The maximum value of quadratic equation s = 12 π X-6 π x ^ 2 Find the maximum value of S = 12 π X-6 π x ^ 2

The maximum value of quadratic equation s = 12 π X-6 π x ^ 2 Find the maximum value of S = 12 π X-6 π x ^ 2

This is the problem of formula
S=-6π（X^2-2X+1)+6π
=-6π(X-1)^2+6π
Because there is no limit to the value of X, when x = 1, the equation has a maximum value of 6 π

If the maximum value of the quadratic function f (x) = (LGA) x2 + 2x + 4lga is 3, find a

For convenience, let LGA = B, then the function of quadratic equation is f (x) = b * x2 + 2x + 4B

On the problem of quadratic equation
Proof: for any real number x, y, inequality
x^2-xy+y^2-2x+y+2.5≥
All set up

The second power of X and the second power of y must be greater than or equal to 0, so it is greater than or equal to 2.5. I believe I am right

Does the motion of pendulum belong to translation or rotation?
The teacher said it's not translation or rotation in class, but I think it's rotation

Rotation

A zero vector has multiple directions, right?

A vector whose module is equal to zero is called a zero vector. Note that the direction of the zero vector is arbitrary. But we stipulate that the direction of the zero vector is parallel to any vector, but not perpendicular. Because the direction is arbitrary, the direction of the zero vector is any direction in space. There are multiple directions, so it is better to say any direction

As shown in the figure, AE ⊥ AB, AF ⊥ AC, AE = AB, AF = AC are known

The following: (1) in △ ABF and △ AEC, \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\8747; ade = 90 °, ∫ ad In △ BDM, ∠ BMD = 180 ° - ∠ ABF - ∠ BDM = 180 ° - 90 ° = 90 ° so EC ⊥ BF

Let m be the set of functions f (x) satisfying the following conditions: when | x1 | ≤ 1, | x2 | ≤ 1, | f (x1) - f (x2) | ≤ 4 | x1-x2 |, if G (x) = x ^ 2 + 2x-1, then the relation between G (x) and M is?
I can't understand the next two. For example, why can | g (x1) - G (x2) | be less than | X1 ^ 2 + 2x1-x2 ^ 2-2x2 | ah?

|When x1 | ≤ 1, | x2 | ≤ 1
|g(x1)-g(x2)|≤|x1^2+2x1-x2^2-2x2|
=|(x1-x2)(x1+x2+2)|
=|(x1-x2)||(x1+x2+2)|①
∵0≤x1+x2+2≤4
Therefore, ① ≤ 4|x1-x2|
That is g (x) ∈ M

In the triangle ABC, ∠ ABC = ∠ C = 2 ∠ a, BD is the bisector of ∠ ABC, find the degree of ∠ A and ∠ ADB
Fast

Let a = x, then ABC = 2x
∠A+∠ABc+∠C=180°
So: x + 2x + 2x = 180 degree
x=36°
∠ABC=2x=72°
And because BD is the bisector of ABC
So: ∠ abd = ∠ ABC ／ 2 = 72 °／ 2 = 36 °
∠ADB=180°-∠A-∠ABD
         =180°-36°-36°=108°
A: the degree of ∠ A is 36 ° and the degree of ∠ ADB is 108 °

A (- 2, - 3) B (2,1) C (1,4) d (- 1, - 4) judge whether AB vector and CD vector are collinear

Vector AB = (4,4), vector CD = (- 2, - 8)
If the vector AB and the vector CD are polygonal, then there is a vector CD with the vector AB = in times (in is a real number)
Obviously, the vector AB and the vector CD are not collinear
Or consider this: because 4 * (- 8) - (- 2) * 4 = - 24 is not equal to 0, the vector AB and the vector CD are not collinear

As shown in the figure, in △ ABC, the bisector of ∠ a = a, ∠ ABC and ∠ ACD intersects at point A1 to get ∠ A1; the bisector of ∠ a1bc and ∠ a1cd intersects at point A2 to get ∠ A2; & nbsp If the bisector of ∠ a2010bc and ∠ a2010cd intersects at the point a2011, we can get ∠ a2011, then ∠ a2011=______ ．

The bisector of ∵ - ABC and ∵ ACD intersects at point A1, ∵ - A1 = 180 ° - 12 ∵ ACD - ∵ acb-12 ∵ ABC = 180 ° - 12 (∵ - ABC + ∵ a) - (180 ° - A - ∵ ABC) - 12 ∵ ABC = 12 ∵ a = A21; similarly, ∵ A2 = 12 ∵ A1 = A22 Therefore, the answer is: a22011