There are three integral operations in the second semester of junior one to help solve 1. If the cube of (x-1 power of a × y + 1 power of B) is 9 power of a and 6 power of B, then X=____ ,y=_____ . 2. If the cube of (- X's Square × a) is known to be the cube of y to the sixth power of X, then a=_____ . 3. Given that the m power of 2 is 3 and the 2m power of a is 4, what is the 2m power of (the 4th power of 4A)? The third question should have a process

There are three integral operations in the second semester of junior one to help solve 1. If the cube of (x-1 power of a × y + 1 power of B) is 9 power of a and 6 power of B, then X=____ ,y=_____ . 2. If the cube of (- X's Square × a) is known to be the cube of y to the sixth power of X, then a=_____ . 3. Given that the m power of 2 is 3 and the 2m power of a is 4, what is the 2m power of (the 4th power of 4A)? The third question should have a process

A ^ 2 is the square of A
A ^ x is the x power of A
1.{[a^(x-1)]*[b^(y+1)]}^3=[a^(3x-3)]*[b^(3y+3)=a^9*b^6
So 3x-3 = 9
3y+3=6
x=4 y=1
2 {[ (-x)^2]*[A}]^3=[(x^2)*A]^3=(x^6)*(A^3)=x^6*y^3
So a = y
3.[(4a)^4]^2m=(4a)^(8m)=[4^(8m)]*[a*(8m)]
Because 4 = 2 ^ 2
So [(4a) ^ 4] ^ 2m = (4a) ^ (8m) = [4 ^ (8m)] * [a * (8m)]
=(2^2)^(8m)*[(a^2m)^4] a^(2m)=4
=2^(16m)*(4^4)
=[(2^m)^16]*(4^4)
=(3^16)*(4^4)

The car runs at a speed of 36 km / h on the straight road at a constant speed, and the acceleration when braking is 4 meters per second square meter. What is the speed of the car at the end of the second second after braking? What is the displacement in 5 seconds?

V0 = 36 km / h = 36000 / 3600 = 10 (M / s), a = - 4 m / s and#178;,
According to the formula: v = V0 + at = 10-4 × 2 = 2 (M / s)
When v = 0, 0 = 10 + at t = 10 / 4 = 2.5 (seconds), the car has stopped in 2.5 seconds, and the displacement in 5 seconds is the displacement at the end of 2.5 seconds,
S = v0t + 0.5at & # 178; = 10 × 2.5-4 × 1.5 & # 178; = 16 (m)
The speed at the end of the second second is 2 m / s, and the displacement is 16 m in 5 seconds

If M > 1 and am-1 + am + 1-am Square-1 = 0.s2m-1 = 39, then M is equal to

In the first formula, am-1 + am + 1 = 2am, we can deduce am = 1
In the second formula, s2m-1 = (2m-1) (a1 + a2m-1) / 2 = (2m-1) * 2 * am / 2 = 39
So, M = 20

The car braked with an acceleration of 5m / S ^ 2 to make a uniform deceleration linear motion, and the last internal displacement before stopping motion was calculated

The last internal displacement before the uniform deceleration linear motion stops = the first internal displacement of uniform acceleration motion with zero initial velocity and equal acceleration
S=(1/2)aT^2=0.5*5*(1*1)=2.5m

Xiao Gang and Xiao Liang practice long-distance running on the 400 meter circular track. Xiao Gang's speed is 180 meters per minute and Xiao Liang's speed is 160 meters per minute. If they start from the same starting point at the same time, they meet for the first time after () minutes, if they start from the same starting point in reverse at the same time, they meet for the first time after () minutes

Set the same starting point and run in the same direction for the first time after X minutes
Because Xiaogang is faster than Xiaoliang, the first time we met, Xiaogang must have run one more lap than Xiaoliang
180X-160X=400 X=20
In the same way, when two people meet in the opposite direction, the distance is exactly the length of the runway
180y + 160y = 400 y = 17 / 20 minutes = 51 seconds later
Study hard and think more~

2 mathematical problems! I want the process! The first day of junior high school!
A 200m long train passes a 1.6km long bridge at a speed of 10m / s. how long does it take for the train to pass the bridge? How long does it take for the train to move on the bridge?
Both a and B are moving in a straight line at a uniform speed. It is known that s a: S B = 2:3; V A: v b = 2:1

1) (1600 + 200) / 10 = 180 (seconds)
(1600-200) / 10 = 140 (seconds)
2) The ratio of exercise time between a and B = (2 / 2): (3 / 1) = 1:3

A point on a number axis that represents an integer is called an integral point. The unit length of a number axis is 1cm. If you draw a line ab of 200cm on the number axis at will
A point on a number axis that represents an integer is called an integral point. The unit length of a certain number axis is 1cm. If a line segment AB with a length of 200cm is randomly drawn on the number axis, how many integral points are covered by this line segment AB?

1 cm covers 2 integral points (e.g. from 0 to 1, the starting point is integral) or 1 integral point (e.g. from 1, the starting point is not integral)
Then 200 cm covers 201 integral points (when the starting point is integral) or 200 integral points (when the starting point is not integral)

The utility model relates to a cloth hat, in which the crown part is a cylinder and the brim part is a ring
It's 1cm. How many square decimeters of cloth does it take to make this hat?
It's an evening assignment. I can't do it,

Crown radius = 1 height = 1 brim radius = 1 + 1 = 2
At least the cloth to be used for hat = 3.14 * 1 & # 178; + 3.14 * (1 + 1) * 1 + 3.14 (2 & # 178; - 1 & # 178;) = 18.84 CM & # 178;
It takes at least 0.1884 square decimeter of cloth to make this hat
Be sure to choose the best answer and encourage me

Let's have a look
We know that the n-3 power of 4Y multiplied by the 3 power of Y is the same as the 4 power of - 3x multiplied by the m power of Y. we can find the value of the 2 power of the algebra 3M + n!

It is known that the n-3 power of 4x multiplied by the 3 power of Y is the same as the 4 power of - 3x multiplied by the m power of Y. we can find the value of the 2 power of the algebra 3M + n
According to the meaning of the title:
N-3 = 4 ｛ original formula = 3M + n & ｛ 178;
∴m=3 =9+49
n=7 =58
Thanks

Write procedures: print out all the "narcissus number", the so-called "narcissus number" refers to a three digit number, the number of cubic sum is equal to the number itself

vb：Private Sub Form_ Click() Dim a,b,c ,x As Integer For a = 0 To 9 For b = 0 To 9 For c = 1 To 9 x=100 * c + 10 * b + a If a ^ 3 + b ^ 3 + c ^ 3 = a + 10 * b + 100 * c Then print x; Next c Next b Nex...