How many tons of sand is equal to one side

How many tons of sand is equal to one side

The bulk density of construction sand is 1350-1450kg / m3, and the weight of one cubic meter of sand is about 1.35-1.45t

Two questions about the complete square formula
1. How to add brackets in order to use the complete square formula to calculate: (- 3x + 2y-z) (3x + 2Y + Z)=______
2. Calculation: (x + Y-Z) (X-Y + Z) - (x + y + Z) (X-Y-Z)

1.(-3x+2y-z）（3x+2y+z）=[2y-(3x+z)][2y+(3x+z)]=4y²-(3x+z)²=4y²-9x²-z²-6xz2.(x+y-z)(x-y+z)-(x+y+z)(x-y-z)=[x+(y-z)][x-(y-z)]-[x+(y+z)][x-(y+z)]=x²-(y-z)²-[x²-(y...

[200 points reward] a question about the volume of a revolving body,
Higher mathematics (1) a problem about the volume of a body of revolution
Find y = x ^ 2, y = x ^ 1 / 2
↑ ↑
(y = the square of x) (y = the root of x)
These two functions, the volume that you rotate around the y-axis
My question: from the graph of function, y = root x is above the square of X,
It's a section more than y = root X
But why do you also rotate around the y-axis in the range of 0 to 1,
Y = the volume converted from the root x (= π / 2) will be larger than the volume converted from the square x (= π / 5)?
Let's talk about it,
Choose whoever you understand,

It's very simple. I think you forgot to convert these two functions. Since we are looking for the volume formed around the Y axis, then y = x ^ 2, in the calculation, we have to replace it with x = y ^ (1 / 2); similarly, y = x ^ (1 / 2) has to replace it with x = y ^ 2
If so, then the volume of y = x ^ (1 / 2) around the Y axis should be:
pi∫_ 0^1 (x^2) dy
=pi∫_ 0^1 (y^4) dy
=pi/5
And the square of X turns out to be pi / 2,
That's it.

The maximum speed of a motorcycle is 30 m / s. when a car with a speed of 10 m / s passes its position, the motorcycle starts immediately. If the motorcycle wants to catch up with the car within one minute after it is stationary, it must at least drive with more acceleration. In the process of catching up with the car, when is the maximum distance between the two cars? What is the maximum distance? If the car is running at 25 m / s, what is the above conclusion?

V steam t = 1 / 2 * amo * t * t, so if you catch up in one minute, amo is equal to 1 / 3 M / s Square
When the motorcycle speed is equal to the vehicle speed, the maximum distance between two vehicles is t = V steam / a friction = 10 / 1 / 3 = 30s
S = V steam * T-1 / 2 * a motorcycle * t * t = 150m, the distance is 150m
If the speed of the car is 25 meters per second, replace the V steam in the previous formulas with 25 m / s
A = 5 / 6 m / s Square maximum distance time t = 30 s maximum distance S = 375 M
Typing is not familiar, many symbols will not be used, this should be a simple physical kinematics problem, to point it!

When 2.5 × 3.65 × 0.8, we can use () to simplify the calculation

5 × 3.65 × 0.8 can be used to simplify the calculation

Given {x-2y + 4Z = 0,2x-3y + 5Z = 0, find the value of X: Y: Z

x-2y+4z=0 （1）
2x-3y+5z=0 （2)
(1) X 2, we get 2x-4y + 8Z = 0 (3)
(2) (3), y-3z = 0 (4)
From (4), y = 3Z
If y = 3Z is replaced by (1), x = 2Z is obtained
So x: Y: z = 2Z: 3Z: z = 2:3:1

2x^3y+4x^2y+2xy
2x^3 y+4x^2 y+2xy

2x^3 y+4x^2 y+2xy =2xy(x+2x)+2xy =2xy(x+2x+1) =2xy(x+1)

Given that the polynomial 2x ^ 4 + ax ^ 3 + BX ^ 2 + 10x + 4 has two factors (x ^ 2 + 2x + 1), (x + 2), find a + B
One of my classmates and I think this question is strange

Suppose the other factor is y, then 2x ^ 4 + ax ^ 3 + BX ^ 2 + 10x + 4 = y (x ^ 2 + 2x + 1) (x + 2) 2x ^ 4 + ax ^ 3 + BX ^ 2 + 10x + 4 = y (x + 1) ^ 2 (x + 2) x = - 1, then x + 1 = 0, so the right side = 0, so the left side is also equal to 0x = - 1, 2x ^ 4 + ax ^ 3 + BX ^ 2 + 10x + 4 = 2-A + B-10 + 4 = - A + B-4 = 0a-b = - 4 (1) similarly, x = - 2, x + 2 = 0, so the right side

Find the maximum and minimum values of the function f (x, y) = x + y + 1 in the bounded domain D: X Λ 2 + y Λ 2 ≤ 4

Let x = rcosa, y = rsina
And X Λ 2 + y Λ 2 ≤ 4
So - 2

① X ^ 4-4x ^ 3Y + 4x ^ 2Y ^ 2 (2x + 3Y) ^ 2 - (2x-y) ^ 2 experts help

①x^4-4x^3y+4x^2y^2
=x^2(x^2-4xy+4y^2)
=x^2(x-2y)^2
②(2x+3y)^2-(2x-y)^2
=[(2x+3y)+(2x-y)][(2x+3y)-(2x-y)]
=(4x+2y)(4y)
=8y(2x+y)