Inequality system: if the solution set of 2x − a ＜ 0x + 3 ＞ 2x − 1 is all negative numbers, then the value range of a is () A. a=0B. a＜8C. a＜6D. a＜4

Inequality system: if the solution set of 2x − a ＜ 0x + 3 ＞ 2x − 1 is all negative numbers, then the value range of a is () A. a=0B. a＜8C. a＜6D. a＜4

From (1) we get: X ＜ A2. From (2) we get: X ＜ 4. Because the solution set of the system of inequalities 2x − a ＜ 0x + 3 ＞ 2x − 1 is all negative numbers, so A2 = 0, a = 0
If the product of integer solutions of inequality 1-2x & lt; 11 (1) x & lt; m (2) is negative, then the value range of M is zero
1-2x＜11,x＜m
x＞-5,x＜m
-5＜x＜m
Because the product of its integer solutions is negative (then negative integers must be odd and cannot have zero solutions)
① When the integer solution is only - 4, m satisfies - 4 < m ≤ - 3
② When the integer solution is only - 4, - 3, - 2, m satisfies - 2 < m ≤ - 1
So the value range of M is {m | - 4 ＜ m ≤ - 3 or - 2 ＜ m ≤ - 1}
Speechless
Given that the solution set of inequality 3 + K (X-2) - 4x > k (x + 3) about X is negative, then the value range of K is ()
A. k≥35B. k≤35C. k＞35D. k＜-35
Inequality 3 + K (X-2) - 4x ＞ K (x + 3), remove brackets to get: 3 + kx-2k-4x ＞ KX + 3k, transfer and merge to get: X ＜ 3 − 5k4, according to the meaning of the problem: 3 − 5k4 ≤ 0, solution: K ≥ 35
If there is no solution for the system of inequalities 9-5x ﹥ 4 ﹥ 1 x ﹥ a, then the value range of a is a: a ﹤ 1 B: a = 1 C: a ﹥ 1 D: a ≥ 1
Be more specific, speed
(9-5x)/4＞1
(9-5x)>4
9-4>5x
If XA has no intersection, then a > = 1
Substituting a = 1 into x > a means x > 1 and X
If a point starts from the origin, it first moves 1 unit to the right, then 2 units to the up, 3 units to the right, and 4 units to the up. After moving 100 times, the position of the point is
One time to the right, one time to the up, a total of 100 times, then moved to the right 50 times, followed by translation 1, 3, 5 The last abscissa is 3 + 1 + 99 +99 = (1 + 99) × 50 △ 2 = 2500, moved up 50 times, followed by translation 2, 4, 6 100 units, so finally, the ordinate is 2 + 4 + 6 + +1...
The product of (a1 + A2 +... + an) (x1 + x2 +. + xn) has several terms
N^2
It is known that, as shown in the figure, EF is the median line of the triangle ABC, and the intersection line EF of the bisector of the outer angle ACG is at point D
prove:
∵ EF is the median of △ ABC
∴EF‖BC
∴∠EDC=∠DCG
∵∠DCG=∠ACD
∴∠ACD=∠EDC
∴FC =FD
∴FA=FC=FD
A right triangle is an ADC
∴AD⊥CD
Extend the extension line of ad-bc to M
Ad = DM is enough
DE‖BC
∠FDC=∠DCG
∠DCG=∠FCD
∠FDC=∠FCD
CF=DF=AF
∠FAD=∠FDA
∠FAD+∠ADF+∠FDC+∠FCD=180°
2∠ADF+2∠FDC=180°
∠ADc=90°
∵ EF is the median line of triangle ABC, and AF = FC ∵ FDC = ∵ DCG ∵ FCD = ∵ DCG ∵ FDC = ∵ FCD ∵ FD = FC ∵ AF = FC = CD. Triangle ADC is a right triangle (according to the inverse theorem that the middle line on the hypotenuse of a right triangle is equal to half of the length of the hypotenuse), that is: ad ⊥ CD
In the expansion of binomial (x + 3x) n, the sum of the coefficients is a, the sum of the coefficients of binomial is B, and a + B = 72, then n=______ .
Let X in the binomial be 1, the sum of the coefficients a = 4N, and the sum of the coefficients B = 2n ∵ a + B = 72 ∵ 4N + 2n = 72, the solution is n = 3, so the answer is: 3
A moving point starts from the origin of the coordinate plane, moves 1 unit to A1 (1,0) to the right, then moves 1 / 2 unit to A2 (1,1 / 2) to the top, and then press left, bottom, right, top The length of each movement is half of the length of the previous movement, and the distance between the limit position of the moving point and the origin is calculated
The coordinates of the limit position of the moving point on the X axis are: 1 - (1 / 2) ^ 2 + (1 / 2) ^ 4 - (1 / 2) ^ 6 + (1 / 2) ^ 8 +. = 1 / [1 - (- 1 / 2) ^ 2] = 4 / 5. The coordinates of the limit position of the moving point on the Y axis are: 1 / 2 - (1 / 2) ^ 3 + (1 / 2) ^ 5 - (1 / 2) ^ 7 +. = (1 / 2) / [1 - (- 1 / 2) ^ 2] = 2 / 5. The distance between the limit position of the moving point and the origin = √ [(4 / 5) ^ 2 + (2 / 5) ^ 2
Known (1 + x) + (1 + x) 2 + + (1 + x )n = a0 + a1x + a2x2 + … +Anxn, if a1 + A2 + a3 + +An-1 = 29-n, then the value of natural number n is ()
Here's what I do: let x = 0, a 0 = n, an = 1
What's next?
Let x = 1
Then 1 + x = 2
So on the left is 2 + +2^n=2^(n+1)-2
Right = A0 + A1 + +a(n-1)+an=n+(29-n)+1=30
That is 2 ^ (n + 1) - 2 = 30
2^(n+1)=32=2^5
n+1=5
N=4