The graph of the linear function y = - 3x + 2 does not pass () A. First quadrant B. second quadrant C. third quadrant D. fourth quadrant

The graph of the linear function y = - 3x + 2 does not pass () A. First quadrant B. second quadrant C. third quadrant D. fourth quadrant

∵ k = - 3 ＜ 0, B = 2 ＞ 0, the image passes through one, two, four quadrants, not through the third quadrant
Given the set a = {x | x2-3x + 2 = 0}, B = {x | x2-2x + A-1 = 0}, B &; a, find the range of A
I can't understand the title, what is a, but I can say how to do it, first use a to find x = 1 or 2, and then use the subordination of B and a, maybe substitute B to find the scope of small a
The image of the first-order function y = - 1 + 3x does not pass through the quadrant, and Y increases with the increase of X?
Passing point (0, - 1) (1 / 3,0)
The image of linear function y = - 1 + 3x does not pass through the second quadrant,
Slope k > 0
Y increases with the increase of X
Three. reduce.
Quadrant 2
k> 0y increases with the increase of X
Let u = R, a = {x ∈ R | a ≤ x ≤ 2}, B = {x ∈ R | 2x + 1 ≤ x + 3, and 3x ≥ 2}. (1) if a = 1, find a ∪ B, (∁ UA) ∩ B; (2) if B ⊆ a, find the value range of real number a
(1) If a = 1, if a = 1, then a = {{x | 1 ≤ x ≤ 2}, B = {x | x ≤ 2, and X ≥ 23} = {x | 23 ≤ x ≤ 2}, if a = 1, then a leta ∪ B = {x | 1 ≤ x ≤ 2}, B = {{x | 1 ≤ 1 ≤ x ≤ 2}, B = {x {x | 1 ≤ 1 ≤ x ≤ x ≤ 2}, B = {{{x ∁ UA) \\\\\\\\\\8705 \\\\\\(2) B = {x | x ≤ 2, and X ≥ 23} = {x | 23 ≤ x ≤ 2}, a = {x ∈ R | a ≤ x ≤ 2}, and ∵ B ⊆ a, ｜ a ≤ 23, that is, real numbers The range of a is: a ≤ 23
How many quadrants does the image of linear function y = -- 3x-1 not pass through
The image of linear function y = -- 3x-1 does not pass through the first quadrant
3x + 7 is greater than or equal to x + 3, 2x + 5-1 is less than 8-x
x+7>=x+3,
Both sides - x-3
4>0,
X is any real number
(2x + 5) - 1
The solution of 3x + 7 ≥ x + 3, 2x / 3-1 < 8 / 3-x is: X ≥ - 2, x < 11 / 5
∴-2≤x＜11/5
Which quadrant does the image of a linear function y = - 3x-2 not pass?
Not through the first quadrant
first quadrant
Not through the first quadrant.
When x = 0, we substitute y = - 2. This point is set as point a (0, - 2)
Similarly, let y = 0, - 3x-2 = 0, then x = (- 3 / 2). Be B (- 3 / 2, 0)
Draw two points a and B on the graph and see that they do not pass through the first quadrant.
It's over.
If 2x power of a = 2 / 3, then (3x power of a + negative 3x power of a) / (x power of a + negative x power of a) is equal to
According to the sum of cubes formula, a 3x + a - 3x = (a x + a - x) (a 2x + a - 2x - 1) can be substituted into the formula in the problem (3x power of a + negative 3x power of a) / (x power of a + negative x power of a) by reduction, a 2x + a - 2x - 1 can be obtained, a 2x = 2 / 3, because a 2x times a - 2x = 1, a - 2x = 3 / 2, so the final result is 2 / 3 + 3 / 2-1 = 7 / 6
If the image of a linear function y = 3x + m-2 does not pass through the second quadrant, then the value range of M is ()
A. m≤2B. m≤-2C. m＞2D. m＜2
If the image of the first-order function y = 3x + m-2 does not pass through the second quadrant, then m-2 ≤ 0 and the solution m ≤ 2
How much is x / (x + 1) = 2x / (3x + 3) + 1!
It's all right,
X / (1 + x) = (5x + 3) / (3x + 3), X ≠ - 1, i.e
3x^2+3x=5x^2+8x+3,
2x^2+5x+3=0,
The solution is x = - 3 / 2 or x = - 1 (rounding),
So, x = - 3 / 2
x=-3/2
x/(x+1)=2x(3x+3)+1
1-1/(x+1)=2/3-2/(3x+3)+1
-2/3=1/(3x+3)
x+1=-1/2
x=-3/2