The function is equal to the root x + 4, and the value range of the independent variable x is

The function is equal to the root x + 4, and the value range of the independent variable x is

Y = √ x + 4, x + 4 ≥ 0, so x ≥ - 4

The image of function f (x) = radical (x-a) (x is greater than or equal to a) has a common point with the image of its inverse function. Find the value range of A

If the image of the function f (x) = root (x-a) (x is greater than or equal to a) has a common point with the image of its inverse function, then there must be an intersection in the image, that is, y = x intersects with (x-a) under y = radical
So x ^ 2-x + a = 0 has a solution

Circle O1 and circle O2 intersect at points a and B, and circle O1 passes through point O2. If angle ao1b = 100 °, find the degree of angle ao2b

130°?
I pass through the circle O1. If the center angle of the circle O1 is 100 degrees, then the circular angle it faces is 50 degrees
From the inner diagonal complementation of the inscribed quadrilateral of a circle, the result is 130 degrees

⊙ O1 and ⊙ O2 cross point a and B, and ⊙ O1 passes through point O2. If ⊙ ao1b = 90 °, then the degree of ⊙ ao2b is______ .

∵∠AO1B=90°，
⊙ when the radius of ⊙ O1 ⊙ O2 ⊙ ao2b = 180 ° - 45 ° = 135 °,
When the radius of ⊙ O1 ⊙ O2 ⊙ ao2b = 45 °,
The degree of ∠ ao2b is 45 ° or 135 °

It is known that O 1 and O 2 intersect at a and B, and 0 o 1 passes 0 O 2. If ∠ ao1b = 90 °, then the degree of ∠ ao2b There are two answers: 135 ° 45 °

∵∠AO1B=90°,
⊙ when the radius of ⊙ O1 ⊙ O2 ⊙ ao2b = 180 ° - 45 ° = 135 °,
When the radius of ⊙ O1 ⊙ O2 ⊙ ao2b = 45 °,
The degree of ∠ ao2b is 45 ° or 135 °

As shown in the figure, circle O1 and circle O2 are inscribed at point a, and their radii are R1 and R2 (R1 > R2). Chord ab of circle O1 intersects circle O2 at point C (O1 is not on AB). It is proved that ab: AC is a constant value

It is proved that: according to ⊙ O1 and ⊙ O2 inscribed at point a, O1, O2, a can be obtained. On a straight line, O1, O2, a are connected to O1f ⊥ AB, o2e ⊥ AB at point F, e, ∵ O1f ⊥ AB, o2e ⊥ AB, ∵ AE = CE, AC = BF, ∵ ABAC = AFAE, ∵ O1f ⊥ AB, o2e ⊥ AB, ⊥ O1f ⊥ O2

As shown in the figure, circle O1 and circle O2 are inscribed at point t, and the chord TATB of circle O2 intersects ⊙ O2 at D and C respectively, connecting AB and CD. Verification: ab ∥ CD

prove:
Making common tangent EF of two circles through t
According to the tangent angle theorem:
∠TAB=∠BTF,∠D=∠CTF
Because ∠ BTF and ∠ CTF are the same angle
So ∠ tab = D
So AB / / CD

Given that ⊙ O1 and ⊙ O2 contain two circles, O1O2 = 3, radius of ⊙ O1 is 5, then the value range of radius r of ⊙ O2 is______ .

According to the meaning of the title, the two circles contain,
It is known that R-5 > 3 or 5-r > 3,
The solution is 0 ＜ R ＜ 2 or R ＞ 8
So the answer is: 0 < R < 2 or R > 8

It is known that the radii of circle O1 and circle O2 are R, R, and R ≥ R, R, R are two of the equations x ^ 2-5x + 2 = 0. Let O1O2 = D When d = 11 / 2, try to determine the position relationship of circle O1 and circle O2 When d = 3 When d = 4.5 If two circles are tangent, what is the value of D? (junior three knowledge)

∵ R, R is two of the equation x ∵ 5x + 2 = 0,
 from Weida theorem: R + r = 5, RR = 2
Then (R-R) 2 = (R + R) 2 - 4rr = 17
∵R≥r
∴R-r=√17
When d = 11 / 2, d > R + R, therefore, the two circles are outward;
When d = 3, D

Given that the radius of tangent circle O is 2R, the radius of circle O1 and circle O2 is r, the radius of circle O3 is calculated The semicircle is O, the largest semicircle O1 O2 is inside the semicircle o, and the circle O3 is above the semicircle O1 O2 [[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[

Let the center of circle o be a, the center of circle O1, and the center of circle O2 are respectively B and C. according to the question, BC = 2R, ab = AC = R, and ab + AC = BC, so a can only be on BC (the sum of the two sides of the triangle is greater than the third side). Therefore, if the radius of circle O3 is x, then: (R + x) ^ 2-r ^ 2 = (2r-x) ^ 2, x = 2R / 3
Learn to draw, hope to help you with geometry!