How to calculate the area of a circle

How to calculate the area of a circle

Area s = π R ^ 2
R is the radius of the circle

Examples of static friction and rolling friction
Let's be clear with examples

Static friction
Friction between the soles of feet and the ground when people walk; friction between the rope and the pile when they pull in; friction between objects resting on the slope and the slope; friction between hands and pens when writing; friction between hands and bottles when holding bottles; friction between ropes when binding objects with ropes
On rolling friction -- why is it difficult to ride a bicycle when it's flat
Everyone may have experienced this. When the bicycle is flat, it's very difficult to ride. What's the reason?
In order to understand why it is difficult to ride a bicycle when there is no air, we must first understand the working principle of the bicycle. When people pedal, they make the chain disk rotate. The flywheel of the rear wheel is driven by the chain to rotate, and the rear wheel also rotates. At this time, the tire has a backward static friction force against the ground, The ground will also have a forward static friction force on the rear wheels. This force is the driving force of the car. As the wheels roll on the ground, there must be rolling friction

What are the similarities and differences of static friction, sliding friction and rolling friction? Can you give some common examples of these friction in life?

Static friction refers to the relative static friction, such as pulling the box, the box does not move. Static friction occurs between the box and the ground. Rolling friction is similar to the friction between the wheel rolling and the ground. Generally speaking, rolling friction is smaller than static friction and sliding friction

What is the limit of IIM (x approaches 0) (1 / X-1 / (e ^ x-1)) and how to find it?

When x approaches zero
1 / x approaches infinity
E ^ x tends to E
So the limit of 1 / X-1 / (e ^ x-1) is + ∞

How to calculate 7.4 × 2.5 + 2.6 △ 0.4

（7.4＋2.6）x2.5÷0.4＝62.5

Decomposition factor: 5m (a-b) + 20n (B-A)

The original formula is 5m (a-b) - 20n (a-b) = 5 (a-b) (m-4n)

26 * 52.7 + 35.2 * 5.27 + 0.222 * 527

Original formula = 42.6 * 5.27 + 35.2 * 5.27 + 22.2 * 5.27 = (42.6 + 35.2 + 22.2) * 5.27 = 527

Given that the circle C is tangent to both axes, the distance between the center of circle C and the line y = - x is equal to the root 2, the equation of circle C is obtained

Let the coordinates of the center of a circle be (m, n)
Because the circle is tangent to the two coordinate axis, then | m | = | n |, radius r = | M|
(1) When m = n, the distance to x + y = 0 is the root 2
That is, | m + m | / radical 2 = radical 2, M = (+ / -) 1
The equation is (x-1) ^ 2 + (Y-1) ^ 2 = 1 or (x + 1) ^ 2 + (y + 1) ^ 2 = 1
(2) When m = - N:
|M-M | / radical 2 = radical 2, no solution

Solve equation 12 divided by 2.5x = 1.2

12/2.5x=1.2
2.5x=12/1.2
2.5x=10
x=10/2.5
x=4

Set a = {(x, y) | y = x ^ 2 + 4x + 6}, B = {(x, y) | y = 2x + a} ask (1) why is a valued set a intersecting B with two elements? (2) why is a valued set a intersecting B at most
Set a = {(x, y) | y = x ^ 2 + 4x + 6}, B = {(x, y) | y = 2x + a} ask (1) why does a value set a intersect B with two elements? (2) Why does set a and B have at most one element when they are valued?

Solution (1): the intersection of set a and B has two elements
That is: y = x ^ 2 + 4x + 6 and y = 2x + a have two different intersections
That is: x ^ 2 + 4x + 6 = 2x + A has two solutions
x^2+2x+6-a=0
>If 0:2 ^ 2-4 * (6-A) > 0:
a>6
(2) The intersection of a and B has at most one element
That is: B ^ 2-4ac