"*" is a kind of algorithm specified in a, a * b = a ^ 3-2b, if 2 * (4 * x) = 2 + X, find the value of X Is a * b = a ^ 2-2b

"*" is a kind of algorithm specified in a, a * b = a ^ 3-2b, if 2 * (4 * x) = 2 + X, find the value of X Is a * b = a ^ 2-2b

2*(4*x)=2*(16-2x)
=4-32+4x=2+x
x=10

*If a * b = a ^ 2-B, find (- 4) * x = X-2

a*b=a^2-b
So (- 4) * x = = (- 4) ^ 2-x = X-2
2x=16+2
x=9

A granary is composed of a cylinder and a cone. The circumference of the cylinder bottom is 10 meters and the height is 6 meters. The height of the cone is 2.1 meters. What is its volume?

The granary is divided into cylinder and cone, and the volume is the sum of them
Cone volume = bottom area * cone height * 1 / 3
Cylinder volume = bottom area * column height
The circumference of the bottom surface of the cylinder is 2 pie and R is 10, so r = 10 divided by 2 pie and R = 5 divided by pie
Base area = pie x square of R = 25 divided by pie
So volume = cylinder volume plus cone volume = 25 divided by pie * 6 + 25 divided by pie * 2.1 * 1 / 3

Given the ellipse 2 / 2 x + y = 1 (1), find the midpoint trajectory equation of parallel string with slope 2

The string is y = 2x + B
Substituting X & sup2; + 2Y & sup2; = 2
9x²+8bx+2b²-2=0
x1+x2=-8b/9
y=2x+b
So Y1 + y2 = 2x1 + B + 2x2 + B = 2 (x1 + x2) + 2B = 2B / 9
Midpoint x = (x1 + x2) / 2, y = (Y1 + Y2) / 2
So y / x = - 1 / 4
9x & sup2; + 8bx + 2B & sup2; - 2 = 0 has a solution if a line and an ellipse intersect
So 64b & sup2; - 72B & sup2; + 72 > = 0
-3

A parallelogram and a triangle have the same area. The bottom of the parallelogram is twice the bottom of the triangle. The height of the triangle is 6 decimeters, and the height of the parallelogram is 6 decimeters

The height of a parallelogram is a quarter of that of a triangle, or 1.5 decimeters

The product of several prime numbers must be (). 1 odd 2 even 3 combined 4 prime numbers

3

If the two sides of a triangle are 5 cm and 8 cm respectively, the length of the third side may be () cm
A. 12 cm B. 13 cm C. 14 cm

According to the trilateral relationship of the triangle, the third side should be greater than 8-5 = 3, and less than 8 + 5 = 13, 3 < the third side < 13

It is known that PA is the tangent of circle O, a is the tangent point, Po is parallel to AC, BC is the diameter of circle o

Connect AB to Q, connect OA
Then OA ⊥ pa
∵PO∥AC,CO＝BO
∴AQ ＝ BQ （1）
∵ BC is the diameter
∴ ∠BAC ＝ 90°
∴ ∠PQA ＝ ∠PQB ＝ 90° （2）
PQ is the common line (3)
∴ ⊿PQA≌⊿PQB（123,SAS）
∴PA=PB
And ∵ OA = ob, Po collinear
∴ ⊿POA≌⊿POB（SSS）
∴ ∠PBO＝∠PAO ＝ 90°（OA⊥PA）
That is Pb ⊥ BC
Ψ Pb is the tangent of circle o

The two sides of a river can be regarded as parallel. On one bank of the river, there is a tree every 4m, and on the other bank, there is a telegraph pole every 50m,
When you look at the other bank 20 meters away from the tree bank, you can see that the two adjacent power poles on the other bank are just covered by the two trees on the other bank, and there are four trees between the two trees to calculate the width of the river

Set recognition station at point a
The two points of the tree are B and C
The poles are D and E
Triangle ABC is similar to triangle ade
AB：AD=BC：DE
AB=30
AD=30+H
BC=2*4=8
DE=40
30：（30+H）=8：40
30：（30+H）=1：5
H = 4 * 30 = 120 mlZ similar triangle

As shown in the figure, O is a point on the straight line AB, ∠ AOC = 13 ∠ BOC, OC is the bisector of ∠ AOD. (1) calculate the degree of ∠ cod. (2) judge the position relationship between od and AB, and give reasons

(1) ∵ - AOC + BOC = 180 °, AOC = 13 ∵ BOC, ∵ 13 ∵ BOC + BOC = 180 °, the solution is ∵ - BOC = 135 °, AOC = 180 ∵ - BOC = 180 ∵ - 135 ∵ 45 °, ∵ OC bisection ∵ AOD, ∵ cod = ∵ AOC = 45 ° (2) od ⊥ ab