1.a square has perimeter of 46cm.What is the length of each side?
2.how many days in february 1998?
3.True or False?Is 1888 a leap year?
2007-01-12 17:51:18 · 11 answers · asked by waiyue 1 in Science & Mathematics ➔ Mathematics
Answer these questions involved maths?
2007-01-12 17:53:15 · update #1
1. Length of each side = 46/4 cm = 11.5 cm.
2. February 1998 had 28 days. (Not a leap year; 98 is not divisible by 4.)
3. True. (1888 is a leap year; 88 IS divisible by 4; second answer on this part is WRONG, and third answer is ambiguous --- to what does "above" refer? --- to the previous responder's part 3, or to the same responder's parenthetical remarks in his/her own part 2? These conflict. As I said, it's ambiguous.)
Live long and prosper.
2007-01-12 17:58:40 · answer #1 · answered by Dr Spock 6 · 0⤊ 0⤋
1. All sides of a square are equal. So let's name a side a. You have: a=46/4=11,5
2.1998 isn't a leap year (because 1998 isn't divided by 4) so February of 1998 has 28 days
3. 1888 is divided by 4 (88/4=22) so 1888 is a leap year. True
2007-01-13 02:28:50 · answer #2 · answered by Bill 1 · 0⤊ 0⤋
1. perimeter of a square is 4 * the length of its side, therefore:
46 / 4 = 11.5cm
2. 1998 isnt a leap year so there were only 28 days on February 1998.
3. Its TRUE that 1888 is a leapyear... If the year is divisible by four, it is a leapyear.
2007-01-13 02:33:43 · answer #3 · answered by Autisteek 2 · 0⤊ 0⤋
1. Perimeter = 4a = 46 so a = 11.5 cm
2. you should find out if 1998 is divisible by 4, if it is then February has 29 days if not then it has only 28 days.
1998/4=499.5 then february 1998 has 28 days
3. 1888/4 = 472 divisible by 4 then yes it is a leap year
2007-01-13 03:06:52 · answer #4 · answered by Anonymous · 0⤊ 0⤋
Yes all of above involved mathhhhh!!!!!!!!!
+++++++1st is geometry basics. side=perimeter/no of side hence it will be 46/4 ie 11.5
+++++++2nd 1998 divisible by 4 then it will be leap year but its is not hence days in feb will be 28 not 29 (29 if it will leapyear)
+++++++3rd one also related to maths as 1888 divisible by 4 then it will be leap year hence true
2007-01-13 02:40:13 · answer #5 · answered by Abu F 2 · 0⤊ 0⤋
1. Look at how you would solve the problem, 46/4= ?
2. Was that a leap year? (The rules of leap years: Every year that is divisible by four, except for centuries, then only those that are also divisible by 4, for example, 1900, since 19 isn't divisible, is not a leap year, but 1600, since 16 is divisible by 4, was a leap year)
3. Look above
2007-01-13 01:57:18 · answer #6 · answered by Dipti 2 · 0⤊ 0⤋
Presidents :
Nixon 1972 - 1974
Gerald Ford 1974 - 1976
Jimmy Carter 1976 - 1980
Ronald Reagan 1980 - 1984
Ronald Reagan 1984 -1988
George Bush Sr. 1988 - 1992
William Jefferson Clinton 1992 -1996
William Jefferson Clinton 1996-2000
George Bush Jr 2000 - 2004
George Bush Jr 2004 -2008
The more times change, the more they remain the same.
2007-01-13 02:28:16 · answer #7 · answered by Anonymous · 0⤊ 0⤋
Leap years are presidential election years. In 1788, George Washington was elected. In 1888, Grover Cleveland, or someone like that, was elected. Who was elected in 1988?
2007-01-13 02:06:57 · answer #8 · answered by ? 6 · 0⤊ 1⤋
1.each side=23/2=11.5 cms
2.28 days
3.yes.
2007-01-13 02:10:21 · answer #9 · answered by Ashley. 3 · 0⤊ 0⤋
46 divided by 4 = 11.5cm
Not a leap year if you can't divide it by 4
yes
2007-01-13 01:56:47 · answer #10 · answered by Anonymous · 0⤊ 0⤋