Let x ∈ (0, π 2), then the minimum value of the function y = 2sin2x + 1sin2x is______ ．

Let x ∈ (0, π 2), then the minimum value of the function y = 2sin2x + 1sin2x is______ ．

∵ y = 2sin2x + 1sin2x = 2 − cos2xsin2x = k, take the left semicircle of a (0, 2), B (- sin2x, cos2x) ∈ x2 + y2 = 1, as shown in the figure, Kmin = tan60 ° = 3

Given that the lines y = x + 3K and y = 2X-4 intersect at a point on the x-axis, then k =

Since two straight lines intersect on the X axis, then y = 0, x = 2, x = 2, y = 0 is brought into the first equation and the solution is k = - 2 / 3

Find the range of F (x) = x ^ 2-3x + 1 / x + 1

f(x)=((x+1)^2-5(x+1)+5)/(x+1)
=(x+1)-5+5/(x+1)
x>1
f(x)≥2√5-5
x

The edge length of a square box without cover is a decimeter, and the surface area of the box is () square decimeter
A. 4a2B. 5a2C. 6a2

A × a × 5 = 5A2 (square decimeter). A: the surface area of this box is 5A2 square decimeter

Given that the equation AX2 + C = 0 (a ≠ 0) has real roots, the relation between a and C is ()
A. C = 0b. C = 0 or a, C different sign C. C = 0 or a, C same sign D. C is an integral multiple of A

If △ = - 4ac ≥ 0 and a ≠ 0, then only C = 0 or a, C is a different sign

The sum of the opposite number of the absolute value of - 1 / 3 and the opposite number of 3 and 2 / 3 is

According to the meaning of the title:
- - 1 / 3 + (- 3 and 2 / 3)
= -1/3-29/3
= -30/3
= -10

Given the function f (x) = (1 / 2) ^ x (x ≥ 2), f (x) = f (x + 1) (x < 2), then the value of F (log2 (3)) is

log3(3)2
So the original formula = (1 / 2) ^ log2 (6)
=2^[-log2(6)]
=2^[log2(1/6)]
=1/6

-If the absolute value of a is equal to - A

Solution; | - a | = - A
a

Let a (- 2, Y1) B (1, Y2) C (2, Y3) be a point on the parabola y = - (x + 1) 2 (square) + m, then the size relation of Y1, Y2, Y3 is

The symmetry axis X = - 1, so the image is symmetrical about the straight line x = - 1, the opening is downward, - 1, the left side increases, the right side decreases,
∴y1＞y2＞y3

Using MATLAB to find the derivative of T for the following function
It is known that X1 = 360 * t; X3 = arctan (200 + 100sin (x1)); S3 = 100cos (x1) / cos (x3); S3 = 100cos (x1) / cos (x3);
Find the derivatives of X3 and S3 to t, and find the derivative function image

Results: DX3 = 36000 * cos (360 * t) / (1 + (200 + 100 * sin (360 * t)) ^ 2) DS3 = - 36000 * sin (360