3 mathematical cylinder and cone application problem formula and answer! (1) The copper cone used by my uncle is a cone with a bottom diameter of 8cm and a height of 9cm. If the copper weight per cubic centimeter is 8.9kg, how many kg is the copper cone? (2) there are eight identical columns in the exhibition hall with a height of 9m and a diameter of 1m. Now we need to paint all the columns. If we use 100g of paint per square meter, how many kg of paint is needed? (3) saw a 10 meter cylindrical wood into four sections, Surface area increased by 24 square decimeters, the original volume of this wood is how many cubic decimeters?

3 mathematical cylinder and cone application problem formula and answer! (1) The copper cone used by my uncle is a cone with a bottom diameter of 8cm and a height of 9cm. If the copper weight per cubic centimeter is 8.9kg, how many kg is the copper cone? (2) there are eight identical columns in the exhibition hall with a height of 9m and a diameter of 1m. Now we need to paint all the columns. If we use 100g of paint per square meter, how many kg of paint is needed? (3) saw a 10 meter cylindrical wood into four sections, Surface area increased by 24 square decimeters, the original volume of this wood is how many cubic decimeters?

1) (8 △ 2) & sup2; × 3.14 × 9 × 1 / 3 × 8.9 = 150.72 × 8.9 = 1341.408 (g) 2) 1 × 3.14 × 9 × 8 × 100 = 84.78 × 100 = 8478 (g) = 8.478 (kg) 3) 10 m = 100 decimeter 24 △ [(4-1) × 2] × 100 = 24 △ 6 × 100 = 400 (cubic decimeter) see how detailed I write, enough

5 times of formula (- 1 / 2)
What is the exponent of degree 5 of equation (- 1 / 2)? Read as a power operation and read as a result of a power-------------------

The fifth power of (- 1 / 2)
=-1 / 2 × 1 / 2 × 1 / 2 × 1 / 2 × 1 / 2
=-One out of 32
The index is 5,
As a power operation, read as: - 1 × - 2 / 1 × - 2 / 1 × - 2 / 1 × - 2 / 1 × - 2 / 1,
As a result of the power read as: - 1 / 2 of the 5th power

Find the limit Lim [(x ^ 5 + 7x ^ 4 + 2) ^ C-X] of high number, the limit exists ≠ 0, find the constant C and the limit value. X -- > ∞

If Lim [(x ^ 5 + 7x ^ 4 + 2) ^ C-X] limit exists ≠ 0, then LIM (x ^ 5 + 7x ^ 4 + 2) ^ C = Lim [(x + m) ^ 5] ^ C = LIM (x + m) ^ (5C) = LIM (x + m) (M is a constant), then 5C = 1 / 5lim [(x ^ 5 + 7x ^ 4 + 2) ^ C-X] = M. because (x + m) ^ 5 = x ^ 5 + 5mx ^ 4 +. + m ^ 5x -- > ∞, the first two terms of the above formula are dominant, and the following

If the expansion of polynomial (MX + 4) (2-3x) does not contain x term, then M=______ ．

∵ (MX + 4) (2-3x) = 2mx-3mx2 + 8-12x = - 3mx2 + (2m-12) x + 8 ∵ after expansion, there is no X term ∵ 2m-12 = 0, that is, M = 6

Simple operation of fraction multiplication
① 15 × 6 / 7-6 / 7
② 17 of 83 × 82 + 17 of 83
③ 5 / 7 × 13 + 5 / 7

1 ( 15-1)x6/7=12
2 ( 82+1)x17/83=17
3 (13+1)x5/7=10

Calculus differential equation problem. To find the general solution of Y '= (y + X LNX) / x, please use formula method. Do not use substitution method. Y' = (Y / x) + LNX
Here P (x) = - 1 / x, q (x) = LNX

Divide both sides by X: y '/ X-Y / x ^ 2 = LNX / X
That is, (Y / x) '= LNX / X
Simultaneous integration: Y / x = ∫ LNX / xdx = 1 / 2 (LNX) ^ 2 + C
y=1/2x(lnx)^2+Cx

If the interval (1,2) satisfies the inequality x2 + MX + M2 + 6m ＜ 0, the value range of real number m is obtained

Let f = x2 + MX + M2 + 6m, then f (1)

3632 △ 28 = vertical calculation

It's 129 plus 20

lim(x→0)(cos(xe^x)-cos(xe^-x))/x^3

The process of quadratic equation of two variables (2) {8x-3y = 11 X-Y = - 8 (3) 1 / 4x-3y = 8 y-2x = 5

{8x-3y=11 （1）,x-y=-8 （2）
From (2), x = - 8 + y (3)
Substituting (3) into (1), 8 (- 8 + y) - 3Y = 11
If we simplify it, we get 5Y = 75
Divide both sides by 5 to get y = 15
Substituting y = 15 into (3) yields x = - 8 + 15 = 7
So {x = 7, y = 15