It is known that SiNx = 2 / 3 and Pai / 2

It is known that SiNx = 2 / 3 and Pai / 2

pai/2sinpi/4=√2/2>0
So sin (x / 2) = (√ 15 + √ 3) / 6
π/2
It is known that sin (x + quarter PAI) = 1 / 3 x ∈ (half Pai, PAI) SiNx =?
Solution;
x∈[π/2,π]
x+π/4∈[3π/4,5π/4]
∵sin(x+π/4)=1/3
∴cos(x+π/4)=-2√2/3
∴sinx
=sin(x+π/4-π/4)
=sin(x+π/4)cosπ/4-cos(x+π/4)sinπ/4
=√2/2(1/3-2√2/3)
=√2/6-2/3
=(√2-4)/6
In response to the call of national fitness issued by the government, a unit plans to build a 60m2 rectangular gymnasium ABCD in a rectangular Hall of 20m and 11m in length and width respectively. Among the four walls of the gymnasium, the old walls of the hall are still used on both sides (as shown in the plan). It is known that the cost of repairing the old walls is 20 yuan / m2, and the cost of building new (including decoration) walls is 80 yuan / m2 The height of the body room is 3M, the length of an old wall AB is XM, and the total investment of building the gym wall is y yuan. (1) find the functional relationship between Y and X; (2) in order to make rational use of the hall, the independent variable x must meet the condition: 8 ≤ x ≤ 12. When the investment is 4800 yuan, what is the total length of using the old wall?
(1) According to the meaning of the question, ab = x, ab · BC = 60, so BC = 60x. Y = 20 × 3 (x + 60x) + 80 × 3 (x + 60x), that is y = 300 (x + 60x). (2) y = 4800 is replaced by Y = 300 (x + 60x), and 4800 = 300 (x + 60x) is obtained .
A problem of inverse proportion function in second grade mathematics
In the inverse scale function y = k △ x, when the value of x increases from 4 to 6, the value of Y decreases by 3. Find the expression of the inverse scale function
Consider y = k △ x as a fraction
Then K / 4-K / 6 = 3
k=36
The expression of this inverse scale function is y = 36 / X
Given a (- 3,0), B (0,6), the area of △ AOB is divided into two parts of 1:2 by the straight line passing through the origin
RT, there are two solutions, there are two solutions,
Let the intersection point of the line and ab be p, then the ratio of AP to BP is 1:2
First: AP: BP = 2:1
Then the coordinates of point P are (- 1,4) and the linear equation is -
y=-4x
The second one is BP = 2:1
Then the coordinate of point P is (- 2,2) and the linear equation is -
y=-x
Let the intersection of the line and ab be (x, y), and know x0. So the area ratio of the separated triangles is - 6x: 3Y = 1:2 (2:1), so the solution is y = - 6x, or y = (- 3 / 2) X
Given a (- 3,0), B (0,6), the area of △ OAB is divided into two parts of 1:3 by the straight line of origin o, and the function analytic formula of this straight line is obtained
Only the answer will do
The analytic expression seems to be negative
Y = 2 / 3x or y = 3 / 2Y
If we know that the line L passes through the point P (2,1) and intersects with the positive half axis of x-axis and y-axis at two points a and B respectively, and O is the origin of the coordinate, then the minimum OAB area of the triangle is ()
A. 1B. 2C. 3D. 4
Let the line l be XA + Yb = 1 (a > 0, b > 0), because the line L passes through the point P (2, 1), then 2A + 1b = 1. The area of △ OAB is s = 12ab. For 2A + 1b = 1, using the mean inequality, we obtain that 1 = 2A + 1b ≥ 22a · 1b = 22ab, that is, ab ≥ 8. Therefore, the area of △ OAB is s = 12ab ≥ 4
Given that the line L passes through the point (2,1), and intersects with the x-axis and y-axis at two points a and B respectively, and O is the coordinate origin, the minimum OAB area of the triangle is obtained
So let the mean area of the triangle (AB) and the triangle (k) = 0.4b + 1 (k) = 0.2K + 1 and (k) = 0.2K + 1 (k) = 0.4b (k) = 0.2K + 1 (k) = 0.2K + 1) and (k) = 0.2K + 1 (k) = 0.2K + 1 (k) = 0.2K + 1 (k) = 0.4b (k) = 0.4b > (k) = 0.2K + 1 (k) and (k) = 0.2K + 1 (k) = 0.2K + 1 (k) = 0.2K + 1 (k) = 0.2K + 1 (k) and (k) = 0.2K + 1 (k) = 0.2K + 1 (k) = 0.2K + 1 (
Hello
Let the linear equation be y = KX + B, and the point (2,1) be substituted
1=2k+b
b=1-2k
The linear equation is y = KX + 1-2k
The intersection points of x-axis and y-axis are
(0,1-2k),((2k-1)/k,0)
S triangle OAB = 1 / 2 │ OA * ob │ = 1 / 2 (1-2k) * (2k-1) / K │ = 1 / 2 (4-4k-1 / k) │
When k = 1 / 2, there is a minimum value of 0
The minimum value of △ OAB area is ()
A. 4B. 42C. 2D. 22
∵ it is known that the positive half axes of the line L and x-axis and y-axis intersect at two points a (a, 0) and B (0, b), respectively. It is easy to get a > 0.b > 0, and ∵ 2A + 1b = 1 ≥ 22ab ∵ ab ≥ 8, and ∵ s △ OAB = 12ab ≥ 4, then the minimum value of △ OAB area is selected as a
If the line L passes through the point P (4,2) and intersects the x-axis and the positive half axis of the y-axis at two points a and B respectively, and O is the origin of the coordinates, then the minimum area of AOB of s triangle is obtained
Let m > 0, n > 0,
From the intersection of X and Y axes to the positive half axis,
Let y = - MX + n
Because the straight line passes through the point P (4,2), so - 4m + n = 2
Set points a (a, 0), B (0, b)
When x = 0, B = n
So, s △ AOB = 1 / 2 * a * b = N2 / 2m = 2m + 1 / 2m + 2
Because m > 0, 2m + 1 / 2m ≥ 2 (2m * 1 / 2m under root)
That is 2m + 1 / 2m ≥ 2
Therefore, s △ AOB ≥ 4
The minimum AOB area of triangle is 4