For help. For help. A math problem in grade one as everyone knows: When two lines intersect, it has at most one intersection; when three lines intersect, it has at most three intersections; when four lines intersect, it has at most six intersections; when five lines intersect, it has at most ten intersections; when six lines intersect, it has at most fifteen intersections Q: when n lines intersect, how many intersections are there at most

For help. For help. A math problem in grade one as everyone knows: When two lines intersect, it has at most one intersection; when three lines intersect, it has at most three intersections; when four lines intersect, it has at most six intersections; when five lines intersect, it has at most ten intersections; when six lines intersect, it has at most fifteen intersections Q: when n lines intersect, how many intersections are there at most

There are 2 points, 3 points, 4 points, 6 points, 5 points and 10 points. You can find that if you add a new point with the most points, you can generate the most focus only when you intersect with each line! If there are three lines, every two lines intersect, there are at most (3) intersections (because every two lines intersect), and four lines, every two lines intersect, there are at most (6) intersections (because three lines intersect, there are three intersections) At most, the fourth one intersects with all the three and produces three intersections By analogy, suppose that if there are n straight lines, every two intersect, and there are at most (n * (n-1) / 2) intersections;

The main difference between snakes and flies and frogs, crocodiles and cranes is that they don't have a back________ .
9.(4)

Spine or vertebrae
I'm the first in science,
The answer is in the book

Simplify sin (K π + a) + sin (A-K π) divided by sin (a + K π) cos (A-K π). (kEz)

When k is even, sin (K π + a) = Sina, sin (A-K π) = Sina,
cos(a-Kπ）=cosa
Sin (K π + a) + sin (A-K π) divided by sin (a + K π) cos (A-K π)
=2sina/sinacosa=2/cosa;
When k is odd, sin (K π + a) = - Sina, sin (A-K π) = - Sina,
cos(a-Kπ）=-cosa
Sin (K π + a) + sin (A-K π) divided by sin (a + K π) cos (A-K π)
=-2sina/sinacosa=-2/cosa;

The area of Xiaowen's room is 10.8m2. The floor of the room happens to be paved with 120 identical square tiles. What is the side length of each tile?

Suppose the side length of each tile is x, then 120X2 = 10.8, and the solution is x = 0.3, that is, the side length of each tile is 0.3m

Find the ratio of 15 cm: 5 / 3 m and 3 / 10: 39 / 4

15 cm: 5 / 3 M
=15: 500 / 3
=45:500
=9:100
=0.09
The ratio of 3 / 10 to 39 / 4
=6:39×5
=2:65
=2 out of 65

If the sequence {an} is an arithmetic sequence and the first term A1 > 0, A2009 + A2010 > 0, the maximum natural number n of A2009 * a20100 is

a1>0,
a2009+a2010>0,
a2009·a20100,a2010|a2010|
a2009+a2011=2*a20100
a2009+a2011=a2008+a2012=-----=a1+a40190
S4019=(a1+a4019)*4019/2

Xiao Li's living room is 5.2 meters long and 4.4 meters wide. Choose a kind of square floor tile to spread. How many centimeters is the maximum side length of this kind of square floor tile?
How much is it altogether?

5.2m = 520cm = 40 × 13
4.4m = 440cm = 40 × 11
So the greatest common factor is 40
So the longest is 40 cm
Total 13 × 11 = 143

29 / 24 - (5 / 24-4 / 9) recursive calculation

29 out of 24 - (5 out of 24-4 out of 9)
=29 / 24-5 / 24 + 4 / 9
=1 + 4 / 9
=1 and 4 / 9

Cos (B / 2) = (2 * radical 5) / 5, find the area s of triangle,
KUAI

cosB=2cos²(B/2)-1=3/5
sin²B+cos²B=1
The inner angle of triangle is SINB > 0
So SINB = 4 / 5
sinA=sin[180-(B+C)]=sin(B+C)
=sinBcosC+cosBsinC
=7√2/10
a/sinA=b/sinB
b=asinB/sinA=16/(7√2)
So s = 1 / 2absin C = 8 / 7

As shown in the figure, the side length of an isosceles right triangle is 3. If you make a square with the largest area inside the triangle, the area of the square is________

9/4
A triangle is an isosceles right triangle
The area of a square is half that of an isosceles right triangle
Area of isosceles right triangle = 3 * 3 / 2 = 9 / 2
So the square with the largest area is 1 / 2 * 9.2 = 9 / 4