1. A cylindrical oil barrel, measured from the inside, has a bottom diameter of 40 cm and a height of 60 cm. Can this oil barrel hold 80 liters of oil? 2. A cylindrical milk powder box, the volume is 5024 cubic centimeters, the bottom radius is 10 cm. How high is it? 3. A cylinder shaped steel part has a bottom area of 5 square centimeters and a length of 8 centimeters. If each cubic centimeter of steel weighs 7.8 grams, how many grams does the part weigh? 4. A cylindrical glass, measured from the inside, has a bottom diameter of 6cm and a depth of 10cm. How many milliliters of water can this glass hold at most? (mark the question number)

1 20 * 20 * 3.14 * 60 = 75260 cm3 75260 cm3 = 75.26l 2 5024 ^ (10 * 10 * 3.14) = 160cm3 5 * 8 * 7.8 = 312kg4 3 * 3 * 3.14 * 10 = 94.2 cm3 94.2 cm3 = 94.2ml

The figure below is the shape of a cylinder after cutting off part of it. Find its volume in cm

Take another identical object and turn it upside down to form a cylinder with a height of 4 + 6 = 10 cm
So the volume of this object is half the volume of a cylinder 10 centimeters high
The radius is
4 △ 2 = 2 (CM)
What is the volume of a cylinder
2 × 2 × 3.14 × 10 = 125.6 (cm3)
What is the volume of this object
125.6 △ 2 = 62.8 (cm3)

When is the sun closest to the earth in a day?
If the sun is almost as close to the earth in the morning and at noon, why is the sun at noon hotter than in the morning?

Your problem is not directly related to the distance from the sun to the earth, or the main factor
The environmental temperature depends on the angle of solar radiation. The more vertical the angle is, the faster the temperature rises. However, it is not necessarily the highest temperature. Generally, around two o'clock in the afternoon is the hottest time of the day. At this time, the heat provided by solar radiation has begun to decrease less than the natural temperature drop, or the heat provided by the sun has begun to fail to maintain the temperature rise
In fact, in winter, the earth is closest to the sun, which is usually called perihelion, but this season is the coldest time of the year, because the sun's angle is not enough

Solving practical problems
It is known that the solution X of the system X-Y = 1 + 3m x + y = - 7-m is a non positive number and Y is a non negative number,
(1) Find the value range of M
(2) Simplify | 2m-6 | + | m + 2|
It's a process
It's a bonus!

（1）
X-Y=1+3M………… （1）
X+Y=-7-M………… （2）
(1) + (2) get: 2x = - 6 + 2m
X=-3+M
(2) (1) 2Y = - 8-4m
Y=-4-2M
X is a non positive number, that is, X

There are two squares with a circumference difference of 12 cm and an area difference of 39 square cm. Are the small squares less in area?

Suppose the side length of a small square is x cm, then the side length of a square is (x + 3) cm
﹙x＋3﹚²－ x²＝39
6x＋9＝39
x＝5
The area of a small square is: 5 × 5 = 25 square centimeters

How to calculate 5.5 * 13.6-0.36 * 55 simply?

5.5*13.6-0.36*55
=5.5*13.6-3.6*5.5
=5.5*(13.6-3.6)
=5.5*10
=55

To find the process! 4cos ^ 2 α - (2 + 2 √ 3) cos α + √ 3 = 0, find the acute angle α

Factorization
(2cosα-√3)(2cosα-1) = 0
Cos α = √ 3 / 2 or 1 / 2
α = 30 ° or 60 °

As shown in the picture, a rectangular grassland is 16 meters long and 10 meters wide. There are two paths in the middle. One is a rectangle and the other is a parallelogram. The area of the grassland is______ Square meters

(16-2) × (10-2) = 14 × 8 = 112 (square meters); answer: the area of grassland is 112 square meters

1 / 2:5 / 7, 2 / 5:1, 9:1 / 11

10/7、5/2、99

[high school mathematics] it is known that the sequence {an} is an equal ratio sequence whose first term is A1 = 1 / 4 and common ratio q = 1 / 4. Suppose that the sequence {BN} satisfies BN + 2 = 3log 1 / 4 an (n ∈ n ×) (1) find the first n terms and Sn of the sequence {an + BN}; (2) if CN satisfies CN = an times BN and CN ≤ 1 / 4 M2 + M-1 holds for all positive integers n, find the value range of real number M

(1) From the meaning of the title, we can get an = (1 / 4) ^ n;
Then: BN + 2 = 3 * log (1 / 4) an = 3N;
So: BN = 3n-2, is the arithmetic sequence;
(2) From the condition CN = an * BN, we can get that:
Cn= (1/4)^n*(3n-2)=3n*(1/4)^n-2*(1/4)^n
Let the sum of the first n terms of cn be SN;
So: SN = 3 [1 / 4 + 2 * (1 / 4) ^ 2 + +n*(1/4)^n]-2*(1/4+(1/4)^2+…… +(1/4)^n);
Let PN = 1 / 4 + 2 * (1 / 4) ^ 2 + +n*(1/4)^n; --------(1)
Then: 1 / 4 * PN = (1 / 4) ^ 2 + 2 * (1 / 4) ^ 3 + +n*(1/4)^(n+1); ------(2)
(1) (2) the results were as follows
3/4 Pn=1/4+(1/4)^2+(1/4)^3+…… +(1/4)^n-n*(1/4)^(n+1) = 1/3*(1-(1/4)^n)- n*(1/4)^(n+1)
So Sn can be deformed as:
Sn=3[1/3*(1-(1/4)^n)- n*(1/4)^(n+1)]-2*[1/3*(1-(1/4)^n)]
=1/3*[1-(1/4)^n]-3n*(1/4)^(n+1);