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The distance between city a and city B is 690 km. A car drives from city a to city B at the speed of 45 km per hour. Two hours later, a motorcycle drives from city B to city A Cars meet 240 kilometers away from city B. how fast is the motorcycle
After 2 hours, the distance between the car and B city is 690-45 * 2 = 600 km
When the car is 240 km away from city B, it takes = (600-240) / 45 = 8 hours
The motorcycle met the car 240 kilometers away from the city B
The speed of motorcycle is 240 / 8 = 30 km / h
As shown in the figure, it is known that ∠ B = ∠ C = 90 °, M is the midpoint of BC, DM bisects ∠ ADC
(1) It is proved that when passing through point m, I make me ⊥ ad, the perpendicular foot is e, ∵ DM ⊥ ADC, ∵ 1 = ∠ 2, ∵ MC ⊥ CD, me ⊥ ad, ∵ me = MC (the distance from the point on the bisector to both sides of the angle is equal), and ∵ MC = MB, ∵ me = MB, ∵ MB ⊥ AB, me ⊥ ad, ∵ am ⊥ DAB (the distance to both sides of the angle is equal)
It takes 30 minutes and 18 minutes for three cars to run a circle on the ring road. Now the three cars are at the same starting point and running at the same time
It takes 30 minutes and 18 minutes for three cars to run on the ring road. Now the three cars are at the same starting point and in the same direction at the same time. Question: at least how long will it take for them to meet?
The least common multiple of 30, 18 and 10 is 90
At least 90 minutes later, they meet
If the real numbers a and B satisfy the equation 2A & sup2; - 6A + 1 = 0, B & sup2; - 6B + 2 = 0 (AB ≠ 1), then (AB + 1) is divided by B =?
2A^2-6A+1=0 1)
B^2-6B+2=0 2)
1) 2-2)
4A^2-B^2-12A+6B=0
(2A+B)(2A-B)-6(2A-B)=0
(2A+B-6)(2A-B)=0
2A + B-6 = 0 or 2a-b = 0
When 2A + B-6 = 0, B = 6-2a
From equation 1, a (2a-6) + 1 = 0
A*(-B)+1=0
AB=1
It does not meet the conditions of the title, so 2a-b = 0, B = 2A
From 1), 2A ^ 2 + 1 = 6A
(AB+1)/B=(2A^2+1)/2A=6A/2A=3
A fast train is 70 meters long and a slow train is 90 meters long. If the two trains run in opposite directions, it takes 20 seconds for the fast train to catch up with the slow train and leave the slow train completely. If the two trains run in opposite directions, it takes 4 seconds for the two trains to meet and leave. How many kilometers per hour can the two trains travel?
Let the speed of the fast train be x m / s and that of the slow train be y m / s
20x－20y＝70＋90
4x＋4y＝70＋90
The solution is: x = 24
y＝16
Speed of express train: 24m / S = 86.4km/h;
Speed of local train: 16m / S = 57.6km/h
Given that the coordinates of two points P1 and P2 on hyperbola are (3, - 42), (94, 5), the standard equation of hyperbola is obtained
Let the hyperbolic equation be mx2-ny2 = 1 (Mn > 0), ∵ P (3, - 42), P2 (94, 5) two points on the hyperbola, ∵ 9m − 32n = 18116m − 25N = 1, ∵ M = - 19, n = - 116, ∵ the standard equation of hyperbola be y216 − X29 = 1
There were 100 tons of cement on the construction site. One fifth of the total amount was used in the first day, and the remaining half was used in the second day. How many tons of cement are left on the construction site
100-100×(1/5)-100×[1-(1/5)]×(1/2)
=100-20-100×(2/5)
=80-40
=40 tons
That's 40 tons left
In triangle ABC, let vector AB = vector B and vector AC = vector C, then the root sign (vector | B | * vector | C |) ^ 2 - (vector B dot multiplied by vector C) ^ 2 =?
A 0
B 2S triangle ABC
C s triangle ABC
D 1 / 2S triangle ABC
Why?
It should be root sign [(vector | B | * vector | C |) ^ 2 - (vector B dot multiplied by vector C) ^ 2], right?
If so, then
Root sign [(vector | B | * vector | C |) ^ 2 - (vector B dot multiplied by vector C) ^ 2]
=Root sign [(vector | B | * vector | C |) ^ 2 - (| vector B | * | vector C | COSA) ^ 2]
=Vector | B | * vector | C | * Sina
=2S triangle ABC
Choose B
When a pile of 2500 tons of Huahuang sand is transported to the construction site, 12% of the total amount is transported away for the first time and 18% for the second time. How many tons are left?
2500 × (1-12% - 18%) = 2500 × 60% = 1500 tons; a: there are 1500 tons left
The relation of the image of the linear function y = 2x-1 with respect to the image of x-axis symmetry (), with respect to the image of y-axis symmetry (),
The relation between the image of the linear function y = 2x-1 and the image of x-axis symmetry (y = - 2x + 1)
If x is constant and Y becomes the opposite number, then: - y = 2x-1 is y = - 2x + 1
The relation (y = - 2x-1) of the image with Y-axis symmetry,
If y is constant and X becomes the opposite number, then y = 2 (- x) - 1 is y = - 2x-1
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