Given the arc length, chord height, how to find the radius or chord length?

Given the arc length, chord height, how to find the radius or chord length?

Given arc length C and chord height h, how to find radius r?
Use the following formula to find r!
Rn+1=(1-(Rn*COS(C/(2*Rn))-Rn+H)/((C/2)*SIN(C/(2*Rn))-H))*Rn
Please give a specific data! I'll ask you to see!

(x-1) ^ n = A0 + a1x ^ 1 + a2x ^ 2 + a3x ^ 3 +... + anx ^ n find A0 + an
(x-1) ^ n = A0 + a1x ^ 1 + a2x ^ 2 + a3x ^ 3 +... + anx ^ n find A0 + an

an = 1,a0 = (-1)^n
When n is odd, A0 + an = - 1 + 1 = 0
When n is even, A0 + an = 1 + 1 = 2

Given (1 + 2x) ^ n = A0 + a1x + a2x ^ 2 +... + anx ^ n, find a1 + 2A2 + 3a3 +... + Nan and give the expression

(1+2x)^n=a0+a1x+a2x^2+...+anx^n
If both sides take derivatives of X at the same time, we get the following result
2n(1+2x)^[n-1]=a1+2a2x+...+nanx^[n-1]
Let x = 1
A1 + 2A2 + 3a3 +... + Nan = 2n (1 + 2) n-1 power = 2n × 3 n-1 power

The function f (x) is defined by the right table: if A0 = 5, a (n + 1) = f (an), n = 0,1,2 Then a2013=
Table: (omit table line)
x 2 5 3 4
f(x) 2 3 4 5

a0=5,a1=f(a0)=f(5)=3
a2=f(a1)=f(3)=4
a3=f(a2)=f(4)=5
a4=f(a3)=f(5)=3
...
Therefore, the above results show that every three are a cycle, so there is 2013 / 3 = 671, and there is no remainder, so a2013 = A3 = 5

The function f (x) is defined by a small table: if A1 = 1, A2 = 5, a (n + 2) = f (an), a2013 =? X 1 2 3 4 5, f (x) 3 4 5 2 1

x 1 2 3 4 5
f(x) 3 4 5 2 1,
a3=f(a1)=f(1)=3,
a4=f(a2)=f(5)=1,
a5=f(a3)=f(3)=5,
a6=f(a4)=f(1)=3,
3 is the period of an,
a2013=a3=3.

The function f (x) is defined by the following table x 2 5 3 1 4 f (x) 1 2 3 4 5
A1 = 5, a (n + 1) = f (an), n = 1,2,3, l, then the value of A2009 is ()
A 1 B 2 C 4 D 5

A1 = 5, A2 = f (A1) = 2, A3 = f (A2) = 1, A4 = f (A3) = 4, A5 = f (A4) = 5, so it is a cycle with a period of 5
a2009=a4=4

The function f (x) is defined by a small table: if A1 = 1, A2 = 5, a (n + 2) = f (an), A2010 =?
x 1 2 3 4 5
f(x) 3 4 5 2 1

Since A1 = 1, let n = - 1, then a (- 1 + 2) = f (- a) = A1 = 1, so f (- a) = 1, let x = - A, then f (x) = 1, then f (5) = 1
So if - a = 5, then a = - 5, then A2010 = - 5 × 2010 = - 10050

It is known that the graph of the function F & nbsp; (x) defined on R is continuous and has the following corresponding value table: X 123 F & nbsp; (x) 6.12.9 - 3.5, then the interval where the function f (x) must have zero is______ ．

∵ the image of function F & nbsp; (x) defined on R is continuous, and f (2) = 2.9 ＞ 0, f (3) = - 3.5 ＜ 0, that is, f (2) · f (3) ＜ 0. According to the existence theorem of function zeros, the interval of function f (x) must have zeros is (2,3), so the answer is: (2,3)

If m, n ∈ {x | x = A2 + a1x10 ^ 2 + a1x10 + A0}, where a1 ∈ {1,2,3,4,5,6,3q}
If m, n ∈ {x | x = A2 + a1x10 ^ 2 + a1x10 + A0}, where a1 ∈ {1,2,3,4,5,6,7} (I = 0,1,2), and M + n = 636, find the real number pair (m, n) representing the number of different points on the plane

Remember a = {x | x = A2 * 10 ^ 2 + A1 * 10 + A0}, that is, to find the number of solutions of M + n = 636 in a, we can investigate it according to the decimal bit. First, look at the individual bit, A0 (m) + A0 (n) = 6, there are 5 possibilities. Then look ahead: A1 (m) + A1 (n) = 3 and A2 (m) + A2 (n) = 6, there are 2 * 5 = 10 possibilities, A1 (m) + A1 (n) = 13 and A2 (m) + A2 (n) = 5, there are 2 * 4 = 8 possibilities, so there are (10 + 8) * 5 = 90 solutions, corresponding to 90 different points on the plane

Let f (x, y) = (1 + m / y) ^ x (M > 0, Y > 0) if f (4, y) = A0 + A1 / y + A2 / y ^ 2 + A3 / y ^ 3 + A4 / y ^ 4 and A3 = 32, find ∑ AI
Let f (x, y) = (1 + m / y) ^ x (M > 0, Y > 0) if f (4, y) = A0 + A1 / y + A2 / y ^ 2 + A3 / y ^ 3 + A4 / y ^ 4 and A3 = 32, find ∑ (4 above, I = 0 below) AI

f(4,y)=(1+m\y)^4
Because A3 = 32
So C34 * m ^ 3 = 32 ----- M = 2
The formula = (1 + 2) ^ 4 = 81