Let ∑ be the plane x + y + Z = 1, then ∫ 2 / (x + y + Z) DS is equal to?

Let ∑ be the plane x + y + Z = 1, then ∫ 2 / (x + y + Z) DS is equal to?

That is, the side length is 2 ^ 0.5, the area of an equilateral triangle multiplied by 2
Question = 2 ^ 0.5 * 2 ^ 0.5 * 3 ^ 0.5 = 2 * 3 ^ 0.5

A 32 decimeter long iron wire is used to form a rectangle with the ratio of length to width of 5:3. The area of the rectangle is () square decimeters

A 32 decimeter long iron wire is used to form a rectangle with a length to width ratio of 5:3. The area of the rectangle is (60) square decimeters

Given that x2 + X-6 is a factor of the polynomial 2x4 + x3-ax2 + BX + A + B-1, then a=______ ；b=______ ．

Let another factor be: 2x2 + MX + N, then (x2 + X-6) (2x2 + MX + n) = 2x4 + (M + 2) X3 + (M + n-12) x2 + (n-6m) x-6n, then: M + 2 = 1m + n − 12 = − an − 6m = Ba + B − 1 = − 6N, the solution is: M = − 1n = − 3A = 16b = 3, so the answer is: 16, 3

What is the maximum value of the function f (x, y) = x ^ 2 + 2Y ^ 2-x ^ 2Y ^ 2 in the region D
I see you have asked this question. What is the answer
Region D = {(x, y) | x ^ 2 + y ^ 2 = 0}, I calculated the answer is 7 / 4,

I remember this seems to be a mathematical problem for postgraduate entrance examination. If I remember correctly, the answer should be 8! I can't calculate without a pen on hand. The owner can solve the problem by himself. The area of D is a semicircle. First calculate the y = 0 straight line, then calculate the Y > 0 arc, and get the maximum value by comparison

Find all integers x that make | 4x ^ 2-12x-27 | prime

|4x^2-12x-27|
=|(2x+3)(2x-9)|
So 2x + 3 = 1 or 2x + 3 = - 1 or 2x-9 = 1 or 2x-9 = - 1
That is, x = - 1 or - 2 or 5 or 4

The absolute value of x plus y minus Z is 0, and X is known to be 4

I see that you have asked me for help, but the question is not complete. You can add it below or ask me. I will answer you if I see it

Given the cubic function f (x) = x & # 179; + BX & # 178; + CX + D passing through the origin, (- 1,0) and (2,0), find the expression of F (x)

Because it's past (00)
d=0
f(-1）=-1+b+c=0
f(2)=8+4b+c=0
∴
b-1=8+4b
8+4b-b+1=0
b=-1
c=-2
f(x)=x3-x2-2x

84 = () × () × () × (). Fill in the prime number in the bracket! And it can't be repeated!

84＝（2 ）×（ 2）×（3 ）×（ 7）.
It can't be repeated. How can it be possible? This problem is clearly qualitative factorization

Let the probability density function f (x) of random variable x satisfy f (x) = f (- x) and f (x) be the distribution function, then for any a > 0, P {x | ≥ a} is equal to
A 2F（a）—1 B 2[1-F(a)] C 2-F(a) D1-2F(a)

P{|X|≥a}＝P|X≥a}＋P{X＜＝－a}＝2P{X＜＝－a}
= 2 integral (from - infinity to - a) f (x) DX
＝2F（－a）
= 2 (1-f (a)) / / because f-even function
Choose B
Or:
P{|X|≥a}＝P｛X≥a}＋P{X＜＝－a}＝2P｛X≥a}
= 2 (1-integral (from - infinity to a) f (x) DX)
＝2（1－F（a））

The range of y = 1 / Sin & # 178; X + 4 / cos & # 178; X

y=1/(sinx)^2+4/(cosx)^2=[(sinx)^2+(cosx)^2]*[1/(sinx)^2+4/(cosx)^2]
>=(√ 1 + √ 4) ^ 2 (Cauchy inequality)
=9,
Therefore, the function range is: [9, + ∞)