AB is the chord in ⊙ o with a radius of 10cm. The concentric circle with a radius of 2 times the root 7cm intersects at two points C. D. if the distance from the center O to AB is 2cm, the length of AC + BD is? It's better to have a picture. I really have no wealth. I'm sorry. I hope you can help me

You can see it by drawing a picture
Make a vertical line from O to AB, and AB to E
AE = root (AO ^ 2-oe ^ 2) = 4 root 6
CE = radical (Co ^ 2-oe ^ 2) = 2 radical 6
AC = ae-ce = 2 root 6
Similarly, BD = AC = 2, radical 6
AC + BD = 4 radical 6
The symmetry axis of the function y = - 2 / 3x ^ 2 is (), the vertex coordinate is (), the vertex is a parabola (), and the parabola is () on the X axis (except the vertex)
Come on, in 15 minutes
The symmetry axis of the function y = - 2 / 3x ^ 2 is (Y axis), the vertex coordinate is ((0,0)), the opening is (down), and the parabola is (below) the X axis (except the vertex)
The parabola y = xsquare - 2x-m (M > 0) intersects the y-axis at point C. the symmetric point of point c about the parabola symmetric axis is point C '
(1) If the point P is on the parabola, the point q is on the symmetry axis of the parabola, and the quadrilateral with points c, C ', P and Q as the vertex is a parallelogram, the coordinates of point P and point Q can be obtained (which can be expressed by an algebraic formula containing m)
(2) Under the condition of (1), find out the perimeter of the parallelogram
(1) The axis of symmetry is: x = 1
C(0,-m),C'(2,-m)
C. A quadrilateral whose vertices are C ', P and Q is a parallelogram
So, CC '/ / PQ, CC' = PQ
Let Q (1, y), then: P (- 1, y) or P (3, y)
So, y = x ^ 2-2x-m = 3-m
The coordinates of P and Q are Q (1,3-m),
P (- 1,3-m) or P (3,3-m)
(2)
CP=√((-1)^2+(3-m-(-m))^2)=√(1+9)=√10
The perimeter of this parallelogram is 2 (2 + √ 10) = 4 + 2 √ 10
Given that the axis of symmetry and X of the parabola y = - x ^ 2 + 2x + C ^ 2 intersect at (m, 0), then the value of M is
y=-x^2+2x+c^2
=-(x-1)^2+1+c^2
X = 1 is its axis of symmetry, so m = 1
When m =, the symmetry axis of the parabola y = 2x square + (3m-1) x + 4m-3 is y-axis
Why is symmetry axis y-axis, B = 0?
The axis of symmetry is the Y axis
So B = 0
therefore
3m-1=0
m=1/3
When B = 0, the function is even
On Y-axis symmetry
X = - B / 2a is the axis of symmetry
B=0
On Y-axis symmetry
As shown in the figure, the parabola y = - x2 + 2x + 3 intersects the X axis at two points a and B, and intersects the Y axis at point C, and point D is a moving point on the symmetry axis L. (1) find out the coordinates of point d when AD + CD is the smallest; (2) make ⊙ a with point a as the center and ad as the radius. (1) prove that when AD + CD is the smallest, the line BD is tangent to ⊙ a. ② write out another coordinate of point d when BD is tangent to ⊙ a______ .
(1) Since the symmetric point of point a with respect to L is point B, connecting BC and intersecting l with point D is the point to be solved. If the parabola y = - x2 + 2x + 3 and X axis intersect at two points a and B, then the axis of symmetry is: x = 1. When - x2 + 2x + 3 = 0, the solution is: x = 3 or x = - 1. Point a (- 1,0), point B (3,0), parabola y = - x2 + 2x + 3. When x = 0, y = 3,  point C (0,3) (2) from the inverse theorem of Pythagorean theorem, ∠ ADB = 90 °, that is, ad ⊥ BD. therefore, when AD + CD is minimum, the straight line BD is tangent to ⊙ a Another coordinate (1, - 2) of point D is obtained
Parabola y = x2 + mx2-2mx-3m, no matter what the value of M is, it always passes through the fixed point____ (urgent)
y=x2+mx2-2mx-3m
The results are as follows
m(x^2-2x-3)+x^2-y=0
No matter what the value of M is, it always passes the fixed point
So there are: x ^ 2-2x-3 = 0 and x ^ 2-y = 0
(x-3)(x+1)=0
X = 3 or - 1
So y = 9 or 1
So the parabola is constant over (3,9) and (- 1,1)
If the axis of symmetry of the parabola y = (m-2) x + 2mx + 1 with downward opening passes through the point (- 1,3), then M =?
The formula of the symmetry axis of the parabola is - B / 2A... Because the symmetry axis passes through (- 1,3), the symmetry axis is x = - 1, that is - (2m) / 2 (m ^ 2-2) = - 1, and the solution is m = - 1 or M = 2.. because the opening of the parabola is downward, so m ^ 2-2
∵ opening down ∵ m ^ 2-2
If the axis of symmetry of the parabola y = (m ^ 2-1) x ^ 2 + 2mx + 1 with downward opening passes through the point (- 1,3), then M =?
∵ opening downward
∴m^2-1
Opening downward, m ^ 2-1
If the parabola y = (m-1) x ^ 2 + 2mx + m + 2 is above the x-axis, then the range of M is
I can understand it
Because it is always above the x-axis, M-1 must be > 0, so m > 1
Calculate the discriminant again
Parabola has several important factors
----1. Opening direction 2. Symmetry axis 3. Intersection with coordinate axis 4. Highest and lowest points
As far as the subject is concerned, the image of the parabola is always above the x-axis
Just open up and the lowest point is above the x-axis
y=（m-1）x^2+2mx+m+2
Then M-1 > 0
Delta = 4m ^ 2-4 (m-1) (M + 2) = - 4m + 82
Parabola has several important factors
----1. Opening direction 2. Symmetry axis 3. Intersection with coordinate axis 4. Highest and lowest points
As far as the subject is concerned, the image of the parabola is always above the x-axis
Just open up and the lowest point is above the x-axis
y=（m-1）x^2+2mx+m+2
Then M-1 > 0
△ = 4m ^ 2-4 (m-1) (M + 2) = - 4m + 82

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