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Car a and car B start from the East and west at the same time and travel in opposite directions. Car a travels 48 kilometers per hour and car B 42 kilometers per hour. The two cars meet 21 kilometers from the midpoint to find the distance between the East and the West______ Kilometers
21 × 2 △ (48-42) × (48 + 42), = 7 × 90, = 630 (km); answer: the distance between East and West is 630 km
A and B start from ab at the same time. A travels 48 kilometers per hour, B 40 kilometers per hour. The two vehicles meet at 32 kilometers away from the destination. How many kilometers is the distance between AB and B
The midpoint
Car a has traveled 32x2km more
Encounter time 32X2 △ 48-40 = 64 △ 8 = 8 hours
Distance (48 + 40) X8 = 88x8 = 704km
A and B vehicles run from AB to ab at the same time. A vehicle runs 70 kilometers per hour, B vehicle runs 65 kilometers per hour. The two vehicles meet at a distance of 20 kilometers from the destination,
A and B are thousands of meters apart? (using the equation) I'm only in grade five!
A and B vehicles travel from AB to ab at the same time. A vehicle travels 70 kilometers per hour, B vehicle travels 65 kilometers per hour. The meeting point of the two vehicles is 20 kilometers away from the midpoint. A and B are thousands of meters apart?
20÷【1/2-65/（70+65）】
=20÷【1/2-13/27】
=20÷1/54
=1080 km
20÷【1/2-65/（70+65）】
=20÷【1/2-13/27】
=20÷1/54
=1080 km
1: (x + 1.7) multiply by 2 = 10.22:3.14x + 8 = 20.56
1, multiply 2 to get 2x + 3.4 = 10.2, shift term to get 2x = 6.8, so x = 3.4
14 x = 12. 56, so x = 4
The speed of car a is 1.5 times that of car B, and the time when car a and car B arrive at station C on the way
It's 5:00 and 16:00 respectively. What's the time when the two cars meet?
Can step by step formula calculation, I am a primary school student, easy to understand
It is known that the time when train a and B arrive at station C on the way is 5:00 and 16:00 respectively,
B = C-11 hours on the way;
Moreover, the speed of car a is 1.5 times that of car B,
It can be concluded that when car a arrives at station C on the way, the time for the two cars to meet is 11 (1.5 + 1) = 4.4 hours;
So the two cars met at 9:24
If there is no linear term of X in the quadratic equation 2x + MX = 5x + 1, then M?
The equation is reduced to 2x ^ 2 + (m-5) X-1 = 0
Because the equation does not contain a linear term of X, m-5 = 0,
The solution is m = 5
A batch of goods will be transported by a number of 8-ton trucks. If each truck only carries 8 tons, there will be 11 tons left. If each truck carries 10 tons, it will be loaded by the last truck
How many cars are there? There are points!
If there are x cars, there are (8x + 11) tons of goods and (x-1) cars full of 10 tons. The weight of goods is more than 10 (x-1), but not more than [10 (x-1) + 2]
8x+11﹥10(x-1) (1)
8x+11≤10(x-1)+2 (2)
The solution to inequality (1) is: x < 10.5
The solution to inequality (2) is: X ≥ 9.5
The solution of inequality system is 9.5 ≤ x < 10.5
Because x is an integer, x = 10
A: there are 10 cars
With X cars
8x+11-10x
1/(x^2-13x+42)+1/(x^2-14x+48)+1/(x^2-15x+56)
1/(x²-13x+42)+1/(x²-14x+48)+1/(x²-15x+56)=1/[(x-6)(x-7)]+1/[(x-6)(x-8)]+1/[(x-7)(x-8)]=[(x-8)+(x-7)+(x-6)]/[(x-6)(x-7)(x-8)]=(3x-21)/[(x-6)(x-7)(x-8)]=3(x-7)/[(x-6)(x-7)(x-8)]=3/[(x-6)(...
Can you tell me what to do?
I suggest you use each one to break through the top students' teaching materials, especially creative! YTT 1) x ^ 2 - answer: X1 = - 10 x2 = - 9 (15) x ^ 2-25x + 156 = 0 answer: X1 = 13 x2 = 12
A and B are walking in the same direction on the same road from A. B, which is 5km away. A's speed is 54m / h, and a's speed is 3km / h. A takes a dog when a's speed is 3km / h
When chasing B, the dog first catches up with B, then returns to meet a, and then returns to catch up with B, again and again, until a catches up with B. It is known that the speed of the dog is 15 km / h, and the total distance of the dog is calculated
FSWLR：
Is the speed of a 5.4 km / h? You may have copied the wrong title
(1) When B = 54000-5000m, it needs to catch up with 4285
It took about 3.57 hours for a and 3.57 hours for a dog
(2) 15000m / h × 3.57h = 53550m
A: the total distance of dog running is about 53550 meters
Good luck and goodbye
In theory, it's the time of a chasing B multiplied by enough speed. The question is how to catch up with 3km / h at 54m / h. I can only give you a formula.
If a pursues B
Dog speed * distance / (a speed - B speed)
If B pursues a
Dog speed * distance / (B-A speed)
It is known that the square of the quadratic equation of one variable x-mx-2 = 0
(1) If x = - 1 is one root of the equation, find the value of M and the other root of the equation;
(2) For any real number m, judge the root of the equation and explain the reason
Substitute x = - 1 into the equation
1+m-2=0
M=1
x^2-x-2=0
(x+1)(x-2)=0
X = - 2 or x = 1
△=b^2-4ac
=m^2-4*(-2)
=m^2+8>0
So there are two unequal real roots of the equation
1. If x = - 1 is a root, substituting into the original formula, M = 1
We get the square of X - X - 2 = 0, and the other root is 2
2. Because B ^ 2-4ac = m ^ 2 + 8 > 0
So for any m, there are two roots X1 = (m ^ 2 + 8) / 2
X2 = (m ^ 2 + 8 under m-radical) / 2
If you have any questions, please take them. Thank you
(1) Substituting x = - 1, we get
1-m-2=0
m=-1
Then the equation is x ^ 2 + X-2 = 0
(x+2)(x-1)=0
x1=-2
x2=1
(2)b^2-4ac=m^2+8>0
Because the square of a number must be greater than or equal to zero, plus 8 must be greater than zero, so the equation has two unequal real roots
1. Substituting x = - 1 into the original equation, 1 + m-2 = 0, M = 1
Substituting M = 1 into the original equation, we get X & sup2; - X-2 = 0, (x + 1) (X-2) = 0, X1 = - 1, X2 = 2
2. Because m is a real number, according to Weida's theorem (- M) & sup2; - 4 * 1 * (- 2), that is (- M) & sup2; + 8 must be greater than 0
So the equation must have two unequal real roots
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