As shown in the figure, AB is the diameter of ⊙ o, extend AB to point P so that ab = 2bp, make a tangent line of ⊙ o through point P, the tangent point is C, connect AC, then ∠ cap=______ ．

As shown in the figure, AB is the diameter of ⊙ o, extend AB to point P so that ab = 2bp, make a tangent line of ⊙ o through point P, the tangent point is C, connect AC, then ∠ cap=______ ．

Connect OC, BC ∵ PC as tangent of circle ∪ OC ⊥ PC ∵ AB = 2bp ∪ ob = OC = BP, that is, Op = 2ocrtocp, we can get ∠ P = 30 ° and ∠ cop = 60 ° in the isosceles triangle OAC, ∠ cob = 2 ∠ OAC = 60 ∪ cap = 30 ° so the answer is 30 °

As shown in the picture, the two sides of a river are parallel. On one bank of the river, there is a tree every 5m, and on the other bank, there is a telegraph pole every 50m. When you look at the other bank 30 m away from the bank, you can see that the two trees on this bank just cover the two adjacent telegraph poles on the other bank, and there are three trees between the two trees, so as to seek the width of the river. (please attach the process. Thank you.)

The two triangles are similar, and the bottom length of the two triangles are: (3 + 1) * 5 = 20m and 50m respectively
Then the height of the two triangles (that is, the distance from the observer to both sides) is proportional to their length
That is: H / h = 20 / 50, and H = 25m
So: H = 50h / 20 = 50 * 25 / 20 = 62.5m
The river width is: H-H = 62.5m - 25m = 37.5m

It is known that, as shown in the figure, the quadrilateral ABCD is inscribed in the circle O, AC is the diameter of the circle O, the extension lines of Ba and CD intersect at P, if the angle P = 30 degrees, ab = 3, DC = 3, root 3, find [1]
【1】 Radius of long circle O of PD
It is known that, as shown in the figure, the quadrilateral ABCD is inscribed in the circle O, AC is the diameter of the circle O, the extension lines of Ba and CD intersect at P, if the angle P = 30 degrees, ab = 3, DC = 3, root 3, find [1]
Radius of long circle O of PD

The quadrilateral ABCD is inscribed with circle O, so ∠ ABC = 90 degrees and ∠ ADC = 90 degrees
∠CAB=∠30°+∠DCA,∠DCB=90°-30°=60
∴∠DCA=30°
｜ CD = PD = 3 radical 3
AB=1/2AC=3
The radius of circle 0 = 3

When x = 1, the value of PX3 + QX + 6 is 2010, and when x = - 1, the value of PX3 + QX + 6 is 2010______ ．

∵ when x = 1, the value of PX3 + QX + 6 is 2010, ∵ P + Q + 6 = 2010, ∵ P + q = 2010-6, ∵ when x = - 1, PX3 + QX + 6, = - P-Q + 6, = - (P + Q) + 6, = 6-2010 + 6, = - 1998, so the answer is - 1998

Parabola C: x ^ 2 = 4Y, straight line m passing through point P (2,2) intersects with C at two points AB, and has vector AP = λ vector Pb
(1) When λ = 1, the equation of straight line m is obtained
(2) When the area of triangle AOB is 4 pieces of 2, the value of λ is calculated

Note that the area can be expressed as 1 / 2 * m longitudinal intercept * (x2-x1) point oblique form. Let m and C be listed simultaneously to obtain the equation (after simplification): (1-k) & sup2; (K & sup2; - 2K + 2) = 2 (it is easy to see if it is a good solution by global substitution). If the solution is k = 0 or K = 2, then λ 1 = 3 + 2 (...)

4X & # 178; - ax + 9 is a complete square, then a=

A: when a = 12,4x & # 178; - ax + 9 = (2x-3) ^ 2
a=-12,4x²-ax+9=(2x+3)^2
So, when a = ± 12, 4x & # 178; - ax + 9 is a complete square

In the cube abcd-a1b1c1d1, m and N are the midpoint of edges AB and ad respectively,
O is the intersection of AC and BD, and Mn ⊥ OC1 is proved
Note: all 1 above are subscripts

Proof: connect a1c1
∵BD⊥AC,BD⊥AA1
Ψ BD ⊥ plane aa1c
OC1 belongs to plane aa1c
∴OC1⊥BD
And ∵ Mn ∥ BD
∴MN⊥OC1
PS. draw a good picture first, it will be easier to understand the picture

Simplify - x square + 9x-20 / x square - 7x + 12

Solution
(x²-7x+12)/(-x²+9x-20)
=(x²-7x+12)/[-(x²-9x+20)]
=-(x²-7x+12)/(x²-9x+20)
=-(x-3)(x-4)/(x-5)(x-4)
=-(x-3)/(x-5)

Let a ^ 2 + 2a-1 = 0, B ^ 4-2b ^ 2-1 = 0, and 1-ab ^ 2 is not equal to 0, then ((AB ^ 2 + B ^ 2-3a + 1) / a) ^ 5 =?

Do you have anything without a radical

Guess idioms 1,2,3,4,5,6,7 () 1,3,5,7,9 () 2,4,6,8,10 () 1,2,5,6,7,8,9 () 3,4,5,6,7,8,9 according to the numbers
( ) 1,2,4,3,5,6,7,8,9( ) 1,4( )

1,2,3,4,5,6,7 (enjoy it) 1,3,5,7,9 (unparalleled in the world) 2,4,6,8,10 (unique) 1,2,5,6,7,8,9 (lost) 3,4,5,6,7,8,9 (clean) 1,2,4,3,5,6,7,8,9 (confused) 1,4 (one after another)