AR 25-2 specifies MINIMUM password length, but the only limitation on MAXIMUM length is how long a password the OS or application can handle; AR 25-2 does not specify a maximum password length, however,
According to AR 25-2, Section IV, paragraph 4-12 b:
The IAM or designee will manage the password generation, issuance, and control process. If used, generate passwords in accordance with the BBP for Army Password Standards.
BBP for Army password standards are contained in 04-IA-O-0001, paragraph 5A:
(1) All system or system-level passwords and privileged-level accounts (e.g., root, enable, admin, administration accounts, etc.) will be a minimum of 15-character case-sensitive password changed every 60 days (IAW JTF-GNO CTO).
(2) All user-level, user-generated passwords (e.g., email, web, desktop computer, etc.) will change to a 14-character (or greater) case-sensitive password changed every 60 days.
Length = 252/18 = 14 feet
3 buses are needed for the fieldtrip, providing maximum capacity for 252 students.
1 yard = 36 inches 7 yards = 252 inches. They are the same length.
Using Pythagoras' theorem: 252-152 = 400 and the square root of this is 20 units in length.
The Tiber is 252 miles or 405km in length.
252 metres( 826.8 ft) in length and 26 metre width
252 square ft. Since the room is rectangular, you find the area by multiplying the length and the width, in this case being 18 and 14. 18x14=252 so the area, which is always in square units, is 252 square feet.
Doesn't seem likely, does it? A diagonal 50 times the length of the sides? Diagonal = sqrt(252 + 252), ie sqrt 1250(!) which is 35.36 to the nearest hundredth.
12 ft
Multiply width by length to get the area. The answer would be 252 square feet.
There are 36 inches to a yard. 36 × 11 = 396 inches Therefore 11 yards is longer than 252 inches.
It is 42 inches.