0.3 is the same as 3/10 (three tenths)
so the digit in the tenths place would be 3
As written ' As written ; 05 & 03 ; just drop the prefixed zeroes, they are trivial , hence 5 > 3. However, I think you may mean 0.5 & 0.3 Since both are to one decimal place, just drop the decimal point and prefixed zero ; 5 & 3 . Since 5 > 3 , then it follows 0.5 > 0.3 05
Three-thousandths .3 - three - Tenths .03 - three - hundredths .003 - three - Thousandths
153 = 13 + 53 + 33 Also: 370 = 33 + 73 + 03 371 = 33 + 73 + 13 407 = 43 + 03 + 73 See the Related Links for "Exponentiated digit sums" to the bottom for more...
To get this answer you can multiply 15 by .03, since any percentage you take the decimal place and move it over to the right twice. 15 times .03 equals 0.45
03
If number in ones place is 5 or more round up, if 4 or less, round down. For example, .09 rounds to 1.0, .03 rounds to zero
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As written ' As written ; 05 & 03 ; just drop the prefixed zeroes, they are trivial , hence 5 > 3. However, I think you may mean 0.5 & 0.3 Since both are to one decimal place, just drop the decimal point and prefixed zero ; 5 & 3 . Since 5 > 3 , then it follows 0.5 > 0.3 05
359702057575605/01
Three-thousandths .3 - three - Tenths .03 - three - hundredths .003 - three - Thousandths
99. But only if you consider the first nine as 01, 02, 03 etc. If not, there are 90 2-digit numbers, starting with 10 and ending with 99.
153 = 13 + 53 + 33 Also: 370 = 33 + 73 + 03 371 = 33 + 73 + 13 407 = 43 + 03 + 73 See the Related Links for "Exponentiated digit sums" to the bottom for more...
-- If 'zero' is always the first digit, then we're pretty darn sure that no other digit can be repeated in the same 2-digit number, since there's only one digit left to be filled in. -- If 'zero' is always the first digit, then there are exactly six possible 2-digit numbers: 00 01 02 03 04 and 05.
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