Let m, M + 1 and M + 2 be the three sides of an obtuse triangle, then the value range of real number m is () A. 0<m<3 B. 1<m<3 C. 3<m<4 D. 4<m<6

Let m, M + 1 and M + 2 be the three sides of an obtuse triangle, then the value range of real number m is () A. 0<m<3 B. 1<m<3 C. 3<m<4 D. 4<m<6

The obtuse angle of M + 2 pair is α,
Then cos α = M2 + (M + 1) 2 − (M + 2) 2 can be obtained from cosine theorem
2m(m+1)=m−3
M < 0, 0 < m < 3
Then according to the sum of any two sides is greater than the third side, M + m + 1 > m + 2, ν m > 1
To sum up, 1 < m < 3 can be obtained,
Therefore, B

What is the probability that an obtuse triangle can be formed by taking any three edges from five segments of length 1,2,3,4,5? Hope to analyze the specific points. I don't know what it takes to form an obtuse angle 3. If you can flip, there should be four obtuse triangles.

Only (2,3,4), (2,4,5), (3,4,5) can form triangles, of which 2 ^ 2 + 3 ^ 2 < 4 ^ 2 indicates that (2,3,4) is an obtuse angle triangle, and 2 ^ 2 + 4 ^ 2 < 5 ^ 2 indicates that (2,4,5) is an obtuse angle triangle, and (3,4,5) is a right triangle, needless to say, a right triangle,
Therefore, from the five line segments with length of 1,2,3,4,5, the probability that any three edges can form an obtuse triangle is
2/C5、3=2/[(5*4)/(2*1)]=2/10=1/5

Given that the three sides of an obtuse triangle are 2,3,4, find the area of the triangle Come on, Pythagorean theorem

According to the problem, the angle of the edge with side length of 4 is obtuse angle. If the height of the edge is calculated, then the side with side length of 4 is divided into two sections. Let these two segments be x and 4-x respectively. Because they are high lines, two right triangle angles are formed. By using Pythagorean theorem, 2 ^ 2-x ^ 2 = 3 ^ 2 - (4-x) ^ 2 is obtained. After solving the value of X, the length of high line is calculated by Pythagorean theorem, Then multiply the length of the bottom edge by the height and divide it by 2 to get the area?

A more comprehensive formula of permutation and combination and probability in high school mathematics

Permutation (in order): man = m * (m-1) *. * (m-n + 1)
Combination (no order): MCN = m * (m-1) *. * (m-n + 1) / (1 * 2 *... * n)

Probability permutation and combination formula I just learn probability, there is a knowledge point more fuzzy Permutation, such as (1,2), (3,4), (2,1), (4,3) P = 4, is that the algorithm of classical probability? The combination becomes (1,2), (3,4), that is to say, the combination formula is not arranged in order like the arrangement?

Man:
You're right. There's no order in the combination. A and B are the same as B and a
But the arrangement has order, a, B and B, a are not the same
I wish you excellent academic performance,

Permutation and combination formula of high school mathematics Now encounter problems, need high school mathematics permutation and combination formula, do not remember the pull, who knows, send down! Thank you!

PN ^ m = [n / (n-m)] P (n-1) ^ m (n, m belongs to N, and M is not big n) PN ^ m = n! / (n-m)! (n, m belongs to N, and M is not greater than N; when m = n, 0! = 1) this is its formula

1、 On the arrangement and combination of comprehensive problems It is known that 4 out of 10 different products are defective, so they are tested one by one until all four defective products are found. If all four defective products are found after the fifth test, what are the number of different testing methods? The more sufficient, the better 2、 On binomial theorem The problem of finding the maximum (small) value of expansion coefficient can be transformed into finding the maximum value of absolute value For example, to find the maximum term of the coefficient of the expansion formula (a + BX) ^ n, let the expansion be a ₀ + a ₁ x Ψ + a Ψ x? 2 + a ₃ x? + + An x^n The coefficients of the expansion are a ₀, a ₁, a Ψ, a ₃ Let K + 1 have the largest coefficient, because the corresponding coefficient of K + 1 is AK A k≥A k-1 A K ≥ a K + 1 [K, K + 1 are subscripts] In this way, K (corresponding to the term K + 1) is the maximum term of coefficient This is the standard solution in the book. So far, I have a question - is the maximum obtained in this way? I think it's just the maximum value, which is larger than the previous term and the latter term. Isn't it the maximum value? Why is it the maximum value? Is it certain that the function image of the absolute value of the coefficient of the expansion formula has only one peak? How to prove it?

If there are several maximum or minimum points, compare the values of these points and the values with the endpoint, and then get the maximum or minimum value

In the senior high school mathematics permutation and combination formula, there are 12 numbers in each group. How many groups are there? (combination)

C3 / 12 * C3 / 9 * C3 / 6 * C3 / 3 divided by A4 / 4C3 / 12 represents the combination of three out of 12. = 12 * 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 / (3 * 2 * 1 * 3 * 2 * 1) / (4 * 3 * 2 * 1) = 15400. Why divide by A4 / 4 is because you have four groups of numbers. Here, taking 1,2,3 first and taking 4,5,6 are repeated calculations

Detailed explanation of permutation and combination formula

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Please explain the formula of permutation and combination in terms of what you can understand in primary school~ RT, please attach some examples,

The formula P refers to arrangement, which is to arrange r elements from n elements
Formula C refers to combination. R elements are taken from n elements without arrangement (i.e. no sorting)
From 1, 2, 3 Any three of these twenty numbers form an arithmetic sequence________ One
Analysis: first of all, the complex life background or other mathematical background should be transformed into a clear permutation and combination problem
Let a, B, C be equal difference,  2B = a + C, we can see that B is determined by a, C,
And ∵ 2b is even,