A and B vehicles run from AB to ab at the same time. A vehicle runs 60 kilometers per hour and B vehicle runs 48 kilometers per hour. The two vehicles meet at 12 kilometers away from the midpoint. At this time, B vehicle is far away from ab How many kilometers are there at point a

A and B vehicles run from AB to ab at the same time. A vehicle runs 60 kilometers per hour and B vehicle runs 48 kilometers per hour. The two vehicles meet at 12 kilometers away from the midpoint. At this time, B vehicle is far away from ab How many kilometers are there at point a

Encounter time: 12 × 2 △ 60-48 = 2 hours
At this time, the distance between B and a is the distance that a will walk when they meet: 60 × 2 = 120 km
12 × 2 / (60-48) = 2 hours
60 × 2 = 120 km
Suppose that the distance between AB and ab is 2x. According to the same time when they meet, there is an equation
（x+12)/60=(x-12)/48
Find x = 108
Therefore, the distance from B to a, that is, the distance from a, is 108 + 12 = 120 km
(60 + 48) x / 2-48x = 12, x = 2 hours, B distance from a120m
Because car a is faster than car B, car a has passed the midpoint, driving 12 × 2km more than car B
Time for two vehicles to meet: 12 × 2 △ 60-48 = 2 hours
The distance between B and a is the driving distance of a vehicle: 60 × 2 = 120 km
Car a and car B respectively travel from AB to each other. Car a travels 40 kilometers per hour, and car B travels 30 kilometers per hour. The distance between the two encounters is 60 kilometers, and the distance between AB and car B is 60 kilometers
The speed ratio of car a and car B is 40 ∶ 30 = 4 ∶ 3. When they meet for the first time, the two cars run a whole journey, and car a runs 4 / (4 + 3) = 4 / 7 of the whole journey, that is, the distance between the first meeting point and ground a is 4 / 7 of the whole journey; when they meet for the second time, the two cars run three whole journey, and car a runs (4 / 7) × 3 = 12 / 7 of the whole journey
Car a travels 40 kilometers per hour, and car B travels 60 kilometers per hour. Car a and car B start from ab at the same time and go in opposite directions
It took another 4.5 hours for car a to reach B. how many kilometers is the distance between AB and B
Meeting time: 40 × 4.5 △ 60 = 3 hours
The distance between the two places is (40 + 60) × 3 = 300 km
Car a is 40 kilometers per hour, and car B is 60 kilometers per hour. After 4.5 hours of meeting, the distance between car a and car B is 40 kilometers per hour
When meeting, B line: 40 × 4.5 = 180 km
Meeting time: 180 △ 60 = 3 hours
AB distance: (40 + 60) × 3 = 300 km
It's very simple. When car a arrived at B in 4.5 hours, it would be 40 * 4.5 = 180
Speed ratio: A: B = 40:60 = 2:3
Whole course: 2 + 3 = 5
It takes 4.5 hours for Party A to travel three times,
Travel time of a line: 4.5 / 3 = 1.5 hours,
The travel time of a line 5: 1.5x5 = 7.5 hours
AB distance: 40x7.5 = 300 (km)
AB distance:
(40 * 4.5 / 60 + 4.5) * 40 = 300km
A and B walk towards each other from a and B at the same time. A walks 5 kilometers every hour. When they meet, B walks another 10 kilometers to a, and a walks another 1.6 kilometers to B. how many kilometers does B walk every hour?
(5 × 1.6) △ 10 △ 5) = 8 △ 2, = 4 (km). Answer: B walks 4 km per hour
(k-1) x & # 178; - (K + 1) x + 8 = 0 is about the unary power of X. first, tell me how much K is equal to, why, be more detailed
(1) Find the value of 2012 (4k-x) (x-2010k) + 2012k
(2) Finding the solution of the equation K y = x about y
(k-1) x & # 178; - (K + 1) x + 8 = 0 is a linear equation of one variable with respect to X
k-1=0
K=1
The linear equation of one variable is
-2x+8=0
X=4
2012（4k-x）（x-2010k）+2012k
=2012
(2) Finding the solution of the equation K y = x about y
|y|=4
y=±4
If it is one variable, then the coefficient of quadratic term k-1 = 0, k = 1. I don't understand the following two questions. Would you like to explain them?
