The following figures must be axisymmetric () A right angle B right triangle C quadrilateral D trapezoid

The following figures must be axisymmetric () A right angle B right triangle C quadrilateral D trapezoid

The following figures must be axisymmetric (a)
A right angle
B right triangle
C quadrilateral
D trapezoid

The shape of a wooden box is cube, 0.8m. How many square meters of wood do you need to make this wooden box? (there is no cover on the top of the wooden box)

0.8 × 0.8 × 5 = 3.2 (M2)
A: at least 3.2 square meters

Given the vector a = (COS θ, sin θ), B = (√ 3, - 1), find the maximum value of ∣ 2a-b ∣
The maximum is 4 and the minimum is 0

According to the title ∣ 2a-b ∣
=√[(2cosθ-√3)^2+(2sinθ+1)^2]
=√[4cos^2θ-4√3·cosθ+3+4sin^2θ+4sinθ+1]
=√(4+4-4√3·cosθ+4sinθ)
=√ [8-8 · (sin60 ° cos θ - cos60 ° sin θ) (according to the formula sin (Φ + θ) = sin Φ · cos θ + cos θ · sin Φ)
=√[8-8·sin(60°-θ)],
It is also obvious that - 1 ≤ sin (60 ° - θ)] ≤ 1 and [8-8 · sin (60 ° - θ)] must be greater than or equal to 0, so the minimum value of ∣ 2a-b ∣ is √ [8-8.1] = 0, and the maximum value of ∣ 2a-b ∣ is √ [8-8 · (- 1)] = 4
I should know

As shown in the figure, in △ ABC, ∠ B = 90 °, ab = 3, BC = 5. Fold △ ABC so that point C coincides with point a, and the trace is De, then the perimeter of △ Abe is______ ．

∵ according to the properties of folding, AE = CE, ∵ Abe perimeter = AB + be + AE = AB + be + CE = AB + BC = 3 + 5 = 8

High school function image drawing, urgent,

As long as you remember the images of the six most basic functions, the others can be drawn on the basis of them. The six basic functions are: first-order function, second-order function, trigonometric function, exponential function, logarithmic function, inverse proportion function. The first-order function is a straight line, the second-order function is a parabola, and the sine and cosine functions in trigonometric function are

A mathematical problem of space vector
It is known that a, B and C are space vectors, 3a-2b = (- 2,0,4), C = (- 2,1,2), the product of a and C is 2, | B | = 4
Then cos =?

Answer: - 1 / 4
Details are as follows:
a(x1,y1,z1),b(x2,y2,z2)
b*c=-2 x2+y2+2 z2 (1)
The quantitative product of a and C is 2
a*c=-2 x1+y1+2 z1=2 (2)
And 3a-2b = (- 2,0,4) has 3 x1-2 x2 = - 2,3 y1-2 y2 = 0,3 z1-2 Z2 = 4,
The results show that X1 = (2 x2-2) / 3, Y1 = 2 Y2 / 3, Z1 = (2 Z2 + 4) / 3, (3)
After substituting (3) into (2) and simplifying, there is: - 2x2 + Y2 + 2z2 = - 3, that is, b * C = - 3,
From C = (- 2,1,2) there is | C | = 3
And | B | = 4, so | B | * | C | = 12
Then cos = (b * c) / (| B | * | C |) = - 1 / 4

In the triangle ABC, AC = AE, ad = AE, if ∠ bad = 20 °, find the degree of ∠ CDE

Solution
x=ABD y=ADE
20+x=y+EDC(1)
x+EDC+y+180-2y=180(2)
x+EDC-y=0
x+EDC=y
x=y-EDC
20+y-EDC=y+EDC
20=2EDC
EDC=10

Factorization of square x-3ax-6ab-4b

If vectors a and B satisfy | a | = | B | = 1, a is perpendicular to B, and (2a + 3b) is perpendicular to (ka-4b)
If vectors a and B satisfy | a | = | B | = 1, a is perpendicular to B, and (2a + 3b) is perpendicular to (ka-4b), the value of real number k is obtained

Because the vectors (ka-4b) and (2a + 3b) are perpendicular, so (ka-4b) * (2a + 3b) = 0,
(ka-4b) * (2a + 3b) = 2kA * a + 3KA * B-8A * b-12b * B, note that the condition | a | = | B | = 1,
So, 2k-12 = 0, k = 6

As shown in the figure, in the triangle ABC, BD and CE are bisectors of angle ABC and angle ACB respectively. Am is perpendicular to CE and an is perpendicular to BD. if AB = 8cm, BC = 9cm and AC = 7cm
Finding the length of Mn

Prolonging am, an to F, G
Because BD is the bisector of angle ABC
Therefore, abd = CBD
And an is perpendicular to BD,
So ∠ anb = ∠ GNB = 90 °,
And BD is the public side,
So △ ABN ≌ △ AGN,
So ad = Gd, ab = BG = 8,
Similarly, am = FM, AC = FC = 7,
So m and N are the median of △ AFG, BF = bc-fc = 9-7 = 2, CG = bc-bg = 1,
So Mn = FG / 2 = (bc-bf-cg) / 2 = (9-2-1) / 2 = 3cm