If the point P (m, 1-2m) is in the fourth quadrant, then the value range of M is

In the fourth quadrant, so the abscissa is greater than 0 and the ordinate is less than 0
That is, M > 0
1-2m1/2

1000 milliseconds is equal to 1 second. How many milliseconds are 10 minutes and 15

Ten minutes is 600000 milliseconds, fifteen minutes is 900000 milliseconds, please accept

X ^ 2 / 2m-y ^ 2 / (m-1) = 1 represents the ellipse with the focus on the Y axis, and the value range of M is obtained

X^2/2M+Y^2/(1-M)=1
∵ ellipse
∴2M>0,1-M>0
∴0

The known vectors a = (cosx, SiNx), B = (- cosx, COS), C = (- 1,0)
① If x = π / 6, find the angle between vector a and vector B
② When x ∈ [π / 2,9 π / 8], find the maximum value of function f (x) = 2A · B + 1

Let's talk about the steps. There are some troubles in the process of writing. I won't ask you again
First question: take x into vectors a and B and find out the specific a and B. then use the angle formula of vectors
The second question: first, take the vectors a and B into f (x), and then simplify them so that f (x) contains only SiNx or cosx, and then find the maximum value of the coincidence function

Given that the intersection of the line y = KX + 2K + 1 and the line y = - x-4 is in the fourth quadrant, try to find the value range of K

It can be seen from the title
kx+2k+1=-x-4 x=-（2k+5)/(k+1)
∵ the intersection is in the fourth quadrant
∴x＜0 -（2k+5)/(k+1)＜0
X ＞ - 1 or X ＜ - 2.5

If f (x) = (- 1 / 2) x & # 178; + bln (x + 2) is a decreasing function on (- 1, ∞), then the value range of B?

f(x)=(-1/2)x²+bln(x+2)
f'(x)=-x+b/(x+2)
The function f (x) is a decreasing function on (- 1, ∞)
Then when x > - 1, f '(x) ≤ 0 is constant
That is - x + B / (x + 2) ≤ 0
b/(x+2)≤x
B ≤ x ^ 2 + 2x
∵x^2+2x=(x+1)^2-1
x>-1
∴(x+1)^2-1>-1
∴b≤-1

If the edge length of an uncovered cube box is a centimeter, the volume of the box is v = cubic centimeter and the surface area is s = square centimeter

The volume is very regular, v = A & # 179;
The surface area is about to take out a surface, because it's a box without a cover
So the surface area s = 5A & # 178;

Given that the range of function y = (2aX + b) / (the square of X + 1) is [- 1,4], find the value of a and B
Those who answer well will score

From y = (2aX + b) / (the square of X + 1), we get
Y * x ^ 2-2ax + (y-b) = 0, which is regarded as the quadratic equation of X,
The discriminant of its root is 4 * a ^ 2-4y * (y-b),
According to the condition, 4 * a ^ 2-4y * (y-b) > = 0,
That is, the solution set of Y ^ 2-B * Y-A ^ 2 is: - 1

One addend is a pair of negative rational numbers whose value is equal to 1 / 3, and the other addend is the opposite of - 1 / 2. The sum of these two numbers is equal to

Negative rational number of absolute value of 1 / 3 = - 1 / 3
-The opposite of 1 / 2 = 1 / 2
-1/3+1/2=1/6

The number of zeros of function f (x) = lgx cos 2x

A:
f(x)=lgx-cos2x=0,x>0
lgx=cos2x

If the point P (m, 1-2m) is in the fourth quadrant, then the value range of M is