As shown in the figure, in the parallelepiped abcd-a1b1c1d1, e is the midpoint of CC1. Let vector AB = a, vector ad = B, and vector Aa1 = C. let a, B, and C denote vector AE

As shown in the figure, in the parallelepiped abcd-a1b1c1d1, e is the midpoint of CC1. Let vector AB = a, vector ad = B, and vector Aa1 = C. let a, B, and C denote vector AE

1001+1003+1005+.+2005+2007+2009=

1001 + 2009 = 3010, 1003 + 2007 = 3010, 1504 + 1506 = 3010, add 1505, 3010 × 504 + 1505 = 1518545

Given the set a = {X / x2 + (M + 2) x + 1 = 0}, B = {X / x > 0}, if a intersects B = an empty set, find the value range of real number m --- this is it
/This is not division. It's a vertical line. X2 is the square of X

Sorry first
I was wrong
Set a = {X / x2 + (M + 2) x + 1 = 0}, B = {X / x > 0}, if a intersects B = empty set,
That is to say, the solutions of x2 + (M + 2) x + 1 = 0 are less than or equal to 0 or a is an empty set
When the solutions of x2 + (M + 2) x + 1 = 0 are less than or equal to 0
According to Weida's theorem, x 1 x 2 = 1 > 0, x 1 + x 2 = - (M + 2) ≤ 0
M ≥ - 2
When a is an empty set
(m+2)^2-4

Look at the numbers and guess the idioms: 1 = 10,1 * 1000010002 = 100 * 100 * 100,1%, 70.86,40 / 6

1. Plenty of food and clothing
2. Do everything possible
3. One in a hundred
4. Scattered
5. Endless

If the length of one side of a parallelogram is 5, the length of its two diagonals can be ()
A. 12 and 2b. 3 and 4C. 4 and 6D. 4 and 8

As shown in the figure, make CF ‖ BD through point C, intersect the extension line AB at point F, the ∪ quadrilateral bfcd is parallelogram, ∪ CF = BD, ∪ in △ AFC: ac-cf ﹤ AF ﹤ AC + CF, that is, ac-bd ﹤ 2Ab ﹤ AC + BD, ∫ AB = 5, only the data in D can satisfy this relationship: 8-4 = 4 ﹤ 5 × 2 ﹤ 8 + 4 = 12, so select D

86 + x = 7 (X-12) to solve the equation
There can only be decimals

86+x=7(x-12)
86＋x=7x－84
7x－x=86＋84
6x=150
x=25

Given sin (XY) = ln ((x + 1) / y) + 1, find y '(0)

The derivation of sin (XY) - ln ((x + 1) / y) + 1 = 0 for X is: (y + XY ') cos (XY) - Y / (x + 1) · [y - (x + 1) y'] / y ^ 2-y / (x + 1) · (x + 1) (- 1 / y ^ 2) y '= 0, substituting Y - [Y-Y'] / y + y '/ y = 02y' = Y-Y ^ 2, and sin (XY) = ln ((x + 1) / y) + 1x = 0, substituting y (0) = e ≈ 2.718y '(0) = (E-E ^ 2) / 2

How to solve 4 / 7x-5 = 3 / 8

4/7x-5=3/8
4/7x=5+3/8
4 / 7X = 5 and 3 / 8
X = 5 and 3 / 8 △ 4 / 7
x=301/32
x=9.40625

(2011 Zhongshan three module) equation |sinx|x=k (k > 0) and only two different real numbers solution theta, phi (theta > phi), then the conclusion of the following two relations is correct.
A. sinφ=φcosθB. sinφ=-φcosθC. cosφ=θsinθD. sinθ=-θsinφ

According to the meaning of the problem, let x ＞ 0 (x cannot be equal to 0) let Y1 = | SiNx |, y2 = KX, and then make the images of two functions respectively. Because the original equation has and only has two solutions, there are only two intersections between Y2 and Y1, and the second intersection is the point where Y1 and Y2 are tangent, that is, the point (θ, | sin θ |) is the tangent point, because (- sin θ) ′ = - cos θ, so the tangent slope k = - cos θ ）On the tangent line y2 = KX = - cos θ x, then substituting the point (φ, sin φ) into the tangent equation y2 = xcos θ, we can get: sin φ = - φ cos θ

How much is a when the product of a number and its reciprocal plus a is 5 / 6?

1+a=5/6 a=-1/6