Fill in the appropriate operation symbol in 0 to make both sides of the equal sign equal 3○3○3○3=8

Fill in the appropriate operation symbol in 0 to make both sides of the equal sign equal 3○3○3○3=8

3*3-3/3=8

[height number] the surface area of a body of revolution is obtained after an arch of cycloid x = a (T - Sint), y = a (1 - cost) and the plane area surrounded by X-axis rotate around x-axis
[advanced mathematics] find the surface area of the body of revolution after an arch of cycloid x = a (T - Sint), y = a (1 - cost) and the plane area surrounded by X-axis rotate around x-axis
Note that it's the surface area, not the volume

If there is a formula, s = integral 2 π x DS
DS is the root X '^ 2 + y' ^ 2

How to make a unary equation,
There is a two digit number. The number on the one digit is twice the number on the ten digit number. After exchanging their positions, the new two digit number is 27 larger than the original one. How much is the original two digit number?

Solution: suppose ten digit is x, then one digit is 2x
2X×10+X-（10X+2X）=27
20X+X-12X=27
21X-12X=27
9X=27
X=3
2X=6,
10x + 2x = 3 × 10 + 6 = 36 (original)

It is known that F1 and F2 are the left and right focal points of the ellipse x ^ 2 / 4 + y ^ 2 = 1, and the point P is on the ellipse C. If | Pf1 | = 4, then | PF2 | = ∠ f1pf2 has the size of

Let me talk about the idea, from the meaning of a = 2, B = 1, F1F2 = 2C
|Pf1 | + | PF2 | = 2A = 4 can find | PF2 | = 4 - | Pf1 | (but I'm very confused, Pf1 | = 4, then | PF2 | =)
Then cosine theorem is used to solve cos ∠ f1pf2 = (f1p ^ 2 + F2P ^ 2-f1f2 ^ 2) / 2 | Pf1 | PF2|
That's OK

Exercises and answers of English pronunciation discrimination in grade five of primary school

First she () 2, fifth there () 3, today Saturday () 4, say says () 5

It is known that the circle C1: X & sup2; + Y & sup2; + 2x + 2y-8 = 0 and the circle C2: X & sup2; + Y & sup2; - 2x + 10y-24 = 0 intersect at two points a and B; 1. Find the equation of the smallest element passing through two points a and B; 2. Find the equation of the circle whose center is on the straight line y = - X and passing through two points a and B

(1) C1: x2 + Y2 + 2x + 2y-8 = 0 and C2: x2 + y2-2x + 10y-24 = 0, C1-C2: 4x-8y + 16 = 0, that is, x-2y + 4 = 0, substituting C1: X (a) = - 4, y (a) = 0, X (b) = 0, Y (b) = 2; if the area of the circle passing through AB is the smallest, then it is the smallest when AB is the diameter, and the equation is (x + 2) ^ 2 + (Y-1) ^ 2 =

