Known: a = 10000, B = 9999, find the value of A2 + b2-2ab-6a + 6B + 9

Known: a = 10000, B = 9999, find the value of A2 + b2-2ab-6a + 6B + 9

∵ a = 10000, B = 9999, ∵ A-B = 10000-9999 = 1, then the original formula = (a-b) 2-6 (a-b) + 9 = 1-6 + 9 = 4

Find the standard equation of an ellipse with the same focus as the square of 9 / x + the square of 4 / y of the ellipse = 1 and passing through the point P (3, - 2)

If the focus of the elliptic equation is on the x-axis and C & # 178; = 9-4 = 5, that is, C = √ 5, then the focus coordinates of the ellipse are F1 (√ 5,0), F2 (- √ 5,0) and the ellipse passes through the point P (3, - 2), then the definition of the ellipse can be obtained as follows: | Pf1 | + | PF2 | = 2A = √ [(3 - √ 5) &# 178; +...]

If a = 12, C = 15, then B=______ If a: B = 3:4, C = 10, then a=______ ，b=______ ．

① ∵ in RT △ ABC, ∠ C = 90 °, a = 12, C = 15, ∵ B = C2 − A2 = 152 − 122 = 9, so the answer is: 9; in RT △ ABC, ∠ C = 90 °, a: B = 3:4, C = 10, ∵ let a = 3x, then B = 4x, ∵ A2 + B2 = C2, namely 9x2 + 16x2 = 102, the solution is x = 2, ∵ a = 3x = 6, B = 4x = 8

Given the function f (x) = X2 (x ∈ [- 2,2]), G (x) = a2sin (2x + π / 6) + 3a, X ∈ [0, π / 2], for any x1 ∈ [- 2,2],
There is always x0 ∈ [0, π / 2], such that G (x0) = f (x1), and the range of a is obtained

The range of F (x) = x ^ 2 in X ∈ [- 2,2] is [- 4,4]. The range of G (x) is d = [- A ^ 2 + 3a, a ^ 2 + 3A]
It can be seen from the meaning that [- 4,4] is a subset of D, that is - A ^ 2 + 3A = 4, then a = 4

What is the ratio of the area of the inscribed circle to the area of the triangle if the length of the hypotenuse of RT △ is C and the radius of the inscribed circle is r?

Make a right triangle and make an inscribed circle inside it. Make a vertical line on three sides through the center of the circle,
Let the three sides of a right triangle be a, B, C respectively,
By using the tangent length theorem, the radius of the inscribed circle r = (a + B-C) / 2 can be obtained,
So a + B = C + 2R,
And because a ^ 2 + B ^ 2 = C ^ 2,
So s triangle = AB / 2 = [(a + b) ^ 2 - (a ^ 2 + B ^ 2)] / 4 = [(c + 2R) ^ 2-C ^ 2] / 4 = (4CR + 4R ^ 2) / 4 = Cr + R ^ 2,
So s circle: s triangle = π R ^ 2: (Cr + R ^ 2) = π R: (c + R)

The concept of complete complement of subsets

If R is the complete set, then R is the premise
Subsets, for example, positive integers are R, and subsets are subsets of big premises. Of course, empty sets are OK
If the complete set is R and a is a positive number, then the complement of a is negative infinity to 0
It's not very complete
I'm too lazy to fight
Turn a math book for senior one

D is a point on the extension line of side BC of triangle ABC, angle a is equal to 96 degrees, angle ABC and angle ACD bisector intersect at A1, angle a1bc and angle a1cd bisector intersect at A2, and so on, angle a4bc and angle a4cd bisector intersect at angle A5, calculate the degree of angle A5
You'd better hurry up and hand in the picture tomorrow
In angle ABC, angle a = 42 ° D and E are the intersection points of angle ABC and angle ACB
The three internal angles a, B and C of angle ABC satisfy that the angle a is greater than the angle B, and the angle c is less than or equal to the angle B. judge the triangle line if it is classified by angle

1. Angle a1bc = 1 / 2 angle ABC angle a2bc = 1 / 2 & # 178; angle ABC Angle a5bc = 1 / 2 ^ 5 angle ABC = 1 / 32 angle ABC the same way: angle a5cb = 1 / 32 angle ACB  angle aba5 = 31 / 32 angle ABC angle aca5 = 31 / 32 angle ACB and angle A5 = angle aba5 + angle a + angle aca5 =

If any two real numbers x, y are taken in the interval (0,1), then the probability of Y & # 178; > x is

Drawing, can be converted into the probability of area possession
The integral can calculate that the area of Y ^ 2 > x in the range of X (0,1) and Y (0,1) is 1 / 3, the interval (0,1) is a square, and the area is 1, so the probability is 1 / 3

Find indefinite integral ∫ (DX) / √ (1 + e ^ 2x) =?
Such as the title, seek detailed explanation of QAQ!

Answer: let t = √ [1 + e ^ (2x)] t ^ 2 = 1 + e ^ (2x) e ^ (2x) = T ^ 2-12x = ln (T ^ 2-1) x = (1 / 2) ln (T ^ 2-1) original formula = ∫ 1 / T d [(1 / 2) ln (T ^ 2-1)] = (1 / 2) ∫ (1 / T) * [1 / (T ^ 2-1)] * 2T DT = ∫ 1 / (T ^ 2-1) DT = ∫ 1 / [(t-1) (T + 1)] DT = (1 / 2) * ∫ 1 / (t-1) - 1 / (t

A pool has two drainage pipes a and B, and one inlet pipe C. if the two pipes a and C are opened at the same time, the full pool water can be emptied in 20 hours; if the two pipes B and C are opened at the same time, the full pool water can be emptied in 30 hours; if the pipe C is opened only, the empty pool can be filled in 60 hours; if the pipe a, B and C are opened at the same time, the water in the pool needs to be emptied______ Hours

1 △ 120 + 160 + 130 + 160-160, = 1 △ 760 − 160, = 1 △ 110, = 10 (hours); answer: it takes 10 hours to empty the water in the pool