The root of equation (x-3) (x + 6) = 10 is______ ．

The root of equation (x-3) (x + 6) = 10 is______ ．

∵（x-3）（x+6）=10∴x2+3x-28=0x=−3±9+4×282=−3±112∴x1=4，x2=-7．

Decomposition factor: 5x & # 178; - 13xy-6y & # 178;

5x²-13xy-6y²
＝(5x+2y)(x-3y)

How to factorize 4xy-5x + 6y = 15

4XY-5X+6Y=15
4XY+6Y=15+5X
2Y(2X+3)=5(3+2X)
2Y=5
Y=5/2

How many seconds is 2000 milliseconds

2 seconds. The unit conversion of time: 1 second = 1000 milliseconds (MS) 1 millisecond = 1 / 1000 seconds (s) 1 second = 1000000 microseconds (μ s) 1 microseconds = 1 / 1000000 seconds (s) 1 second = 1000000000 nanoseconds (NS) 1 nanoseconds = 1 / 1000000000 seconds (s) 1 second = 1000000000 picoseconds (PS) 1 picoseconds = 1 / 10000000000 seconds (s)

If the line y = KX + 1 (K ∈ R) and the ellipse X25 + y2m = 1 always have a common point, then the value range of M is ()
A. [1，5）∪（5，+∞）B. （0，5）C. [1，+∞）D. （1，5）

If y = KX + 1x25 + y2m = 1, then (M + 5k2) x2 + 10kx + 5-5m = 0, (m ＞ 0, m ≠ 5) ∵ the line y = KX + 1 (K ∈ R) and the ellipse X25 + y2m = 1 have a constant common point, that is, 100k2-20 (1-m) (M + 5k2) ≥ 0, then M2 + 5mk2-m ≥ 0, ∵ m ＞ 0, ∵ m ≥ - 5k2 + 1 ≤ 1, ∵ m ≥ 1 (m ≠ 5)

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

If we know that the intersection of the images of two linear functions y = x + 3K and y = 2x-6 is on the Y axis, then the value of K is ()
A. 3B. 1C. 2D. -2

∵ the intersection point of the image of the first-order function y = x + 3K and y = 2x-6 is on the y-axis, ∵ 3K = - 6, and the solution is k = - 2

Finding the range of F (x) = 3x-1 / 2-x

First of all, you are missing the domain. For the moment, the default value is x, which is not equal to 2
f(x)=3*(x-2)+5/2-x
=3*(x-2)/2-x+5/2-x
=-3+5/2-x
Because x is not equal to 2, so the range is, not equal to minus 3

To make a cuboid box with a volume of 396cm cubic, a height of 6cm and a bottom surface length 5cm more than the width, how many centimeters should the bottom surface be?

Let the width be x, then the length be x + 5
x(x+5)*6=396
X = 6 (x = - 11, rounded off)
x+5=11
A: the bottom is 11cm long and 6cm wide

It is known that the equation (a square + 1) x square - 2 (a + b) x + b square + 1 = 0 has a real root. When - 3 is less than a and less than - 1, the value range of B is obtained