How much is Tan 60? How much is Tan 45

How much is Tan 60? How much is Tan 45

Root three, one
Tan 60 ° is much better?
Tan60? = root 3? Why,
=Sin60 / cos60 = 2 / 2 root divided by 1 / 2 root 3
Draw a right triangle so that an angle equals 60 ° and you'll see
One eighteenth is equal to a fraction plus a fraction plus a fraction plus a fraction plus a fraction plus a fraction plus a fraction
1 1 1 1 1 1
——=——+——+——+——+——
18 324 162 108 54 54
On the number of subsets of a set
How to deduce that the number of subsets of the set a = {A1, A2, A3. An} is the nth power of a?
Sorry, I have the wrong number. It should be the nth power of 2
First of all, correct what you said, which is the nth power of 2. Permutation and combination can be learned as a sophomore in high school. You can look ahead. Now there is no need to study deeply, and it will be understood naturally in the future. First, classify all subsets of a: 1. Contain 0 elements (empty set) C (0, n) = 12
2 ^ n
In each subset, AI can appear or not, so it's 2 * 2 * 2... = 2 ^ n
(1/?)+(1/?)+(1/?)=11/12
A fraction plus a fraction plus a fraction is twelve out of eleven
Let a fraction be x, then x + X + x = 12 / 11, 3x = 12 / 11, x = 4 / 11
Set {0,1,2,3,4,5 What is the number of subsets of N}?
The result is to the (n + 1) power of 2
Combination principle:
An empty set is also a subset, that is, one element has no cN0
A set contains one element, all of which have cN1
The set containing two elements is cn2
Assemble
If a set has n elements, it has CNN
The N + 1 power of cN0 + cN1 +. + CNN = 2
There should be proof in the textbook
For example, the set {0,1} has four subsets, which are empty sets, {1}, {2}, {1,2}, with a total of four exactly 2-th power
For example, the subsets of the set {0,1,2} are empty sets, {0}, {1}, {2}, {0,1}, {0,2}, {1,2}, {0,1,2} respectively. There are eight kinds of subsets, that is, the cubic power of 2
From this, we can get our conjecture from special cases, and finally we have to prove our conjecture
This is a good mathematical method
Cn1+Cn2+Cn3…… Cnn
(2 ^ n) + 1
If there are n elements in a set, the number of subsets is 2 ^ n, and the number of proper subsets is 2 ^ n-1
There are 2 ^ (n + 1) in this problem
Specific derivation should use binomial theorem and permutation and combination knowledge. If you have learned it, you can tell me that I can deduce it.
If there are n elements
If there is one element in the subset, there are cN1 elements in total (equivalent to asking how many ways to take one out of n elements)
Similarly, there are two elements, a total of cn2
The number of nonempty subsets of a set: cN1 + cn2 +... Expansion
If there are n elements in a set, the number of subsets is 2 ^ n, and the number of proper subsets is 2 ^ n-1
There are 2 ^ (n + 1) in this problem
Specific derivation should use binomial theorem and permutation and combination knowledge. If you have learned it, you can tell me that I can deduce it.
If there are n elements
If there is one element in the subset, there are cN1 elements in total (equivalent to asking how many ways to take one out of n elements)
Similarly, there are two elements, a total of cn2
The number of nonempty subsets of a set: cN1 + cn2 +... + CNN
Plus an empty set:
1 + cN1 + cn2 +... + CNN = cN0 + cN1 + cn2 +... + CNN = (1 + 1) ^ n = 2 ^ n
Subset 2 ^ n (n power of 2)
2 ^ n-1 nonempty subsets
2 ^ n-2 nonempty proper subsets
2 N + 1
One element is n + 1
The two elements are n + (n-1) + (n-2) +... +3+2+1
……
Do mathematical induction
071400225
The third answer is correct and detailed,
If there are n elements in a set, the number of subsets is 2 ^ n, and the number of proper subsets is 2 ^ n-1
There are 2 ^ (n + 1) in this problem
Empty set: 1, C (n + 1,0)
One element: C (n + 1,1)
Two elements: C (n + 1,2)
....
N + 1 elements: C (n + 1, N + 1)
Total: C (n + 1,0) + C (n + 1,1) + C (n + 1,2) +... + C (n + 1, N + 1)
How to sum: (1 + 1) ^ (n + 1) = 2 ^ (n + 1)
Binomial theorem, because the power of 1 is 1, the sum of coefficients is: (1 + 1) ^... Expansion
Empty set: 1, C (n + 1,0)
One element: C (n + 1,1)
Two elements: C (n + 1,2)
....
N + 1 elements: C (n + 1, N + 1)
Total: C (n + 1,0) + C (n + 1,1) + C (n + 1,2) +... + C (n + 1, N + 1)
How to sum: (1 + 1) ^ (n + 1) = 2 ^ (n + 1)
Binomial theorem, because the power of 1 is 1, the sum of coefficients is: (1 + 1) ^ (n + 1) = C (n + 1,0) + C (n + 1,1) + C (n + 1,2) +... + C (n + 1, N + 1)
So there are 2 ^ (n + 1) subsets, including empty sets
For each element, there are two cases: belonging to this subset and not belonging to this subset
So there are two different subsets
Contains 2 ^ n empty sets
1 / 12 = 1 / () + 1 / () + 1 / () + 1 / (), 1 / 12 is equal to a fraction plus a fraction plus a fraction plus a fraction
If it's a different number
The first 1 / 12 = 10 / 120 = 1 / 120 + 4 / 120 + 5 / 120 = 1 / 120 + 1 / 30 + 1 / 24
The second one is 1 / 12 = 10 / 120 = 1 / 120 + 3 / 120 + 6 / 120 = 1 / 120 + 1 / 40 + 1 / 20
If the numbers can be the same, it's simple: 1 / 12 = 3 / 36 = 1 / 36 + 1 / 36 + 1 / 36
Why is the number of subsets of a set with n elements 2 ^ n?
Because the elements contained in the subset are selected from the original set,
Each element in the original set can be selected or not; any subset of the set containing n elements can be regarded as the final result of selecting each element separately, and N choices are made;
Therefore, the number of its subsets is N 2 times, that is 2 ^ n
[for example: how many ways can n different balls be taken out at one time
The protagonist is traditional Chinese medicine, practicing medicine in modern cities to save people and so on ~ the best protagonist will point martial arts ~ single pick more than a dozen people~
How much is five sixties
Five sixties is one twelfth
If you agree with my answer, please click the "select as satisfied" button below,
Junlang lieying team answers for you
Five out of sixty equals a fraction = 1 / 12 (1 / 12)
5/60=1/12
1/12
Divide the numerator and denominator by 5
5/60=1/12
Name the number of subsets of a set
p=｛x：2x＜9,x∈N｝
That colon is a vertical
There are five elements in P, 0,1,2,3,4,
So it's the five permutations with one to five permutations, as well as the sky set
LZ estimates that senior one has not learned permutation and combination, I use the popular method to say
There are five ways to choose one
There are five ways to choose four
2 out of 10
Take 3 out of 10
Take five, there is one way
Empty set, a way to get
The total is 10 + 10 + 5 + 5 + 1 + 1 = 32 species = 2 ^ 5
2 to the fifth power.. Because the number of elements of P is five