Xiao Li has measured that her normal average speed is 1.2 meters per second, how many kilometers per hour

Xiao Li has measured that her normal average speed is 1.2 meters per second, how many kilometers per hour

You can see that 1 hour 3600 seconds is 1.2 * 3600 = 4320 meters
4320m = 4.32km
That is 4.32 km / h

Xiao Li thinks that "average speed is the average value of speed". Is Xiao Li right?

No. the average speed refers to the ratio of the displacement of an object in a period of time to the time used. It reflects the average speed of an object in a period of time and corresponds to a period of displacement or a period of time

Xiaoli pushes the wooden box with a horizontal force of 100 N and moves in a straight line at a speed of 2 m / s
If you want to make the wooden box move in a straight line at a speed of 5 meters per second, the human thrust should be cattle
I know that the first empty should be 100N, which can be obtained by the balance of two forces. But what's the answer to the second question?

Then the friction force on the wooden box is 100 n. if the wooden box is to move forward in a straight line at a speed of 5 m / s, the human thrust force should be 100 n
As long as the motion is uniform, thrust = friction. If thrust > friction, it is uniform acceleration

Xiaoli has 15 more stamps than Xiaolan, which is exactly 10:3 of Xiaolan's. how many stamps does Xiaoli have

15 △ 3 / 10 + 15 = 65 sheets

Help me to solve some math problems
The first: given x ^ 2 + y ^ 2-2x-6y + 10 = 0, find the value of x ^ 2006 * y ^ 2
Second: given x ^ 2 + 4x-3 = 0, find the value of 2x ^ 3 + 9x ^ 2-2x-3
The first question has been answered. Is there a meeting in the second question
Given x ^ 2 + 4x-3 = 0, find the value of 2x ^ 3 + 9x ^ 2-2x-3
This question nobody can, really really really really really really really really really really really really really really really really really really really really really really really really really really really really really really really really really really really really really really really really really really really really really really really really sorry, the card has! Really nobody can, hee hee hee~

X ^ 2 + y ^ 2-2x-6y + 10 = (x ^ 2-2x + 1) + (y ^ 2-6y + 9) = (x-1) ^ 2 + (Y-3) ^ 2 = 0. Because (x-1) ^ 2 and (Y-3) ^ 2 are greater than or equal to 0, the sum of them should be 0, so: (x-1) ^ 2 = 0 and (Y-3) ^ 2 = 0, that is, x = 1, y = 3, x ^ 2006 * y ^ 2 = 1 ^ 2006 * 1 / (y ^ 2) = 1 / 9

On the number axis, the point whose distance from point 3 is equal to root 3 represents the number

3 + radical 3 or 3-radical 3

Solution of ternary system of equations, urgent
Formula 1, x + y

If this problem is to find the value range of X, y, Z single variable, it can tell you that this problem has no solution. If it is to find the value range of X + y + Z or X-Y-Z relations with parameters, it can be found. Excel equations are very troublesome, not to mention inequality. If you are a college student, it is recommended to learn matlab

In the equal ratio sequence {an} with common ratio 2, Sn is the sum of the first n terms. If s99 = 56, then A3 + A6 + A9 + +The value of A99 is ()
A. 4B. 8C. 16D. 32

In {an} which has a common ratio of 2, the sequence {a3n} is an equal ratio sequence with a common ratio of 8. Let s = A3 + A6 + A9 + +Then s = A3 (1-833) 1-8 = A1 · 4 · (1-299) - 7, ① ∵ s99 = 56, ∵ A1 (1-299) 1-2 = - A1 (1-299) = 56, ② comparing the two formulas, s56 = 47 and S = 47 × 56 = 32 are obtained

360 square centimeter = () square decimeter 3.4 cubic decimeter = () cubic centimeter

3.6
three thousand and four hundred

An automobile transportation company plans to use 20 vehicles to transport 36 tons of vegetables (a, B and C) to other places for sale (each vehicle is limited to be fully loaded, and can only carry the same kind of vegetables, and each vegetable is not less than one vehicle). How should the transportation make the company obtain the maximum profit? What is the maximum profit?
A, B and C
2.5 tons per vehicle
The profit per ton of vegetables (100 yuan) 57.4 the answer should be detailed, and the plan should be listed, and then the plan with the highest profit should be calculated carefully. I won't choose the answers that are wrong in the process or copied as the best answers, and the ones that are good,

If x tons of type a vegetables are set, X / 2 vehicles will be required for shipment;
The quantity of C vegetables is y tons, which needs to be transported by Y / 1.5 vehicles;
Then the vegetable B is (36-x-y) tons, which needs (36-x-y) / 1 vehicle for shipment
X/2+(36-X-Y)/1+Y/1.5=20
If both sides of the equation are multiplied by 6, we get:
3X+216-6X-6Y+4Y=120
3X+2Y=96
X=(96-2Y)/3
Then: Total Profit = 5x + 7 (36-x-y) + 4Y
=252-2X-3Y
=188-16Y/3
Because each vegetable is not less than 1 truck, then y is an integral multiple of 1.5, and the smallest y = 1.5
When y = 1.5, x = 31, that is, a vegetable can not turn full car, so y = 1.5 is not the problem
When y = 3 and x = 30, the maximum profit is 188-16 * 3 / 3 = 17200 yuan = 17200 yuan
At this point:
A, B and C
Weight: 30 3
Number of vehicles required: 15 3 2