A focus f passing through the hyperbola x2a2 − y2b2 = 1 (a > 0, b > 0) leads to the perpendicular FM of its asymptote, the perpendicular foot is m, and the intersection Y axis is e. if M is the midpoint of EF, then the eccentricity of the hyperbola is () A. 2B. 3C. 3D. 2

A focus f passing through the hyperbola x2a2 − y2b2 = 1 (a > 0, b > 0) leads to the perpendicular FM of its asymptote, the perpendicular foot is m, and the intersection Y axis is e. if M is the midpoint of EF, then the eccentricity of the hyperbola is () A. 2B. 3C. 3D. 2

As shown in the figure, take the right focus f (C, 0), asymptote y = Bax. ∵ FM ⊥ OM, ∪ to get the equation of straight line FM as y = − AB (x − C), let x = 0, solve y = ACB, ∪ e (0, ACB).. the midpoint m (C2, ac2b) of line Fe, and the midpoint m is on asymptote y = Bax, ∪ ac2b = BA × C2, solve a = b.. The eccentricity of the hyperbola is e = CA = 1 + b2a2 = 2

What is the minimum integer solution that makes the inequality - 2x-2 / 3 ≤ 6 / 11x + 1 / 2 hold? (to write the process) I have urgent need, fast~~~~~~~~

-2x-2 / 3 ≤ 11x / 6 + 1 / 2
transposition
23x of 6 ≥ - 7 of 6
Divide both sides by 23 / 6 at the same time
X ≥ - 7 / 23
Minimum integer = 0

The sum of all integer solutions of inequality-1 ≤ 3-2x < 6 is______ The product of all integer solutions is______ ．

From - 1 ≤ 3-2x, the solution is: X ≤ 2, from 3-2x < 6, the solution is: − 32 < x, so the solution of the inequality system is: − 32 < x ≤ 2, and the integer solution is: - 1, 0, 1, 2. So the sum of all integer solutions is: - 1 + 0 + 1 + 2 = 2, and the product of all integer solutions is: - 1 × 0 × 1 × 2 = 0

Solving the system of inequalities - 1 less than or equal to 3 molecules 2x - 1 less than 6

It is divided into two inequalities
-1≤(2x-1)/3.（1）
(2x-1)/3