Add operational symbols or brackets to the four 5S so that the result is 8

Add operational symbols or brackets to the four 5S so that the result is 8

5!/(5+5+5)=120/15=8
! is factorial 5! Is 5 * 4 * 3 * 2 * 1

Add operation symbol or bracket to 5 5, make the result of calculation 0 1 2 3 respectively

（5－5）×5×5×5=0
（5－5）×5＋5÷5=1
（5×5÷5＋5）÷5=2
（5＋5＋5）÷5=3

2 / 3, 0.75, 1, 3 / 5 and 3 / 8 are added with operation symbols and parentheses to maximize the calculation results

1 and 3 / 5 ／ 3 / 8 ／ (0.75-2 / 3)
=8/5*8/3*12
=(8*8*12)/(3*5)
=768/15
I hope you can understand, you can understand, hope to adopt, agree

How to write the mixed operation of rational number addition and subtraction in the form of omitting plus sign and bracket?

Formula: negative get positive, positive and negative get negative, positive and positive do not change, such as: a + (+ B + C) = a + B + C, a - (+ B + C) = a-b-c, a - (+ B-C) = A-B + C

A particle starts to make uniform acceleration motion from static state, and the acceleration is A1. After time t, it makes uniform acceleration motion in reverse direction, and the acceleration is A2. If the particle just reaches the starting point after time 2T, what is the ratio of the two accelerations A1: A2?

When the direction of initial velocity is defined as positive direction, the displacement in 2T second is: s ′ = V · 2T − 12 & nbsp; A2 & nbsp; (2t) 2 because s = - s', so: 12 & nbsp; a1t2 = - [v · 2T − 12 & nbsp; A2 & nbsp; (2t) 2] the solution is: A1A2 = 45 A: the ratio of two accelerations A1: A2 is 4:5

Is the algebraic formula (a + 1) (a + 2) (a + 3) (a + 4) + 1 a complete square? Please state your reasons

The original formula = (a + 1) (a + 4) (a + 2) (a + 3) + 1 = (A2 + 5A + 4) (A2 + 5A + 6) + 1 = (A2 + 5a) 2 + 10 (A2 + 5a) + 25 = (A2 + 5A + 5) 2. Then the algebraic formula is a complete square formula

As shown in the figure, a uniform plate with mass m and length L is placed on a long smooth horizontal table. There is an object block with mass m at the left end of the plate, which is connected with the object block

(1) Under the action of friction, the plate accelerates to the right uniformly until the speed is the same as that of the block. At this time, the block just reaches the midpoint of the plate. Let the acceleration of the board be A1, and the motion time be T1. For the board, μ 1mg = ma, v = a1t1
∴t1 = Mv/u1mg
In this process, if the block forward displacement is S1 and the plate forward displacement is S2, then S1 = VT1 and S2 = vt1g2
And because S1-S2 = L / 2, it can be obtained from the above formulas
The dynamic friction coefficient between the block and the plate μ 1 = MV & # 178; MGL, the displacement of the plate S2 = L / 2
(2) Let the dynamic friction coefficient between the board and the table be μ 2, the sliding time of the block on the board be T2, and the acceleration of the board be A2. For the board, there is μ 1mg ― μ 2 (M + m) g = ma2, and V = a2t2. The solution is T2 = MV / μ 1mg ― μ 2 (M + m) g
Let the time from the left end to the right end of the plate be T3, then vt3-vt3 / 2 = L, T3 = 2L / v
In order to reach the right end of the board, T2 ≥ T3 must be satisfied
Therefore, in order to reach the right end of the board, the friction coefficient between the board and the table is μ 2 ≥ MV & # / 2 (M + m) gl

How to prove that 100 yuan is one point

Verification: 1 yuan = 1 point
Because 1 yuan = 100 points
=10 points × 10 points
=1 angle × 1 angle
=RMB 0.1 × RMB 0.1
=01 yuan
=1 point
The proof is complete

What is the main content of Newton's second law

Newton's second law: the acceleration of an object is directly proportional to the combined external force F, and inversely proportional to the mass of the object. The direction of acceleration is the same as that of the combined external force, that is, f = ma. Newton's second law of motion can also be expressed as "the rate of change of momentum of an object with time is directly proportional to the sum of the external forces."

The radius of the outer circle is 1cm and the radius of the inner circle is 0.5cm. What is the area of this ring?

Outer circle area = square of radius * π - square of inner circle radius * π
1*1*3.14-0.5*0.5*3.14
=3.14-0.785
=2.355