Omega, it seems to have something to do with the plural plane.

Omega, it seems to have something to do with the plural plane.


x^3=1
x1=1,x2=w,x3=w^2
5 + (radical 3 / 2) I



I () complex——


we



It is known that the length of the hypotenuse of a right triangle is C, the two right angles are a, B (b > A), and a, B, C are in an equal ratio sequence. The value of a: C is obtained


The length of the hypotenuse of a right triangle is C, and the two right angles are a and B respectively. According to the Pythagorean theorem, we can get a & # 178; + B & # 178; = C & # 178; and if a, B and C are in equal ratio sequence, B & # 178; = AC. therefore, a & # 178; + AC = C & # 178; both sides of the equation are at the same time △ C & # 178;, and we can get (A / C) &# 178; + (A / C) - 1 = 0 to (A / C)



Two pieces of wood a and B with mass of M1 and M2 are stacked on a smooth horizontal plane. The dynamic friction coefficient between a and B is μ. If a and B are kept relatively stationary, what is the maximum horizontal tension f applied to a? If a and B are kept relatively stationary, what is the maximum horizontal tension f applied to B? If a is pulled out from the upper surface of B, what is the minimum horizontal tension f applied to a?


The maximum horizontal tension f applied to a is (M1 + m2) * M1 * g * μ / m2
The maximum horizontal tension applied to B is (M1 + m2) * g * μ
If a is pulled out from the upper surface of B, the minimum horizontal tension applied to a is (M1 + m2) * M1 * g * μ / m2



How to calculate the value-added rate from negative growth to positive growth?


The so-called negative growth is reduced. If you calculate the output, then in the period of negative growth, the initial value * (1-negative growth rate) (1-negative growth rate). Calculate year by year or month by day. If you know that the output is calculated as the value-added rate, then calculate (positive growth + reduced output) / time / initial value



A car is braked with an acceleration of 2m / S ^ 2 to make a uniform deceleration linear motion, and the displacement within 1s before the car stops is calculated
Isn't V0 in v0t speed? And the stopping speed is the final speed. Where is the VOT in vot-1 / 2at ^ 2?


Because the speed at stop is 0, and it was 2 a second ago, so s = 1 / 2 * at ^ 2 = 1m
You can think the other way around. We can start from the opposite direction, which is how much our car moves after accelerating from 0 to 1 second
And you're right, but a is negative. Bring it in
When decelerating, V0 = 2 T = 1 a = - 2 original formula = v0t + 1 / 2at ^ 2 = 2 * 1 + (- 2 * 1 * 0.5) = 1



A 1 = 5 / 6. D = - 1 / 6, Sn = - 5, the problem of finding N and an. Higher one sequence


Because A1 = 5 / 6, d = - 1 / 6, Sn = - 5, so - 5 = 5 / 6N + n (n-1) * - 1 / 6
It is known that the square of n is - 11n-60 = 0, and N = 15 and N = - 4 are obtained (because n > 0, it is rounded off)
So n = 15, an = 5 / 6 + 14 * (- 1 / 6) = - 3 / 2



When the car starts to move at an acceleration of 1 m / S2 from a standstill, the displacement of the car in the first 5 seconds and the average speed in the second second second will be calculated!


s=1/2a*t^2=1/2*1*5*5=12.5m
s2-s1=1/2a*t2^2 - 1/2a*t1^2=1/2*1*2*2--1/2*1*1*1=1.5
a=(s2-s1)/(t2-t1)=1.5/m



The circumference of the circular track on the school playground is 400 meters. Xiaohua's running speed is three times that of Xiaohong. If two people start from the same starting point along the track in the same direction at the same time, Xiaohua meets Xiaohong for the first time five minutes later, and Xiaohong's running speed is 3 times that of Xiaohong


Let Xiaohong's speed be x km / min and Xiaohua's speed be 3x km / min
According to the meaning of the title:
5(3X--X)=400
The solution of this equation is as follows
X=40
A: Xiaohong's speed is 40 km / min



In hiking activities, some people walk, some people take a car, and two people start at the same time. The speed of the car is 60km / h, and the walking speed is 5km / h. walking starts one hour earlier than the car. After driving to the destination, they return to pick up the walkers, and the starting point is 60km to the destination. Q: how many hours after the walkers start, do they meet the car that picks them up? (equation solution)


Suppose the walkers set out and meet the car that came back to meet them via XH
5x+60(x-1)=120
The solution is x = 2.77