As shown in the figure, the known point a is a trisection point on the semicircle with Mn as the diameter, point B is the midpoint of an, and point P is the point on the radius on. If the radius of ⊙ o is l, then the minimum value of AP + BP is () A. 2B. 2C. 3D. 52

As shown in the figure, the known point a is a trisection point on the semicircle with Mn as the diameter, point B is the midpoint of an, and point P is the point on the radius on. If the radius of ⊙ o is l, then the minimum value of AP + BP is () A. 2B. 2C. 3D. 52

Make a symmetric point a 'of point a about Mn, connect a' B, intersect Mn at point P, then PA + Pb is the smallest, connect OA ', AA', ob, ∵ point a and a 'are symmetric about Mn, point a is a trisection point on semicircle, ∵ a' on = ∠ AON = 60 °, PA = PA ', ∵ point B is the midpoint of arc an ^, ∵ Bon = 30 °, and ∵ a' ob = ∠ a 'on + ∠ Bon = 90 °, and ∵ OA = OA' = 1, ∵ a 'B = 2. ∵ pa + Pb = PA' + Pb = a B = 2

If the solution of the equation 2x + ax − 1 = 1 of X is positive, then the value range of a is ()
A. A ＞ - 1B. A ＞ - 1 and a ≠ 0C. A ＜ - 1D. A ＜ - 1 and a ≠ - 2

If the denominator is removed, the solution of the equation 2x + a = X-1 〉 x = - 1-A ∵ is a positive number 〈 - 1-A ＞ 0, that is, a ＜ - 1. Because X-1 ≠ 0 〉 a ≠ - 2, then the value range of a is a ＜ - 1 and a ≠ - 2, so D is chosen

Let p be a moving point on hyperbola x24-y2 = 1, o be the origin of coordinates, and m be the midpoint of line OP, then the trajectory equation of point m is______ ．

Let m (x, y), then p (2x, 2Y) be substituted into hyperbolic equation to obtain x2-4y2 = 1, which is the trajectory equation of point m x2-4y2 = 1. Answer: x2-4y2 = 1

Polynomial ax & # 178; + BX + C, when x = 1, its value is 5; when x = 2, its value is 7; when x = - 2, its value is 11, try to find the value of a, B, C

Let y = ax & # 178; + BX + C
X = 1, y = 5; X = 2, y = 7; X = - 2, y = 11, respectively
a+b+c=5 (1)
4a+2b+c=7 (2)
4a-2b+c=11 (3)
(2)+(3)
8a+2c=18
4a+c=9
c=9-4a
Substituting (2)
4a+2b+9-4a=7
2b=-2
b=-1
B = - 1, C = 9-4a substituting (1)
a-1+9-4a=5
-3a=-3
a=1
c=9-4a=9-4=5
a=1 b=-1 c=5

In the tetrahedral ABCD, let AB = 1, CD = 2 and ab ⊥ CD. If the distance between AB and CD is 2, the volume of the tetrahedral ABCD is ()
A. 13B. 12C. 23D. 43

∵ AB is perpendicular to CD, ∵ we can make a plane α containing AB so that plane α is perpendicular to line CD. In this way, α cuts the tetrahedron into two small tetrahedrons. If the cross section is regarded as the bottom and CD is regarded as the sum of the heights of the two tetrahedrons, then the sum of the volumes of the two small tetrahedrons is the volume of the tetrahedron ABCD: v = 13 × (12 × 2 × 1) × 2 = 23, so we choose C

If the cube of polynomial K (K-2) x - (K-2) x squared-6 is a quadratic polynomial about X, when x is equal to 2, the value of the polynomial is obtained

If the polynomial K (K-2) x & # 179; - (K-2) x & # 178; - 6 is a quadratic polynomial about X
Then K (K-2) = 0, K-2 ≠ 0
So k = 0
So the quadratic polynomial is 2x & # 178; - 6
When x = 2, the value of the polynomial is 2 * 2 & # 178; - 6 = 2

Given the set a = {X - 2 ≤ x ≤ 5}, B = {x m + 1 ≤ x ≤ 2m - 1}. If B is contained in a, find the value range of real number M
Why is not m + 1 < 2m-1 in empty set

When B is an empty set: M + 1 > 2m-1, M = 2 and - 2 are derived

Judge whether the points a (1, - 2), B (2, - 3), C (3,10), D (- 3, - 2) are on the curve X & # 178; - XY + 2Y + 1 = 0

Let f (x) = x & # 178; - XY + 2Y + 1
Substituting point a (1, - 2), we get f (x) = 0, and point a is on the curve;
Substituting point B (2, - 3), we get f (x) = 5, and point B is not on the curve;
Substituting point C (3,10), we get f (x) = 0, and point C is on the curve;
Substituting point d (- 3, - 2), we get f (x) = 0, and point D is on the curve

The heights of triangles and parallelograms are equal. The lengths of the bottom sides are 4cm and 8cm respectively. What is the area ratio of triangles and parallelograms?

If the height is h, then:
Triangle area = (1 / 2) × 4 × H = 2H
Area of parallelogram = 8h
Therefore, triangle area: parallelogram area = (2H): (8h) = 1:4

X Times - (5 / 6 + 5 / 8) = 7 / 20 how to solve the equation? Thank you for your help

Divide by half on both sides of the equation and multiply by 60 on both sides
We get x = 84 / 175