English translation 180 degree honeymoon 2F 180 degree sea view honeymoon room 3-5f Super sea view Honeymoon Room 6F Sea view leisure mahjong room 2F 280 degree all sea view romantic honeymoon suite 6F
180 degree honeymoon 2F
180 degree viewing honeymoon room 2 F
180 degree sea view honeymoon room 3-5f
180-degree sea view honeymoon room 3-5 F
Super sea view Honeymoon Room 6F
Super seaview honeymoon big bed room 6 F
Sea view leisure mahjong room 2F
Seaview leisure mahjong room 2 F
280 degree all sea view romantic honeymoon suite 6F
280 degrees seaview romantic honeymoon suites 6 F
English translation
Exquisite big bed room
Deluxe Twin / King Room
Urban landscape Twin / King Room
Executive Room
Executive Suite
We don't have any culture. Thank you first
Exquisite bed rooms
Deluxe Twin / King Room
Urban landscape Twin / King Room
Executive Room
Executive Suite
How to calculate 222 times 999?
Test papers
222*(1+999)-222=221778
Take 1000 222s, add them and subtract one 222
Solve the equation x (3x-4) - 7 = 2 (x ^ 2-2) + 2x
x(3x-4)-7=2(x^2-2)+2x
Remove the brackets to get 3x ^ 2-4x-7 = 2x ^ 2-4 + 2x
Transfer and merge similar items to get x ^ 2-6x-3 = 0
The formula is: (x-3) ^ 2 = 12
Both sides square at the same time: x-3 = ± √ 12
X = 3 + 2 √ 3 or x = 3-2 √ 3
Conversion of degrees, minutes and seconds
How to convert degrees, minutes and seconds into degrees?
Using 56 ° 25 ′ 12 ″
Divide by 60, (56 + 25 / 60 + 12 / 3600) degrees
Chord length AB = under the root sign (x2-x1) ^ 2 + (y2-y1) ^ 2 = under the root sign (1 + K ^ 2) (x2-x1) ^ 2 please explain in detail how to get the one after the equal sign? K is
Because X1 = kx1, X2 = kx2
So substitute X1 = kx1, X2 = kx2 into (x2-x1) ^ 2 + (y2-y1) ^ 2 = (x2-x1) ^ 2 + (kx2-kx1) ^ 2 = (1 + K ^ 2) (x2-x1) ^ 2
So chord length AB = under root sign (x2-x1) ^ 2 + (y2-y1) ^ 2 = under root sign (1 + K ^ 2) (x2-x1) ^ 2
-How much is 6 times 11 and 11 out of 12
Let a = (3 / 2, SiNx), B = (cosx, 1 / 3), and a / / B, then the acute angle X is?
The necessary and sufficient condition for a / / B is 3 / 2 × 1 / 3-sinx × cosx = 0 (1 / 2) sin2x = 1 / 2
sin2x=1 x=45°
o. What's 55
It's 11 / 20
The point P is a moving point on the parabola C1: x ^ 2 = 2PY. Two tangent lines of circle C2: x ^ 2 + (Y-3) = 1 intersect Y-axis at two points a and B. It is known that the distance from the fixed point Q (1,13 / 4) to the collimator of parabola C1 is 7 / 2
(1) It is proved that the line PA and Pb are not perpendicular. (2) if the line AB is bisected by the line PQ, the coordinates of point P are obtained
(1) The distance between Q (1,13 / 4) and parabola C1: y = - P / 2 is 13 / 4 + P / 2 = 7 / 2, P = 1 / 2,
Let P (T, T ^ 2) be a moving point on parabola C1: x ^ 2 = y,
Make the tangent of circle C2: x ^ 2 + (Y-3) ^ 2 = 1 through P: Y-T ^ 2 = K (x-t), that is, kx-y + T ^ 2-kt = 0,
The distance from the center C2 (0,3) to the tangent = | - 3 + T ^ 2-kt | / √ (k ^ 2 + 1) = 1,
The square is (- 3 + T ^ 2-kt) ^ 2 = k ^ 2 + 1,
(T ^ 2-1) k ^ 2-2t (T ^ 2-3) K + (T ^ 2-3) ^ 2-1 = 0, ①
If the product of slopes of two tangent lines k1k2 = [(T ^ 2-3) ^ 2-1] / (T ^ 2-1) = - 1,
Then (T ^ 2-3) ^ 2-1 = 1-T ^ 2,
The results show that T ^ 4-5t ^ 2 + 7 = 0, there is no real solution,
The line PA is not perpendicular to Pb
(2) The tangent intersects Y-axis at a (0, T ^ 2-k1t), B (0, T ^ 2-k2t),
From (1), K1 + K2 = 2T (T ^ 2-3) / (T ^ 2-1), the,
The midpoint of AB is m: XM = 0, YM = T ^ 2-T (K1 + K2) / 2 = T ^ 2-T ^ 2 (T ^ 2-3) / (T ^ 2-1) = 2T ^ 2 / (T ^ 2-1),
P, m and Q are collinear when line AB is bisected by line PQ,
The slopes of PQ and MQ are equal, that is, (T ^ 2-13 / 4) / (t-1) = 13 / 4-2t ^ 2 / (T ^ 2-1),
Multiply both sides by 4 (T ^ 2-1) to get (4T ^ 2-13) (T + 1) = 13 (T ^ 2-1) - 8t ^ 2,
4t^3+4t^2-13t-13
-5t^2 +13=0,
4T ^ 3-T ^ 2-13t = 0, T1 = 0, t2,3 = (1 Soil √ 209) / 8,
The coordinates of point P are (0,0), ((1 + √ 209) / 8, (105 + √ 209) / 32), (1 - √ 209) / 8, (105 - √ 209) / 32)