(k-1) x & # 178; - (K + 1) x + 8 = 0 is a one variable linear equation about X, that is to say, the coefficient of quadratic term is 0,
That is: k-1 = 0, k = 1, substituting into the original equation to calculate x = 4,
(1)2012（4k-x）（x-2010k）+2012k
=2012（4×1-4）（4-2010×1）+2012×1
=0×（-2006）+2012
=6
(2) K ︱ y = x is: 1 ×︱ y =... Expansion
(k-1) x & # 178; - (K + 1) x + 8 = 0 is a one variable linear equation about X, that is to say, the coefficient of quadratic term is 0,
That is: k-1 = 0, k = 1, substituting into the original equation to calculate x = 4,
(1)2012（4k-x）（x-2010k）+2012k
=2012（4×1-4）（4-2010×1）+2012×1
=0×（-2006）+2012
=6
(2) K y = x is 1 x y = 4, y = ± 4. Follow up: Er, the first question is wrong
A question about the value range of inequality
The positive integer solutions of X + 1 ＜ a are 1 and 2, and the value range of a is obtained
Write the reason
X+1＜a
X＜a-1
Because the positive integer solutions of X + 1 < A are 1 and 2
SO 2 < A-1 ≤ 3
3＜a ≤4
Which score is higher, 4 out of 14 or 9 out of 21
4/14＝2/7
9/21＝3/7
So 4 / 14
A and B start from a at the same time and drive along the same road to B
A and B start from place a at the same time and drive along the same road to place B. car a is 5 kilometers faster than car B every hour. Car a arrives at place C on the way 0.5 hours earlier than car B. when car B arrives at place C, car a just arrives at place B. It is known that the road from place C to place B is 30 kilometers long
(1) How many kilometers does car B travel per hour?
(2) How long is the highway between a and B?
Wait!
1) It takes 30 / 5 = 6 hours for a to drive the whole course
Speed A: 30 / 0.5 = 60 km / h
Speed B: 60-5 = 55 km / h
2) Total length: 60 * 6 = 360 km
(1) A's speed is 30 / 0.5 = 60 km / h
So, B's speed is 60-5 = 55 km / h
(2) A is 5 kilometers faster than B per hour, and at the end of AB, it is 30 kilometers more than B
Therefore, it takes 30 / 5 hours to get to the end
Therefore, the distance between AB and ab is 60 * 6 = 360 km
The topic says that when a arrives at C, it's half an hour earlier than B, and then when B arrives at C, a arrives at B, so within half an hour, a drives a distance of 30 km, so a's speed is 30 / 0.5 = 60, and B's speed is 60-5 = 55 km. The second question requires the full length of ab. we already know the length of BC. We only need to calculate the length of AC. the first question has found that the speeds of a and B cars are 60 and 55, then AC / 55-ac / 60 = 0.5, and the solution is AC = 330
The topic says that when a arrives at C, it's half an hour earlier than B, and then when B arrives at C, a arrives at B, so within half an hour, a drives a distance of 30 km, so a's speed is 30 / 0.5 = 60, and B's speed is 60-5 = 55 km. The second question requires the full length of ab. we already know the length of BC. We only need the length of AC. the first question has found that the speeds of a and B cars are 60 and 55, so AC / 55-ac / 60 = 0.5 and AC = 330
For the equation (K & # 178; - 2) x & # 178; + 2 (k - √ 2) x + K + 3 = 0, when K (), it is a linear equation of one variable with respect to X
The above √ is the root sign
In order to make the original equation a linear equation with one variable, then
k∧2-2＝0
∴k＝±√2
When k = √ 2, the coefficient of the first term is equal to 0, so it is rounding off!
∴k＝-√2
In other words, the coefficient of quadratic term in the original equation is 0,
That is: K & # 178; - 2 = 0, k = ± √ 2
So, when K (= ± √ 2), it is a linear equation of one variable about X
From the meaning of the question, K ∧ 2 = 2 and K ≠ 2 from K ∧ 2 = 2, we can get k = ± √ 2, because K ≠ 2, so k = - √ 2 does not know how to ask