A. B are two points on the number axis, the number corresponding to a is - 10, and the number corresponding to B is 90. & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; (1) please write the number corresponding to m point whose distance is equal to a and B. & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp;                     (2) If the electron ant P starts from point B and moves to the left at a speed of 3 units per second, and the other electron ant Q starts from point a and moves to the right at a speed of 2 units per second, how long does it take for the two electron ants to be 35 units apart on the number axis?

A. B are two points on the number axis, the number corresponding to a is - 10, and the number corresponding to B is 90. & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; (1) please write the number corresponding to m point whose distance is equal to a and B. & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp;                     (2) If the electron ant P starts from point B and moves to the left at a speed of 3 units per second, and the other electron ant Q starts from point a and moves to the right at a speed of 2 units per second, how long does it take for the two electron ants to be 35 units apart on the number axis?

(1) 90 - (- 10) = 100, 100 △ 2 = 50. According to the number axis, the corresponding number of m point with the same distance between a and B is 40. (2) before meeting: (100-35) / (2 + 3) = 13 (seconds), after meeting: (35 + 100) / (2 + 3) = 27 (seconds), after 13 or 27 seconds, two electronic ants are 35 unit lengths apart on the number axis
As shown in the figure, we know that there are three points a, B and C on the number axis, AC = 2Ab, and the number corresponding to point a is 400. (1) if AB = 600, find the distance from point C to the origin;
(2) Under the condition of (1), the moving points P, Q and R start from C and a at the same time, where P and Q move to the right and R to the left. As shown in the figure, it is known that the speed of point q is 2 times that of point R, less than 5 unit length / s, and the speed of point P is 3 times that of point R. after 20 seconds, the distance between points P and Q is equal to that between points Q and R, and the speed of moving point q is calculated;
(3) Under the condition of (1), O is the origin, and the moving points P, t and R start from C, O and a respectively, where P and t move to the left and R move to the right, as shown in the figure. The points P, t and R are 20 unit length / s, 4 unit length / s and 10 unit length / s respectively. During the movement, if point m is the midpoint of line Pt and point n is the midpoint of line or, does the value of (PR + OT) / Mn change? If not, the value of (PR + OT) / Mn will not change, Find the value; if it changes, explain the reason
1.AC = 2AB = 1200
Origin of distance from point C: 400-1200 = 800
Speed set point Q
[1200 +20（2X-5?） -20 * 3X = 20（2X-5?）+20×
-80X = -1200
X = 15
alpha
Alpha
As shown in the figure, the numbers represented by a, B and C on the number axis are a, B and C respectively, ab = BC. If | a | > C | > b |, then the origin o of the number axis should be in ()
A. Left side of point a B. between points a and B C. between points B and C D. right side of point C
∵|||||||||||||||||||||||||||||||||||||||||||
Given the function f (x) = cosx cos (x + Π / 2), x = R, if f (x) = 3 / 4, find sin2x
f(x)=cosx-cos(x＋∏/2)=cosx+sinx=3/4
(cosx+sinx)^2=1+2sinxcosx=9/16
sin2x=2sinxcosx=-7/16
Given that the function f (x) = x ^ 2 + (loga + 2) x + logb satisfies f (- 1) = - 2 and has f (x) > = 2x for all real numbers x, find the value of a and B
1.f(-1)=-2,1-(loga+2)+logb=-2
We get loga logb = 1
2.x^2+(loga+2)x+logb-2x>=0
x^2+logax+logb>=0
(x + loga / 2) ^ 2 > = loga ^ 2 / 4-logb
Then loga ^ 2-4logb
Given the function FX = cosx cos {x + π / 2}, X belongs to R. if FX is equal to three fourths, find the value of sin2x
f(x)=cosx-cos(x+π/2)=cosx+sinx=3/4
sin^2x+cos^2x+2sinxcosx=9/16
2sinxcosx=sin2x=9/16-1=-7/16
six-seventeenths
Given the function f (x) = x ^ 2 + (2 + LG a) x + LG B, and f (- 1) = - 2, if the function f (x) = 2x has two equal real roots, find the values of real numbers a and B
F (- 1) = 1 - (2 + LGA) x + LGB = - 2, so: LGA LGB = LG (A / b) = 1F (x) = 2x = x & sup2; + (2 + LGA) x + LGB has two equal real roots, so the discriminant △ = [LGA] & sup2; - 4lgb = 0, substituting LGB = lga-1 into: [LGA] & sup2; - 4lga + 4 = [lga-2] & sup2; = 0, so lga-2 = 0, so LGA = 2
f(-1) = 1 - (2+lg a) + lg b = -2,
So - LG a + LG B = - 1...... （1）
That is LG B / a = - 1, B / a = 1 / 10
F (x) = 2x, that is, x ^ 2 + (2 + LG a) x + LG B = 2x has two equal real roots
Then x ^ 2 + LG a x + LG B = 0 has two equal real roots
Discriminant (LG a) ^... Expansion
f(-1) = 1 - (2+lg a) + lg b = -2,
So - LG a + LG B = - 1...... （1）
That is LG B / a = - 1, B / a = 1 / 10
F (x) = 2x, that is, x ^ 2 + (2 + LG a) x + LG B = 2x has two equal real roots
Then x ^ 2 + LG a x + LG B = 0 has two equal real roots
Discriminant (LG a) ^ 2 - 4LG B = 0
So LG B = (LG a) ^ 2 / 4 substituting (1)
The solution is LG a = 2, a = 100
LG, B = 1, B = 10, LG, B = 1
Binary linear equations
If | cosx | = cos (- x + Π), then the value range of X is?
cos(-x+π）=cos(x-π）=cosx*cosπ+sinx*sinπ=-cosx=|cosx|
So, cosx=
From the meaning of the title, | cosx | = cos (- x + Π) = - cosx.
So cosx
Function f (x) = x ^ 2 + 2x, if there is a real number T, when x ∈ [1, M], f (x + T) ≤ 3x holds, find the range of real number M
F (x + T) ≤ 3xx ^ 2 + (2t-1) x + T ^ 2 + 2T ≤ 0g (x) = x ^ 2 + (2t-1) x + T ^ 2 + 2T in X ∈ [1, M] constant ≤ 0g (1) = T ^ 2 + 4T ≤ 0 → - 4 ≤ t ≤ 0g (m) = T ^ 2 + (2m + 2) t + m ^ 2-m = (T + m + 1) ^ 2-3m-1 ≤ 0 → - √ (3m + 1) - M-1 ≤ t ≤ √ (3m + 1) - m-1t, there is ■ - √ (3M + 1) - M-1 ≤ 0 √ (3m + 1) - M-1
|Cosx | = cos (- x), then the value range of X is______ (in Yuan)
That is | cosx | = cosx
So cosx ≥ 0
So 2K π - π / 2 ≤ x ≤ 2K π + π / 2