As shown in the figure, given the line segments a and B, make a line segment that is equal to 3a-2b Drawing

As shown in the figure, given the line segments a and B, make a line segment that is equal to 3a-2b Drawing

As shown in the figure, ① draw ray am, ② take points B, C and D successively on ray am, so that ab = BC = CD = a, ③ take points E and f successively on line Da (from D to a direction), so that de = EF = B, and line AF is the line segment

Square of solving inequality 6x_ x_ twelve

6x²－x－12

To draw a right triangle with an area of 9 square centimeters, make the ratio of right angle sides of its two sides 2:3, and what are the lengths of its two right angle sides?

Let one side be 2x cm and the other side be 3x cm
According to the meaning of the title:
2x*3x/2=9
The solution is: x = √ 3
∴2x=2√3 3x=3√3
A: 2 √ 3cm on one side
The other side is 3 √ 3cm

High school quadratic function
If f (x) = 4x ^ - 2 (P-2) x-2p ^ - P + 1 has at least one point (C, 0) in [- 1,1] such that f (x) is less than 0, then the value range of real number P is

1 to 8

Find the partial derivative z = (1 + XY) ^ (x + y)!
And the total differential u = x / Y (e ^ z)

Is Z = (1 + XY) ^ (x + y)! Followed by a factorial sign?
Factorial is not a continuous function, it is not differentiable
If the factorial symbol is ignored
z=(1+xy)^(x+y)
lnz=(x+y)*ln|1+xy|
(∂z/∂x)/z=(1+y)ln|1+xy|+y(x+y)/(1+xy)
∂z/∂x=[(1+y)ln|1+xy|+y(x+y)/(1+xy)]*[(1+xy)^(x+y)]
=(1+y)ln|1+xy|[(1+xy)^(x+y)]+y(x+y)[(1+xy)^(x+y-1)]
In the same way
∂z/∂y=(1+x)ln|1+xy|[(1+xy)^(x+y)]+x(x+y)[(1+xy)^(x+y-1)]
u=x/y(e^z)=x[e^(-z)]/y
∂u/∂x=1/y(e^z)
∂u/∂y=-x/y²(e^z)
∂u/∂z=-x[e^(-z)]/y=-x/y(e^z)
du=( ∂u/∂x)dx+(∂u/∂y)dy+(∂u/∂z)dz
=[1/y(e^z)]dx-[x/y²(e^z)]dy-[x/y(e^z)]dz

As shown in the figure, put the same size of black chess pieces on the edge of the regular polygon, according to this rule, then the nth figure needs the shape of the black chess pieces
The number is -
The first figure: each vertex of an equilateral triangle has a chess piece (3 pieces in total)
The second figure: there are three pieces on each side of the square (a total of 8 pieces)
The third figure: regular pentagon with 4 pieces on each side (15 pieces in total)
The fourth figure: a regular hexagon with 5 pieces (25 pieces in total)
...

The fourth figure should have a total of 24 pieces
The rule is (n + 2) × n, which can also be written as N & sup2; + 2n

An elementary third quadratic function problem
(4) When x = 2, the minimum value of function y is 1. When x = 3, y = 1.5, find the analytic formula of function

From the minimum value, let y = a (X-2) ^ 2 + 1
Substituting x = 3, 1.5 = a + 1
A = 0.5
So y = 0.5 (X-2) ^ 2 + 1

Given the point P (2,2) on the image of inverse scale function y = K / X (K ≠ 0). (1) when y = - 3, find the value of X; (2) when 1

Point P (2,2) on the image of inverse scale function y = K / X (K ≠ 0)
SO 2 = K / 2
We get k = 4
So y = 4 / X
When y = - 3, x = - 4 / 3
(2) Y decreases monotonically on the domain of definition
Therefore, when 1

Factorization of 2a2-3ab-2b2 = 0

Cross phase multiplication, 2A2 and - 2B2 are expanded respectively, 2A2 = a * 2A, - 2B2 = b * (- 2b)
Then - 3AB = a * B + 2A * (- 2b) gives (2 * a + b) (A-2 * b) = 0
Just look for more rules, expand the two sides separately, and then cross and multiply them to form the middle term. Try it several times

In △ ABC, if the vertical bisector of BC = 8cmab intersects AB at point D and AC at point E, and the circumference of △ BCE = 18cm, what is the length of AC

Because be = AE,
Therefore, the circumference of △ BCE = BC + CE + AE = BC + AC = 18cm,
So AC = △ BCE perimeter - BC = 10cm