The scale of a paper drawing of precision parts is 20:1. The length of a part measured on the drawing is 3cm. What is the actual length of the part?

The scale of a paper drawing of precision parts is 20:1. The length of a part measured on the drawing is 3cm. What is the actual length of the part?

3 divided by 20 = 0.15 cm

As shown in the figure, the pointer rotates 90 ° anticlockwise from "1" around point O and points to______ ．

The analysis shows that the pointer rotates 90 ° clockwise from "1" around point O and points to 4

Why do I do the flash clock second hand will reverse back? Itself clockwise to the number 9, I do not know why and counter clockwise back

These letters control the deformation

The distance between the end point a of the second hand of a clock and the center point O is 5cm, and the second hand rotates uniformly around the point O. when the time t = 0, point a coincides with point B marked 12 on the clock face, and the distance D (CM) between two points a and B is expressed as a function of T (s), then d (CM)=______ Where t ∈ [0,60]

∵ - AOB = T60 × 2 π = π T30 ∵ according to the side length of right triangle, d = 2 × 5 × sin12 ∵ AOB = 10sin π T60, so the answer is: 10sin π T60

It is known that F 1 and F 2 are the two focal points of the ellipse. A straight line passing through F 1 and perpendicular to the major axis of the ellipse intersects the ellipse at two points a and B. If △ ABF 2 is an equilateral triangle, then the eccentricity of the ellipse is ()
A. 33B. 23C. 22D. 32

From the question | AF1 | = 33 | F1F2 |, | B2A = 33.2c, that is, A2 − C2 = 233ac | C2 + 233ac − A2 = 0, | E2 + 233e − 1 = 0, the solution is: e = 33 (negative value is rounded off)

It is known that F 1 and F 2 are the two focal points of the ellipse. A straight line passing through F 1 and perpendicular to the major axis of the ellipse intersects the ellipse at two points a and B. If △ ABF 2 is an equilateral triangle, then the eccentricity of the ellipse is ()
A. 33B. 23C. 22D. 32

From the question | AF1 | = 33 | F1F2 |, | B2A = 33.2c, that is, A2 − C2 = 233ac | C2 + 233ac − A2 = 0, | E2 + 233e − 1 = 0, the solution is: e = 33 (negative value is rounded off)

Let F1 and F2 be the focal points of the ellipse x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0), and the line passing through F1 and perpendicular to the X axis intersects with the ellipse at two points a and B. If △ abf2 is an acute triangle, then the value range of eccentricity of the ellipse is

hbnhjm,
If △ abf2 is an acute triangle, then ∠ af2b

Given a point P (6,8) on the ellipse x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0), F1 and F2 are the two focuses of the ellipse, and Pf1 ⊥ PF2, the elliptic equation is solved

Pf1 ⊥ PF2, then p is on the circle with the diameter of F1F2, F1 (- C, 0), F2 (C, 0). Therefore, the equation of the circle with the diameter of F1F2 is: X & # 178; + Y & # 178; = C & # 178; substituting p (6,8) into: C & # 178; = 100, then: B & # 178; = A & # 178; - 100, let a & # 178; = m, then: B & # 178; = M-100, M > 100, the elliptic equation is: X

It is known that the two focal points of the ellipse are F1 (- 1,0), F2 (1,0), P is a point on the ellipse, and 2f1f2 = Pf1, PF2 is the equation for solving the ellipse

PF1+PF2=2F1F2
Defined by ellipse
PF1+PF2=2a
F1F2=2c
So 2A = 4C
Obviously, C = 1
So a = 2
b^2=a^2-c^2=3
The focus is on the x-axis
So x ^ 2 / 4 + y ^ 2 / 3 = 1

It is known that the left and right focal points of ellipse x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 (a > b > 0) are F1 and F2 respectively, the eccentricity e = √ 2 / 2, and the right quasilinear equation is x = 2,
Find: (1) the standard equation of the ellipse, (2) the intersection of the line L passing through point F1 and the ellipse with two points m and N, and ▏ F2m + f2n ▏ = 2 √ 26 / 3, find the equation of the line L

(1) e=c/a=√2/2
Right guide line x = a ^ 2 / C = 2
So a = √ 2, C = 1, B ^ 2 = a ^ 2-C ^ 2 = 1
The standard equation of circle is x ^ 2 / 2 + y ^ 2 = 1
(2) F1 (- 1,0), let the L equation be y = K (x + 1), and substitute it into the elliptic equation to eliminate y
x^2+2k^2(x+1)^2=2
(2k^2+1)x^2+4k^2*x+2k^2-2=0
Let m (x1, Y1), n (X2, Y2), then
x1+x2=-4k^2/(2k^2+1),(1)
x1*x2=(2k^2-2)/(2k^2+1) (2)
From | F2m | + | f2n | = | F2m | + | f2n | + | MF1 | + | NF1 | - | Mn | = 4A - | Mn | = 2 √ 26 / 3
|MN|=4√2-2√26/3 (3)
And | Mn | = √ (k ^ 2 + 1) * √ [(x1 + x2) ^ 2-4x1 * x2] (4)
Therefore, from (1) (2) (3) (4), k =?
(you can substitute (1) (2) (3) into (4) and do it yourself. The number is too irregular.)