It is known that, as shown in the figure, in △ ABC, the bisector of ∠ ABC and ∠ ACB intersects at point o

It is known that, as shown in the figure, in △ ABC, the bisector of ∠ ABC and ∠ ACB intersects at point o

It is proved that: the bisector of ∵ ABC and ∵ ACB intersects at point O, ∵ OBC = 12 ∵ ABC, ∵ OCB = 12 ∵ ACB, ∵ OBC + ∵ OCB = 12 (∵ ABC + ∵ ACB). In △ OBC, ∵ BOC = 180 ° - (∵ OBC + ? OCB) = 180 ° - 12 (? ABC + ? ACB) = 180 ° - 12 (180 ° - a) = 90 ° + 12 ? a

If you haven't learned equation, you can't use equation to solve it, you can only use arithmetic
Xiaohua spent 300 yuan on his coat, hat and shoes. The coat was 80 yuan more expensive than the shoes. He spent 240 yuan more on his coat and shoes than the hat. How much did he spend on his shoes
Because we can't set unknowns without learning equations, we can only use arithmetic formulas

300-240=60
60÷2=30 …… This is a hat
300-30=270 …… This is the sum of coat and shoes
270-80=190
190÷2=95 …… This is the price of the shoes

A vector coordinate parallel to vector a = (1, - 2,3) is ()
A.1,3,2
B.-1,-3,2
C.-1,3,-2
D.1,-3,-2

C, a = K (,) are parallel, k = - 1

A triangle ABC ad bisector angle a be bisector angle B A is 120 degrees find the degree of angle bed

The ∠ AEB is the outer angle of ∠ bed ∠ AEB = 180 ° - ∠ Abe - ∠ EAB ∫ BAC = 120 ° ad bisection ∠ BAC ∫ EAB = 60 ° (the size of ∠ ABC is not given, but ∠ Abe = 1 / 2 ∠ ABC can be obtained by the same principle), then ∠ AEB = 180 ° - ∠ Abe - ∠ EAB = 120 ° - 1 / 2 ∠ ABC (if △ ABC is an isosceles triangle, then ∠ Abe = 1

If x = 16, y = 18, then the value of (2x + 3Y) 2 - (2x-3y) 2 is______ ．

(2x + 3Y) 2 - (2x-3y) 2, = (2x + 3Y) 2 - (2x-3y) 2, = (2x + 3y-2x + 3Y) (2x + 3Y + 2x-3y), = 6y · 4x, = 24xy, when x = 16, y = 18, the original formula = 24 × 16 × 18 = 12. So the value of algebraic formula (2x + 3Y) 2 - (2x-3y) 2 is 12

Given that P is a point on the ellipse X25 + y24 = 1, and the area of the triangle with P and focus F1, F2 as vertices is equal to 1, then such a point P has ()
A. 1 B. 2 C. 3 d. 4

The standard equation of∵ellipse is X25 + y24 = 1, ∵ F1F2 | = 2; let the coordinates of point p be (x, y), ∵ p be a point on the ellipse, and the area of the triangle with the vertex of point P and focus F1, F2 is equal to 1, ∵ y = ± 1. Substituting y = ± into the elliptic equation, the coordinates of point P are (152,1), (152, - 1), (- 152,1) and (- 152, - 1)

As shown in the figure, in △ ABC, ∠ B = ∠ C = 50 °, BD = CF, be = CD, then the degree of ∠ EDF is______ ．

As shown in the figure, in △ BDE and △ CFD, BD = CF, B = C = 50 ° be = CD, ≌ BDE ≌ CFD (SAS), ≌ BDE = CFD, ≌ EDF = 180 ° - (≌ BDE + CDF) = 180 ° - (≌ CDF) = 180 ° - (≌ C) = 50 ° and ≌ EDF = 50 °, so the answer is: 50 °

If x is a real number, y is a pure imaginary number and satisfies 2x-1 + 2I = y, then x =?, y =?
How to write it

Y is a pure imaginary number
Let y = AI, a be a real number and not equal to 0
Then 2x-1 + 2I = y = AI
So 2x-1 = (A-2) I
The left is real, so the right is real
Then only A-2 = 0, a = 2,
Then 2x-1 = (A-2) I = 0
So x = 1 / 2, y = 2I

Let f (x) = LNX ax, a ∈ R. when a = - 1, the equation 2mf (x) = x ^ 2 (M > 0) with respect to X has a unique real solution

If 2mf (x) = x ^ 2 (M > 0) has a unique real solution, it can be seen that the image of Y1 = x ^ 2 and y2 = 2m (LNX ax) has a unique intersection x0, where the two functions have the same tangent. The derivative function of Y1 is 2x, the derivative function of Y2 is 2m / x-2ma, 2m / x-2ma = 2x has a unique solution, that is, x ^ 2 + max-m = 0 has a unique real solution

It is known that the waist length of an isosceles triangle is 10 cm and the height of its bottom edge is 8 cm

1 / 2 bottom = root (10 ^ 2-8 ^ 2) = 6
Bottom = 12
Area = 1 / 2 * 8 * 12 = 48