（x+x-2.5)*6=1290
2x-2.5=215
x=108.75
y=108.75-2.5=106.25
6 x = 652.5
Second, 6y = 637.5
1+2+3+4+… +96 + 97 + 98 + 99 + 100 =? Simple algorithm
（1+100）×50
A car runs 100km with 9L gasoline. How many liters of gasoline does it use per kilometer?
9 △ 100 = 0.09l
A: the average gasoline consumption per kilometer is 0.09l
In △ ABC, if ∠ a - B = 90 °, then the triangle is______ Triangle; if ∠ a = 12 ∠ B = 13 ∠ C, the triangle is______ A triangle
(1) The results show that ∵ a - ∵ B = 90 °, a ∵ a ＞ 90 ° and ∵ triangle is obtuse angle triangle; (2) ∵ a = 12 ∵ B = 13 ∵ C, ∵ a + ∵ B + ∵ C = 180 °, C = 90 ° and ∵ triangle is right angle triangle
A and B leave each other from a and B at the same time. They meet in ten hours. When they meet, car a travels five seventh of car B's journey. How many hours does it take car a to complete the whole journey?
(don't just list the formula, but tell me the reason and the way to solve the problem, so I will get more points.)
Speed sum of two cars: 1 △ 10 = 1 / 10
The speed of vehicle B is "1", and the speed of vehicle B is: 1 / 10 (1 + 5 / 7) = 7 / 120
The speed of car a is 1 / 10-7 / 120 = 1 / 24
Time for car a to complete the whole journey: 1 / 24 = 24 (hours)
Given a ^ 2 + B ^ 2-4a + 6B + 13 = 0, find a + B ^ 2
The original formula = (a ^ 2-4a + 4) - 4 + (b ^ 2 + 6B + 9) - 9 + 13 = 0
(a-2)^2+（b+3)^2-4-9+13=0
(a-2)^2+(b+3)^2=0
Because (A-2) ^ 2 is greater than or equal to 0, (B + 3) ^ 2 is greater than or equal to 0
So A-2 = 0, B + 3 = 0
Then a = 2, B = - 3
A + B ^ - 2 = 2 + (- 3) ^ - 2 = 2 + (- 1 / 3) ^ 2 = 2 + 1 / 9 = 19 / 9
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