LG (x + y) + LG (2x + 3Y) - Lg3 = LG4 + lgx + lgY for X / Y

LG (x + y) + LG (2x + 3Y) - Lg3 = LG4 + lgx + lgY for X / Y

LG (x + y) + LG (2x + 3Y) - Lg3 = LG4 + lgx + lgylg (x + y) + LG (2x + 3Y) = Lg3 + LG4 + lgx + lgylg [(x + y) * (2x + 3Y)] = LG [3 * 4 * XY) (x + y) (2x + 3Y) = 12xy 2x ^ 2 + 5xy + 3Y ^ 2 = 12xy2x ^ 2-7xy + 3Y ^ 2 = 0, (2x-y) (x-3y) = 02x = y or x-3y = 0x / y = 1 / 2 or X / y = 3

The general solution of dy / DX = 1 / (x – y ^ 2)

∵ dy / DX = 1 / (x – y ^ 2) = = > DX XDY + y ^ 2dy = 0 = = > e ^ (- y) DX Xe ^ (- y) dy + y ^ 2E ^ (- y) dy = 0 (multiplication of both ends of the equation E ^ (- y)) = > d (Xe ^ (- y)) - D ((y ^ 2 + 2Y + 2) e ^ (- y)) = 0 = = > Xe ^ (- y) - (y ^ 2 + 2Y + 2) e ^ (- y) = C (C is constant) = = > x = y ^ 2 + 2Y + 2 + CE ^ y ∵

There is a big square made up of nine small squares. How many different ways to black two of them?
(if several coating methods can be overlapped by rotation, these coating methods can only be regarded as one)

If not considered (if several coating methods can be overlapped by rotation, these coating methods can only be regarded as one), then 36 coating methods are considered (if several coating methods can be overlapped by rotation, these coating methods can only be regarded as one), then one coating method can be changed into four after rotation
However, if there are only two kinds of diagonal squares, then (36-2 * 2) / 4 + 2 = 10

There are 8 Pillars in the school corridor. The bottom radius of each pillar is 4 decimeters, and the height is 30 decimeters. Now, if you want to paint these pillars, if you use 0.25 kg paint per square meter, how much paint do you need?

3.14x4x30x8 = 3014.4m2 = 30.144m2
0.25x30.144 = 7.536kg
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In the following five 13s, add appropriate operation symbols and brackets to make the calculation result equal to 29.13 & nbsp; & nbsp; 13 & nbsp; & nbsp; 13 & nbsp; & nbsp; 13 & nbsp; & nbsp; 13 = 29

According to the stem analysis, we can get: (13 + 13) × 13 + 13-13, = 23 × 13 + 13 − 13, = 29

Let f (x) have continuous first derivative and satisfy the expression of F (x) = ∫ (upper limit is x, lower limit is 0) (x ^ 2-T ^ 2) f ^, (T) DT + x ^ 2 to find f (x)

F (x) = ∫ (the upper limit is x, and the lower limit is 0) (x ^ 2-T ^ 2) f '(T) DT + x ^ 2, so f (0) = 0, and the function f (x) has continuous first-order derivatives, which is derived from both sides of the above formula; f' (x) = ∫ (the upper limit is x, and the lower limit is 0) 2xf '(T) DT + 2x = 2xf (x) + 2x = 2x (f (x) + 1) dy / (y + 1) = 2xdx, the solution is f (x) = e ^ x ^ 2-1

[cube of 12 (x + y) + square of 3 (x + y) (X-Y)] 6 (x + y)=

[cube of 12 (x + y) + square of 3 (x + y) (X-Y)] 6 (x + y)
=2(x+y)²+1/2（x-y）²
=2x²+4xy+y²+0.5x²-xy+0.5y²
=2.5x²+3xy+2.5y²
If you don't understand this question, you can ask,

Five fives add, subtract, multiply and divide by 12
5 5 5 5 5=12

(5+5)/5+5+5=12

It is known that a (a, A-2) is a point in the first quadrant, and the difference between its distance to the Y-axis and the x-axis of the trace is A-3

a-(a-2)=a-3
a=5
A(5,3)
On Y-axis symmetric coordinates (- 5,3)

Dig a cylindrical pool on the flat ground. The depth of the pool is 4 meters and the diameter is 6 meters. The pool covers an area of () square meters and needs to dig () cubic meters?
Urgent, for help, it's due tomorrow

Dig a cylindrical pool on the flat ground. The depth of the pool is 4 meters and the diameter is 6 meters. The pool covers an area of (28.26) square meters and needs to excavate (113.04) cubic meters of soil
3.14*6/2*6/2=28.26
28.26*4=113.04