Inequality-x + 2y-4 Why must it be a plus sign? Don't you just draw directly?

Inequality-x + 2y-4 Why must it be a plus sign? Don't you just draw directly?

Let's turn the coefficient of X into a positive number
X-2y + 4 > 0
Greater than zero on the lower right and less than zero on the upper left
So choose B
If the plane region represented by the inequality system X − y + 5 ≥ 0y ≥ A0 ≤ x ≤ 2 is a triangle, then the value range of a is______ .
The feasible region satisfying the constraint condition x − y + 5 ≥ 00 ≤ x ≤ 2 is shown in the following figure. It can be seen from the figure that if the plane region represented by the inequality system X − y + 5 ≥ 0y ≥ A0 ≤ x ≤ 2 is a triangle, then the value range of a is: 5 ≤ a < 7, so the answer is: 5 ≤ a < 7
The coordinates of the three vertices of △ ABC are a (4,1), B (2,3) C (6, - 3), respectively. The median line parallel to AB is MB, and the equation of choosing the line Mn is?
Because a (4,1), B (2,3), C (6, - 3) and Mn are the median lines parallel to AB, the coordinates of Mn are (4,0), (5, - 1) respectively, so the equation of Mn is: Y - (- 1) X-5 -------- = ----- 0 - (- 1) 4-5, so x + y-4 = 0
Can MB parallel AB?
If LIM (an * BN) = 0, is Liman = 0 or limbn = 0 right or wrong?
Why?
No, an = 0, n is odd
Even number is one
BN = 1, n is odd
0, n is even
Then an * BN = 0, but an and BN have no limit
Let me give you a straightforward example
If an = n, BN = 1 / N ^ 2
An/Bn=1/n=0
Obviously, none of them is equal to zero
The real number range of M can be obtained by adding 4m-2m-6 under the root sign and 2m greater than or equal to 0
First of all, if we look at the domain of definition, 4m & # 178; - 2m-6 ≥ 0, the solution is m ≥ 3 / 2 or m ≤ - 1;
The square of the inequality is: 4m & # 178; - 2m-6 > 4m & # 178;, and the solution is m
M = 2. Question: 4m-2m-6 is under the root, while 2m is outside the root...
Let f (x) be continuous on [a, b], a
According to the intermediate value theorem of continuous function on closed interval, a continuous function on closed interval must be able to take any value between the maximum value and the minimum value
min(x∈[a,b]){f(x)}
In △ ABC, D is the midpoint of BC. Prove that vector ad = 1 / 2 (vector AB + vector AC)
Take the midpoint e of AB and connect De
So De is the median line of the triangle
So: (vector ED) = (1 / 2) * (vector AC) [vector has directivity, so the alphabetic order can't be reversed here]
And: (vector AE) = (1 / 2) * (vector AB)
So: (vector AE) + (vector ED) = (1 / 2) * [(vector AB) + (vector AC)]
That is, (vector AD) = (1 / 2) * [(vector AB) + (vector AC)]
If LIM (an BN) = 0, then Liman = limbn. Why wrong? N tends to infinity
An is {1, - 1,1, - 1,1, - 1,1, - 1,1, - 1,1, - 1,...}
BN is {- 1,1, - 1,1, - 1,1, - 1,1, - 1,1,...}
an+bn=0
Liman limbn does not exist
For example, take an = BN = (- 1) ^ n
It doesn't necessarily converge.
For a simple example, if the expressions of an and BN are the same and divergent,
If an = BN = n, it can satisfy LIM (an BN) = 0, but Liman and limbn do not exist. Because it tends to infinity.
The square of 3-A under the root sign = the square of 3-A, how much is a?
The square of 3-A under the root sign
=│3-a│≥0
The square of 3-A ≥ 0
The square of 3-A under the root sign = the square of 3-A
3-A = the square of 3-A = 0
A=3
Let v 1 and V 2 be vector spaces over real number fields, and prove that v 1 ∩ V 2 is also vector spaces over real number fields