What is the speed of light and sound

What is the speed of light and sound

The speed of light is 3 × 10 8 meters per second, and the speed of sound is 340 meters per second

Let a > 1,0 < B < 1, then the range of loga (b) + logb (a) is?

loga(b)+longb(a)=lg(a)/lg(b)+lg(a)/lg(b)=-[lg(a)/lg(1/b)+lg(1/b)/lg(a)

How to convert degrees, minutes and seconds? For example, how to convert 36 ° 17'42 "into degrees

Because 1 / 60 of 1 ° is 1 point, it is recorded as 1 ', i.e. 1 ° is 60'
1 / 60 of 1 'is 1 second, which is recorded as 1 ″, i.e. 1 ′ = 60 ″
36 ° 17'42 "= 36 * 60 * 60 + 17 * 60 + 42 = (the answer is calculated by yourself) is in seconds
Hope to adopt

All odd sums within 20 are () and even numbers are ()

All odd numbers within 20 are (100) and even numbers are (110)

How much is the square of 2 + the square of 4 + the square of 6 +... + the square of 50?
The formula for the sum of the squares of continuous even numbers!

n*(n+1)(2*n+1)/6
In fact, you propose a square of 2 for each term, then n = 25, and then multiply the result by 4, that is:
25*（25+1）*（2*25+1）*4/6=22100

Simple calculation of 0.24 * 0.25

0.2*0.25+0.04*0.25-=1.5

Split term elimination method
1 / (2n-1) (2n + 1) = 1 / 2 [1 / (2n-1) - 1 / (2n + 1)] (3) 1 / N (n + 1) (n + 2) = 1 / 2 [1 / N (n + 1) - 1 / (n + 1) (n + 2)] how are these equations solved? Or what are the rules
The last sentence of the broken sword is not very clear. Put the molecule forward?
What I don't understand most is, for example, how to get the 1 / 2 after 1 / N (n + 1) (n + 2) = 1 / 2 [1 / N (n + 1) - 1 / (n + 1)]

Split term cancellation
as
An = 1 / N * (n + 1) so an = ((n + 1) - n) / N * (n + 1) = 1 / N - 1 / (n + 1)
An = 1 / N * (n + k) k is a constant
Multiply the numerator and denominator by K, that is, an = K / k * n * (n + k) = (1 / k) * (n + k - n) / (n * (n + k))
=(1/k)*(1/n - 1/(n+k) )
An=1/n*(n+k)(n+2k)
K is a constant
Multiply the numerator and denominator by 2K
That is, an = 2K / 2K * n * (n + k) (n + 2K)
=(1/2k)*(n+2k - n)/n*(n+k)(n+2k)
=(1/2k)*(1/n*(n+k) - 1/(n+k)(n+2k)
In the next four and five, we'll see less
For other split terms
as
When (an + 1 - an) / an + 1 appears, it can also be considered to change it into 1 / an + 1 - 1 / an, and then 1 / an can be regarded as a new sequence
There is also a forced split term
An=n*(2^n)
Let an = BN + 1 - BN, then Sn = a1 + A2 +... + an = (b2-b1) + (b3-b2) + (BN + 1 - BN)
=Bn+1 - Bn
If we observe that there is a 2 ^ n after an, we can be sure that there is also a 2 ^ n after BN
Let BN = (KN + T) 2 ^ n, then BN + 1 = (K (n + 1) + T) 2 ^ (n + 1)
Write 2 ^ (n + 1) as 2 * 2 ^ n and multiply 2
Bn+1 = (2K(n+1)+2T)2^n=（2Kn+2K+2T)2^n
An=Bn+1 - Bn =(2Kn+2K+2T -Kn - T)2^n=(Kn+2K+T)2^n
Compared with an
K = 1, 2K + T = 0, so t = - 2
Bn=(n-2)*2^n
Sn=Bn+1 - B1 =(n-1)2^（n+1）+2

The monotone increasing interval of function y = 2x ^ 2-3 | x | is

When x ≥ 0, y = 2x ^ 2-3x, then x = 3 / 4
When x

The greatest common divisor of two numbers is 6, and the least common multiple is 36. How many pairs of such numbers are there

12,18
6,36

a> 0, b > 0, ab = a + B + 3, find the minimum value of ab

ab-a-b=3
(a-1)(b-1)
=ab-a-b+1
=4
ab=(a-1)+(b-1)+5≥2√［（a-1)(b-1)］+5
Because (A-1) (B-1) = 4
If and only if A-1 = B-1 = 2
There is a minimum of 4 + 5 = 9
reference resources:
If a and B are positive real numbers and satisfy AB = a + B + 3, find the range of ab
, ∵ positive numbers a, B
∴a+b≥2√ab∵ab=a+b+3
∴ab≥2√ab+3
The solution of the inequality about √ AB is √ ab ≥ 3
∴ab≥9
Similarly, ab ≤ (a + b) ^ 2 / 4 can be obtained by using the mean inequality
The inequality of a + B + 3 ≤ (a + b) ^ 2 / 4 solution about (a + b) is a + B ≥ 6, that is, the minimum value of a + B is 6
reference resources:
∵a>0,b>0,∴ab=a+b+3>3.
Let AB = u, then B = u / A, and substitute AB = a + B + 3
u=a+u/a+3=(a²+3a+u)/a
So a & # 178; + (3-U) a + U = 0
Since a is a real number, its discriminant is as follows
△=（3-u）²-4u=u²-10u+9=(u-9)(u-1)≥0
That is, u ≥ 9 or u ≤ 1 (rounding off, because u > 3 is known)
When u = AB = 9, a + B = 6, and a = b = 3
The range of AB is [9, + ∞)
The range of a + B is [6, + ∞)