It is known that the area of the triangle formed by the image of the first-order function y = 2x + B and the coordinate axis is 12, so B can be obtained

It is known that the area of the triangle formed by the image of the first-order function y = 2x + B and the coordinate axis is 12, so B can be obtained

y=2x+b
Let x = 0, then y = B
Let y = 0, then x = - B / 2
1/2*|b|*|-b/2| = 12
b^2=48
b=±4√3
If the intersection of the image and x-axis and y-axis is (- B / 2,0) (0, b), then according to the meaning of the title, b * (B / 2) * 1 / 2 = 12, B = plus or minus 4 3
The area of the triangle formed by the image and the coordinate axis of the linear function y = 2x + B is B ^ 2 / 2K = 12,
That is, B ^ 2 / 4 = 12,
So B = ± 4 √ 3.
The intersection of image and x-axis is B / 2, and the intersection of y-axis is B; 12 = B x (B / 2) 1 / 2; b = ± 4 √ 3
The area of triangle surrounded by (0,3), (1.5,0) is × 3 × 1.5 = 2.25, and the first-order function is y = - 2x + 3, so the image with area of 9 / 4 y = KX + B is parallel to y=-
Let y = 0, x = - B / 2 - B / 2 absolute value multiplied by B absolute value divided by 2 = 12, B = plus or minus 4 times 3 square root
Let m and p be two nonempty sets, M - (M-P) = {x ∈ m, and X &; P}. According to this rule, M-P (M-P) =?
M-P is the part that belongs to m but not to P
M - (M-P) is the part that belongs to m but does not belong to (M-P), that is, M-P (M-P) = m-p
The area of the triangle formed by the line y = 2x + 4 and the two axes is ()
A. 2B. 4C. 8D. 16
When x = 0, y = 4; when y = 0, x = - 2; so the area of the triangle formed by the line y = 2x + 4 and the two coordinate axes is 12 × 4 ×| - 2 | = 4
Inequality: log2 (x ^ 2-2x + 2) > log2 (2x-1)
It can be seen from the meaning of the title:
The two functions are both increasing functions with the same base. As long as the comparison of true numbers is satisfied, they are equivalent to:
x^2-2x+2>2x-1；
Solve the inequality;
It is worth noting that the true numbers on both sides should be greater than zero;
Finally, we take the intersection of three inequalities;
The solution is over
X3
Find the area of the triangle formed by the image and two coordinate axes of the linear function y = 1 / 2x + 1
Let x = 0 get y = 1, let y = 0 get x = - 2, so the lengths of the two right angles of the triangle enclosed by the image and the coordinate axis are 1 and 2 respectively, and the area is 1 / 2 (1 * 2) = 1
x=0, y=1
y=0, x=-2
Area = 1 / 2 * 1 * 2 = 1
Because the image of the linear function y = 1 / 2x + 1 intersects the X axis, so the Y axis is 0, 0 = 1 / 2x + 1, x = - 2, intersecting the Y axis X=
0y=1/2×0+1=1
So area = 2 × 1 △ 2 = 1
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Solving inequality log2 (4x + 2) ≤ 8
log2(4x+2)≤8=log2(256)
4x+2>0
x>-1/2
4x+2≤256
4x≤254
x≤127/2
-1/2
Domain: X ＞ - 1 / 2
(4x+2)≤256
4x≤254
x≤127/2
∴-1/2＜x≤127 /2
㏒2(4x+2)≤8→㏒2(4x+2)≤㏒2(2^8)→4x+2≤2^8→x≤32
Find the area of the triangle formed by the image and the coordinate axis of the linear function y = 1 / 2x + 1
Let x = 0 get y = 1, let y = 0 get x = - 2, so the lengths of the two right angles of the triangle enclosed by the image and the coordinate axis are 1 and 2 respectively, and the area is 1 / 2 (1 * 2) = 1
Solving inequality log2 (x2-1)
Zero
log2(x2-1)
If we know that the images of the linear functions y = 3 / 2 x + m and y = - 1 / 2 x + n pass through the points a (- 3,0) and intersect with the Y-axis and B and C respectively, then △ ABC
the measure of area
Y = 3x / 2 + m and y = - X / 2 + n both pass a (- 3,0)
That is, 0 = - 3 × 3 / 2 + M = 9 / 2
0=3/2+n n= —3/2
Then the intersection points of y = 3x / 2 + 9 / 2 and y = - X / 2-3 / 2 and Y axis are B (0,9 / 2), C (0, - 3 / 2) respectively
△ABC=3×（9/2+3/2）×1/2=9
Given the set a = {x | y = x + 2}, B = {y | y = x ^ 2 + X + 1}, then the intersection of a and B is equal to
A = {x y = x + 2} = R
B = {y y = x & # 178; + X + 1} = {Y Y ≥ 7 / 3}
A ∩ B = {Y Y ≥ 7 / 3}