Take some examples of the constant motion of molecules in life The more the better, the more detailed = = |

Take some examples of the constant motion of molecules in life The more the better, the more detailed = = |

Sound waves are produced by the vibration of air molecules

（-2x-1)_______ =Square of 1-4x

2x-1

The zero vector is parallel to any vector. So, is the zero vector parallel to the zero vector

Parallel, but we generally do not compare, because there is no significance
The simplest understanding is that any vector contains zero vectors
In fact, the zero vector can be in any direction, so no matter what direction the known vector is, the zero vector is parallel to it, no matter whether the known vector is non-zero or not

In triangle ABC, ad is perpendicular to BC, intersects BC at point D, passes through point a to make ef parallel BC, BF parallel AC, CE parallel ab
Verification: de = DF
Draw a picture, otherwise you can't see it (triangle ABC is an acute triangle)
Please answer in terms

EF parallel BC, BF parallel AC, quadrilateral acbf is parallelogram, AF = BC
Similarly, the quadrilateral abce is a parallelogram AE = BC, so AE = AF
And because ad is vertical BC.EF Parallel BC, so ad vertical EF, that is, ad is EF vertical bisector
So de = DF (the distance from any point on the vertical bisector to both ends is equal)

A cube, a sphere and an equilateral cylinder have the same volume. The surface area of these three geometric bodies is in the order of size

Cube a, cube a = 4 / 3 π r cube = 2 π r cube, first look at the last two, get 2 / 3R cube = R cube, get r = XXX, substitute 4 π r square and 3 π r square (π RL + π r square, l = 2R), get the first conclusion that the sphere is larger than the cylinder, the second comparison between the sphere and the cube, finally get the cube is less than the surface area of the sphere, the same reason can get the cube is larger than the cylinder

In the triangle abd, D is the midpoint of AB, e is the point on AC, CE = one third of AC, be / CD intersects o, be = 5cm, then OE =?

Make DF / / AC through D and turn AC to F
Because D is the midpoint of AB, f is the midpoint of AE, that is, EF = FA = 1 / 2ae
Meanwhile, DF = 1 / 2be = 5 / 2
And: CE = 1 / 3aC, so: CE = 1 / 2ae
That is: CE = EF, then E is the midpoint of CF
So: OE = 1 / 2DF = 1 / 2 * 5 / 2 = 5 / 4

Given the point a (3, - 4) B (- 2,3), find the coordinates of AB vector and Ba vector

AB vector = B (- 2,3) - A (3, - 4) = (- 5,7) BA vector = a (3, - 4) - B (- 2,3) = (5, - 7)

It is known that, as shown in the figure be, CF is the height of the edges AC and ab of △ ABC. Intercept BP = AC on be and CQ = AB on the extension line of CF. verify: AP ⊥ aq

It is proved that: ∵ CF ⊥ AB, be ⊥ AC, ∩ AEB = ∠ AFC = 90 °, ∩ Abe = ∠ ACQ = 90 ° - BAC. ∵ BP = AC, CQ = AB, in △ APB and △ QAC, BP = AC ∠ Abe = ∠ acqcq = AB, ≌ APB ≌ △ QAC (SAS). ∩ BAP = ∠ CQA. ∩ CQA + ∠ qaf = 90 °, ∩ BAP + ∠ QA

If the distance from a point P to the right focus on the ellipse x2 / 25 + Y2 / 9 = 1 is 6, then the distance from P to the left quasilinear is 6

Because a = 5, B = 3, C = 4, the Quasilinear equation is x = ± A & # 178;; C = ± 25 / 4, the left quasilinear equation is x = - 25 / 4, let p be (n, m), the right focus is O (4,0) (PO) &# 178; = (n-4) &# 178; - M & # 178; = 36, and there is n & # 178 / / 25 + M & # 178 / / 9 = 1, 34n & # 178; - 200n-725 = 0n = 50 / 17 ± 15 (√ 154) / 3

Known: as shown in the figure, in △ ABC, ab = AC, BC = BD, ad = de = EB, then the degree of ∠ A is ()
A. 30°B. 36°C. 45°D. 50°

In the case of EBD (EBD = x °, and the "be = De,, \\\\\\\\\\\\\ \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\1