Score mixed operation exercises urgent to 50

1、 2/3÷1/2－1/4×2/5 2、 2－6/13÷9/26－2/3
3、 2/9＋1/2÷4/5＋3/8 4、 10÷5/9＋1/6×4
5、 1/2×2/5＋9/10÷9/20 6、 5/9×3/10＋2/7÷2/5
7、 1/2＋1/4×4/5－1/8 8、 3/4×5/7×4/3－1/2
9、 23－8/9×1/27÷1/27 10、 8×5/6＋2/5÷4
11、 1/2＋3/4×5/12×4/5 12、 8/9×3/4－3/8÷3/4
13、 5/8÷5/4＋3/23÷9/11
9/7 - （ 2/7 – 10/21 ）
5/9 × 18 – 14 × 2/7
4/5 × 25/16 + 2/3 × 3/4
14 × 8/7 – 5/6 × 12/15
17/32 – 3/4 × 9/24
3 × 2/9 + 1/3
5/7 × 3/25 + 3/7
3/14 ×× 2/3 + 1/6
1/5 × 2/3 + 5/6
9/22 + 1/11 ÷ 1/2
5/3 × 11/5 + 4/3
45 × 2/3 + 1/3 × 15
7/19 + 12/19 × 5/6
1/4 + 3/4 ÷ 2/3
8/7 × 21/16 + 1/2
101 × 1/5 – 1/5 × 21
50＋160÷40
（58+370）÷（64-45）
120-144÷18+35
347+45×2-4160÷52
（58+37）÷（64-9×5）
95÷（64-45）
178-145÷5×6+42
420+580-64×21÷28
812-700÷（9+31×11）
（136+64）×（65-345÷23）
85+14×（14+208÷26）
（284+16）×（512-8208÷18）
120-36×4÷18+35
（58+37）÷（64-9×5）
(6.8-6.8×0.55)÷8.5
0.12× 4.8÷0.12×4.8
（3.2×1.5+2.5）÷1.6
3.2×（1.5+2.5）÷1.6
6-1.6÷4= 5.38+7.85-5.37=
7.2÷0.8-1.2×5= 6-1.19×3-0.43=
6.5×（4.8-1.2×4）= 0.68×1.9+0.32×1.9
10.15-10.75×0.4-5.7
5.8×（3.87-0.13）+4.2×3.74
32.52-（6+9.728÷3.2）×2.5
（7.1-5.6）×0.9-1.15] ÷2.5
5.4÷[2.6×（3.7-2.9）+0.62]
12×6÷（12-7.2）-6 （4）12×6÷7.2-6
2/3÷1/2－1/4×2/5
2－6/13÷9/26－2/3
2/9＋1/2÷4/5＋3/8
10÷5/9＋1/6×4
1/2×2/5＋9/10÷9/20
5/9×3/10＋2/7÷2/5
1/2＋1/4×4/5－1/8
3/4×5/7×4/3－1/2
23－8/9×1/27÷1/27
8×5/6＋2/5÷4
1/2＋3/4×5/12×4/5
8/9×3/4－3/8÷3/4
5/8÷5/4＋3/23÷9/11
0.6×（1.7－0.9）÷0.24+1.25
5.4×[（2.73＋1.85）÷2.29]-3.56
9405-2940÷28×21
920-1680÷40÷7
690+47×52-398
148+3328÷64-75
360×24÷32+730
2100-94+48×54
51+（2304-2042）×23
4215+（4361-716）÷81
（247+18）×27÷25
36-720÷（360÷18）
1080÷（63-54）×80
（528+912）×5-6178
8528÷41×38-904
264+318-8280÷69
（174+209）×26- 9000
814-（278+322）÷15
1406+735×9÷45
3168-7828÷38+504
796-5040÷（630÷7）
285+（3000-372）÷36
546×（210-195）÷30
I searched everywhere, too

If the central angle of the sector is enlarged by 2 times and the radius is reduced by 12, the area of the sector is ()
A. Unchanged B. 2 times larger C. 4 times larger D. 12 times smaller

If the sector area is n π r2360 and the changed sector area is n π (R2) 2360 = n π R22 × 360, the changed sector area will be reduced to 12 times of the original area

The direction relation of simple harmonic motion displacement should be classified and analyzed

(1) The displacement X in the vibration starts from the equilibrium position, and its direction points from the equilibrium position to the end position. Its size is the linear distance between the two positions. It is the largest at the two "ends" and zero at the equilibrium position
(2) The change of acceleration a is consistent with the change of KX, which is the largest at the two "ends", zero at the equilibrium position, and always points to the equilibrium position
(3) The change of velocity V and acceleration a is just opposite. It is zero at two "end points" and the maximum at equilibrium position. Except for two "end points", there are two possible directions of velocity at any position

Cut a piece of plastic plate 120 cm long and 80 cm wide into a square with the same area and no surplus, which can be cut at least

Finding the greatest common factor of 120 and 80
120=40*3
80=40*2
Greatest common factor = 40 cm
120/40=3 80/40=2
3 * 2 = 6
The maximum side length of the square is 40 cm, 3 knives, 6 pieces in total