Virtual module length, how to find ah. Use mathematical expressions bar
For example, in the example you gave, you must first multiply the denominator of the conjugate (1-I) into reality. Then you get the length of the circuit chip (square root of 2) / 2 with * (1 - Ⅰ) / (1 +) (1 - Ⅰ) = 1 / 1 + 1 = 1 / 2 + / 2
Finding the value of an expression
five
How much is this?
r=0
It seems that the answer is 112 ~ I don't know what the situation is 0 - 0
Original formula = 0x1 + 1x2 + 2x3 + 3x4 + 4x5 + 5x6 = 2 + 6 + 12 + 20 + 30 = 70
If r = 6, the result is 112
Find a mathematical expression
A variable a represents 1,2,3, equal number, a constant 46, how to get the expression of 1,46,92138184, equal number
n=1,a1=1 n≥2,an=46(n-1)
The number of intersections between the square of y = x - ax + A-2 and the coordinate axis is ()
Please tell me why, thank you!
There is no sign on the keyboard, so use triangle instead of triangle = b2-4 * a * C = A2-4 * 1 * (A-2) = A2-4 * a + 8. In the formula A2-4 * a + 8, use the Vader's theorem again to get triangle = - 160, so A2-4 * a + 8 is always greater than 0, that is, the number of intersections between y = x2-ax + A-2 and the coordinate axis is 2
S=2πr²+2πrh
V=2πrh
These two are s, R, h are unknowns. Which of them is a quadratic function······
The second V is also unknown
First
Because the quadratic function refers to the polynomial function of which the highest degree of the unknown is quadratic. The quadratic function can be expressed as f (x) = ax ^ 2 + BX + C (a is not 0)
How to write the mathematical expression of n-th power with root sign in C language
#Include # include void main() {int i, J; double x; scanf (% d ", & J); / / enter the number of square root scanf (% F", & x); / / enter the number of square root for (I = 0; I
On the sum formula of the first n terms of equal difference
The sum formula of the first n terms of the arithmetic sequence {a (n)} is
S(n)={n[a(1)+a(n)]}/2=na(1)+[n(n-1)]d/2=na(n)-[n(n-1)]d/2
All in parentheses are subscripts
Here I would like to ask the second and third formula under what circumstances? Why do I use the first formula for the question? The answer is the same as the standard answer, the second and third formula (that is, s (n) = Na (1) + [n (n-1)] d / 2 = Na (n) - [n (n-1)] d / 2) is wrong, and the answers calculated by the second and third formulas are the same
Take the following question for example
If {a (n)} is an arithmetic sequence and a (2) = 4 / 5, a (5) = 8 / 5, then s (6)=___________
Using the first formula s (n) = {n [a (1) + a (n)]} / 2, the result is 36 / 5, which is the same as the standard answer
But using the second and third formula, the answer is 108 / 5
What's the matter, please?
It's all the same. It's all 36 / 5
It's your own miscalculation!
Using the second formula:
S(6)=6*(8/15)+6*5*(4/15)/2
=16/5+4
=36/5
Using the third formula:
S(6)=6*(28/15)-6*5*(4/15)/2
=56/5-4
=36/5
Derivation of the product formula of the first n terms of arithmetic sequence
Derivation of the sum of the first n terms of the arithmetic sequence:
Sn = a1 + A2 +. An-1 + an
Sn=an+an-1+.a2+a1
By adding the two formulas, 2Sn = (a1 + an) + (A2 + an-1) + (an + A1)
=n(a1+an)
So Sn = [n (a1 + an)] / 2
If it is known that the first term of the arithmetic sequence is A1, the tolerance is D, and the number of terms is n, then an = a1 + (n-1) d is substituted into formula (1)
Sn=na1+ [n(n+1)d]/2(II)
No,
Product formula of the first n terms of arithmetic sequence
Given A1 = 1, a (n) - 3A (n) &; a (n-1) = 0, find the general formula of a (n)
If a (2) = 3A (2) a (1), then a (2) = 3A (2), then a (2) = 0
1 n=1
a(n)={
0 n>1
Given A1 = 1, a (n) - 3A (n)? A (n-1) = 0, find the general formula of a (n)
Given a (1) = 1, a (n) - 3A (n) &; a (n-1) - A (n-1) = 0, the general term formula of a (n) is obtained
It's best to write the process
Because a (n) - 3A (n) &; a (n-1) - A (n-1) = 0, so a (n-1) = a (n) / (1 + 3A (n)), so 1 / a (n-1) = 1 / a (n) + 3, so 1 / a (n-1) - 1 / a (n) = 3, so 1 / a (n) is an arithmetic sequence with 1 / A2 as the first term - 3 as the tolerance