A cylindrical timber with a radius of 20cm. After sawing it into three equal small cylinders, how many square centimeters has the surface area increased

A cylindrical timber with a radius of 20cm. After sawing it into three equal small cylinders, how many square centimeters has the surface area increased

14 × 20 & sup2; × 4 = 5024 (cm2)

The bottom diameter of a cylindrical timber is 16cm and its height is 20cm. Cut it into two equal blocks along its bottom diameter. What is the surface area of each block?
Quick!

Bottom radius: 16 / 2 = 8 cm
Bottom area: 3.14 * 8 * 8 = 200.96 square centimeter
Side area: 3.14 * 16 * 20 = 1004.8 square centimeter
Surface area: 200.96 * 2 + 1004.8 = 1406.72 square centimeter
Area of each block: (1406.72/2) + 16 * 20 = 1023.36 square centimeter

The phases of the moon change regularly: what is the phase of the first half of the moon and what is the phase of the second half of the moon?

The phases of the moon are calculated by the degree of the difference between the meridians of the sun and the moon (the following degree is the difference between the meridians of the sun and the moon)
New moon: 0 degrees;
On Emei month (generally around the second night of the lunar new year --- around the seventh day of the Lunar New Year): 0 degrees --- 90 degrees;
The first quarter of the moon (around the eighth day of the lunar calendar): 90 degrees;
Convex moon (about the ninth day of the lunar calendar --- about the fourteenth day of the lunar calendar): 90 degrees --- 180 degrees;
Full moon (watch the sun, on the 15th or 16th of the lunar calendar): 180 degrees;
Waning moon (around the 16th of the lunar calendar --- around the 23rd of the lunar calendar): 180 degrees --- 270 degrees;
The last quarter of the moon (around the 23rd lunar calendar): 270 degrees;
Next Emei month (around the 24th lunar month --- the end of the month): 270 degrees --- 360 degrees;
In addition, the last day of the lunar month is called the dark day, that is, there is no moon;
There are four main phases of the moon: the new moon (the first day of the lunar calendar), the first string (about the eighth day of the lunar calendar), the full moon (about the fifteenth day of the lunar calendar), and the last string (about the twenty third day of the lunar calendar). They all have a definite occurrence time, which is calculated by precise orbit;

Midterm test questions fifth grade math questions

Name of midterm paper of fifth grade mathematics volume I
1、 Fill in the blanks:
1. Keep 5.2819 to three decimal places as (), and keep two decimal places as ()
2. Four times the number a is equal to five times the number B. It is known that the number B is 0.8 and the number a is ()
3、3.6×1.9+0.36×81=3.6×(1.9+ ）
4. The smallest two digits are (), the largest three digits are (), and their sum multiplied by 0.01 product is ()
5. If you reduce the largest three digits by 1000 times, it will be () and then add 0.01, divide by 0.5, and the quotient will be ()
6. A two place decimal "rounded" to retain a decimal can get 8.5, the two place decimal maximum is (), the minimum is ()
7. The product of 0.98 × 0.46 is () decimal. 100 times of 0.365 is ()
8. 75 is () times of 2.5
9. When the quotient of 5.03 △ 0.4 is 12.5, its remainder is ()
10. Fill in ()

Guangming primary school campus has a 15 meter diameter circular flower bed, surrounded by a 50 cm wide cement path, can you calculate the area of this cement path?

15÷2=7.5
3.14 × (7.5 + 0.5) × (7.5 + 0.5) - 3.14 × 7.5 × 7.5 = 24.335m and 178;
The area of this cement road is 24.335 meters;

4+5+6+7+8+… +80 9+19+29+39+49+59+69 465+476-365． 11+12+13+… +20 987-491-187．

（1）4+5+6+7+8+… +80，=（4+80）×77÷2，=84×77÷2，=3234；（2）9+19+29+39+49+59+69，=（9+69）×7÷2，=78×7÷2，=273；（3）465+476-365，=465-365+476，=100+476，=576；（4）11+12+13+… +20，=（11+20）×...

It is known that there are two points a and B on the parabola y = ax ^ 2 (a > 0), whose abscissa are - 1 and 2 respectively. When we find out the value of a, the triangle AOB is a right triangle

The coordinates of a and B can be solved, a (- 1, a), B (2,4a); then | OA | ^ 2 = a ^ 2 + 1; | ob | ^ 2 = 16A ^ 2 + 4; | ab | ^ 2 = 9A ^ 2 + 9, AOB is a right triangle, and the equation can be listed according to Pythagorean theorem
① If the angle AOB is a right angle, then | OA | ^ 2 + | ob | ^ 2 = | ab | ^ 2 is brought into the solution and a = √ 2 / 2 (a = - √ 2 / 2 is rounded off)
② If the angle ABO is a right angle, then | ab | ^ 2 + | ob | ^ 2 = | OA | ^ 2, the equation has no solution
③ If the angle Bao is a right angle, then | OA | ^ 2 + | ab | ^ 2 = | ob | ^ 2 is brought into the solution and a = 1 (a = - 1 is rounded off)
In conclusion, when a = √ 2 / 2 or 1, AOB is a right triangle

The original price of a dress is 200 yuan. First, the price is reduced by 10% and then increased by 10%. How much is the price of this dress now?

200 × (1-10%) × (1 + 10%), = 200 × 90% × 110%, = 198 yuan; a: the current price of this dress is 198 yuan

Given the function f (x) = loga1 + x1 − x (where a ＞ 1); (I) find the domain of definition of F (x); (II) judge the parity of F (x) and give the proof; (III) find the range of value of X to make f (x) ＞ 0

(I) from the function f (x) = loga1 + x1 − x (where a ＞ 1), we can get 1 + x1 − x ＞ 0, that is, x + 1x − 1 ＜ 0, that is, (x + 1) (x-1) ＜ 0, the solution is - 1 ＜ x ＜ 1, so the definition domain of the function is (- 1,1). (II) because the definition domain of the function is symmetric about the origin and satisfies f (- x) = loga1 − X1 + x = - loga1 + x1 − x = - f (x), the function is odd. (III) from F (x) ＞ 0 We can get 1 + x1 − x ＞ 1, that is, 2XX − 1 ＜ 0, 2x (x-1) ＜ 0. The solution is 0 ＜ x ＜ 1, so the value range of X is (0, 1)

The ratio of length, width and height of a cuboid is 3:2:1. If the sum of cuboid's edge lengths is equal to the sum of cuboid's edge lengths, the ratio of cuboid's surface area to cuboid's surface area is (), and the ratio of cuboid's volume to cuboid's volume is ()

Let the cuboid be 3x, 2x, X
It can be found that the side length of the cube is 2x
Cuboid surface area = 2 (3x * 2x + 2x * x + 3x * x) = 22x & sup2;, volume = 3x * 2x * x = 6x ^ 3
Cube surface area = 6 * 2x * 2x = 24x & sup2; volume = 2x * 2x * 2x = 8x ^ 3
The surface area ratio is 22x & sup2;: 24x & sup2; = 11:12
The volume ratio is 6x ^ 3:8x ^ 3 = 3:4