Sincerely solve the math problem Thank you, math problem find the correct solution Thank you

The speed of car A is 72m h, and the speed of car B is 5 6 of car A, that is, 72 * 5 6 = 60m h, since it is relative driving, the time required for the two cars to meet should be h = 330000 (72 + 60) = 2500h, that is, after 2500 hours of meeting.

Also, are you sure that the speed of car A is 72 meters per hour and not 72 kilometers per hour? If it's 72 kilometers an hour, the answer is hours.

If you say that the driving speed is 72 meters per hour, then the answer is as follows:

Solution: The speed of car B is 72*5 6=60 meters per hour.

330,000 (72+60)=2,500 hours.

But I always think that there is a problem with your question, if you should drive at a speed of 72 km/h, then the speed of car B is 60km/h.

330 (72+60) = hours.

72x5 6 = 60 km (B speed).

Set x hours for two cars to meet.

72x+60x=330

x = A: Hour when two cars meet.

Solution: Let the two cars meet in x hours, and list the equation according to the problem

x*(72+72*5/6)=330

Solution: x = Answer: Two cars meet in hours.

The speed of car B is 72 * 5 6 = 60 meters per hour.

The encounter time is 330 (72+60) = hours.

Solution: Derived from the question:

The velocity of B is 5 6 72 = 60 (meter hours).

330km=330000m

2500 (hours).

Therefore, at 2500 hours the two cars met.

The above is the answer to this question, I want to ask, usually people's questions don't come out like this, right? Could it be that the landlord made a mistake?

First of all, the conversion unit: 330 kilometers = 330,000 meters.

B car speed: 72 6 * 5 = 60

330,000 (72+60) = hours

Solution: Set the two cars to meet after x hours.

72x+5/6*72x=330

10 million * 1000 = 10 10 (total sum insured) 10 10 * 50 10000 = 50000000 (income premium) 500000000 * 70% = 350000000 (payment of accompaniment) 10 million * 15 10000 = 15000 (risk score) 15000 * 1000 = 15000000 (total payout) 350000000-15000000 =20000000 (remaining funds) 20000000 * (1-33%) = 13400000 (profit.

Simultaneous equation, substituting y=(k+1)-3x into equation 2

Simplified: x=3 8*k

So the value of x can range from 3 to 4

2x2+ax-y+6-2bx2+3x-5y-1=(2-2b)x2+(a+3)x-6+5

If it is not related to x, the coefficient is 0

So 2-2b = 0, a+3 = 0

a=-3,b=1

So the original formula = 1 12a3 + b2 = -9 4 + 1 = -5 4

Solution: Original formula = (2-2b) x + (a + 3) x - 6y+5

The value of the formula is independent of a, b, a= -3, b=1

1/3a³-2b²-1/4a³+3b²=1/3x(-3)³-2x1-1/4x(-3)³+3x1=-5/4

The value has nothing to do with the value of the letter, then it is transformed into that no matter what value a and b take, the algebraic value is unchanged, and a, b are regarded as independent variables, x and y are regarded as constants, and then the similar terms are combined, and the coefficients in front of a and b are 0, which are solved.

Find an identical one. Hope it helps. Don't know how to ask me.

It's just that the letters are different. It's the same when it comes to it.

Extend CP, cross AB and M. When pm=pq, pc+pq=cm is the shortest. Whereas, when cm is vertical AB, cm is the shortest. Both ab sides on the high. Calculate cm = by area