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A 24 cm long and 8 cm wide rectangular sheet iron, can you cut it into 5 pieces and weld it into a cuboid container with a cube at the bottom? Draw a picture of the scissors. What's the volume of the container?
A factory wants to use a 240 cm long and 120 cm wide rectangular sheet iron to weld a 30 cm high cuboid uncovered water tank. Please design a reasonable plan
(it is required to save materials and ensure the maximum volume of water tank)
Tell me Danson: Hello
Cut a 120cm × 120cm square iron sheet for the bottom
Cut four pieces of 120cm × 30cm rectangular iron sheet for four sides
Volume = (120cm) &# 178; × 30cm = 432000cm & # 179; = 432dm & # 179;
In this way, there is no leftover material at all, and the volume is 432 cubic decimeters (liters)
Good luck and goodbye
A piece of 8 decimeters long and 4 decimeters wide iron sheet is welded into a 1 decimeter deep uncovered rectangular water tank. You can design several schemes, which one has the largest volume?
Cut out a 1-decimeter-long square from each corner, then fold it up and weld it with 6 乗 2 times 1, that is 12 cubic decimeters
Xiaowan's mobile phone received the following message: "think of a number, add 52.8 to it, multiply by 5, subtract 3.9343, divide by 0.5, and finally subtract ten times of the number in mind. The answer is very romantic!" According to the above instructions, let the number in Xiaowan's mind be x, and find out the romantic number
According to the meaning of the title, we get 10.5 [5 (x + 52.8) - 3.943] - 10x, = 2 (5x + 264-3.943) - 10x, = 520.1314
Ax + by = 2A + 3 2x-3y = 5 and X + 9y = - 8, ax-by = 5b-1 find the value of a and B
In this paper, we first solve 2x-3y = 5 (1) x + 9y = - 8 (2) (1) * 3 + (2) 6x-9y + X + 9y = 15-87x = 7x = 1, y = (- 8-x) / 9 = - 1 into ax + by = 2A + 3ax by = 5b-1a-b = 2A + 3 (3) a + B = 5b-1 (4) from (3), we get a + B = - 3 (5) into (4) 5b-1 = - 3B = - 2 / 5A = - 3-B = - 13 / 5
It should be accurate to 20 digits
3.14159265358979323846
Calculate the value of the fourth-order determinant, the first line: 1.21 2, the second line: 5.24 3, the third line: 0.01 2, the fourth line: 0.04 1 (PS:
Method 1: 121 25 224 300 1 200 41 = 122 2 [multiply the first row by - 5 and add it to the second row] 0 - 8 - 1 - 7 001 200 41 = 12 - 6 2
51+52+53+…… 79+80=( )
1965
When x approaches 0, what is the limit of [(3-E ^ x) / (2 + x)] ^ (1 / SiNx),
lim(x→0) ln[(3-e^x)/(2+x)]^(1/sinx)
=lim(x→0) ln[(3-e^x)/(2+x)]/sinx
=lim(x→0) [ln(3-e^x)-ln(2+x)]/sinx
=lim(x→0) [ln(3-e^x)-ln(2+x)]/x (0/0)
=lim(x→0) -e^x/(3-e^x)-1/(2+x)
=-1
therefore
lim(x→0) [(3-e^x)/(2+x)]^(1/sinx)
=lim(x→0) e^ln[(3-e^x)/(2+x)]^(1/sinx)
=e^(-1)
=1/e
3 / 4 / 6 / 11 + 3 / 17 + 5 / 8
Original form
=3/4x11/6+3/17+5/8
=11/8+5/8+3/17
=2+3/17
=2 and 3 / 17
RELATED INFORMATIONS
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