What is the circumference and diameter of the earth?

What is the circumference and diameter of the earth?

Unit: km
The equatorial radius is 6378.137km
The polar radius is 6356.752km
The average radius is 6371.012km
Flatness 1 / 298.257
The equatorial perimeter is 40075.7km
The circumference of meridian is 40008.08km
The surface area is 5.101 × 108km2
Volume 10832 × 108km3

Given that the solution set of the inequality kx-2 ＞ 0 (K ≠ 0) about X is x ＜ - 3, then what is the coordinate of the intersection of the line y = - KX + 2 and the X axis? Why?

Kx-2 ＞ 0 means: KX ＞ 2 and the solution set is x ＜ - 3, so K ＜ 0, so x ＜ K / 2, so k = - 2 / 3
When y = 0, x = - 3
So the coordinates are (- 3,0)

See, the distance from the intersection a of KX + B and X axis to the origin o is 7. Oh, the solution of the equation KX + B = 0 is

The answer is: ± 7

The distance from the intersection of the line y = KX + B and the Y axis to the origin is B
Is that right~

No
The distance is a positive number when B

In Cartesian coordinate system, if the line y = KX (k is not equal to 0) passes through point a and the distance between point a and two coordinate axes is equal, then K=

A is (a, - a) or a (a, a)
y=kx
-a=ak
k=-1
a=ak
k=1
k=±1

It is known that a straight line y = - x + 2 intersects the x-axis and y-axis at a and B. another straight line y = KX + B (k is not equal to 0) passes through the point (1,0) and divides the triangle AOB into two parts
It is known that: a straight line y = - x + 2 intersects with the x-axis and y-axis at a and B, another straight line y = KX + B (k is not equal to 0) passes through the point (1,0) and divides the triangle AOB into two parts. If the area of triangles is equal, calculate the values of K and B. if the area of triangles Ao is 1:5, calculate the values of K and B
Hope as soon as possible, online and so on

k=9
b=6

⊥ 65 [1 / 2] straight line y = - x + 2 intersects with X axis and Y axis at a and B respectively, another straight line y = KX + B (k is not equal to 0) passes through C (1,0), and divides the triangle AOB into faces
⊥ 65 [1 / 2] straight line y = - x + 2 intersects with X axis and Y axis at a and B respectively. Another straight line y = KX + B (k is not equal to 0) passes through C (1,0), and divides the triangle AOB into two parts with an area of 1:5

A（2,0）,B（0,2）,S△AOB=2*2（1/2）=2
Straight line: y = KX + B, via (1,0), bring in: 0 = K + B, k = - B
Then y = kx-k
There are two cases when a line intersects the y-axis: k = 2 / 3 and K = 10 / 3 respectively
When a line intersects a line y = x + 2, there are two cases: k = 2 and K = (- 10 / 7) (rounding off)
B is the opposite of K

If △ AOB is an isosceles triangle, what is k equal to?

According to the meaning of the title, the coordinate of point a is a (- 2 / K, 0), and the coordinate of point B is B (0,2);
AOB is a right triangle
There is only one condition to be a high triangle: | OA | = | ob|
∴ |2/k| = 2 ==> k = ±1;
When k = 1 or K = - 1, AOB is an isosceles triangle

If the solutions of the equation Z-2 / x = y and X + 3y-2 / 5Z = 1 / 2 also satisfy the equation X-Y + 5 / z = 2, find the values of X, y and Z

A:
The solutions of the equation Z-2 / x = y and X + 3y-2 / 5Z = 1 / 2 also satisfy the equation X-Y + 5 / z = 2
The equations are as follows
z-x/2=y,x/2+y-z=0………… （1）
x+3y-5z/2=1/2……………… （2）
x-y+z/5=2……………………… （3）
(2) - (1) * 2
3y-2y-5z/2+2z=1/2,y-z/2=1/2……………… （4）
(3) 2
-y-3y+z/5+5/2z=3/2,-4y+27z/10=3/2………… (5)
(4) * 4 + (5)
-2z+27z/10=2+3/2
7z/10=7/2
z=5
By substituting (4), Y-5 / 2 = 1 / 2, y = 3
Substituting (3) we get: x-3 + 1 = 2, x = 4
So the solution of the equations is as follows:
x=4
y=3
z=5

Given the plane π: 2x + y-3z + 2 = 0, then the equation of the line passing through the origin and perpendicular to π is?

S=(2,1,-3)
x/2=y/1=z/(-3)