Well, let's see . . .
0.777 has no whole-number part at all, so it's less than ' 1 '.
' 7 ' is a whole number, so it's at least ' 1 ', or maybe even more.
Can you see now that ("less than 1") can't be greater than ("at least 1 or maybe more") ?
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No.
Is this what you really meant to ask?
All numbers of the form .abc (meaning 0.abc) with a,b,c digits are between 0 and 1, by definition of the decimal writing system. E.g.,
0.777 = 7/10 + 7/100 + 7/1000 < 10/10 + 10/100 + 10/1000 = 1110/1000 = 1.11 < 2 < 7
77 / 7 - 777 / 7 - 777 / 7 - 777 / 7 - 7
Yes, 777 is in the 7 times tables. To determine this, you would divide 777 by 7. If the result is a whole number, then 777 is in the 7 times tables. In this case, 777 divided by 7 equals 111, which is a whole number, confirming that 777 is indeed in the 7 times tables.
0.009
There are: 777/7 = 111
-5
The answer is greater than, because 7 divided by 9 is .777777, and 2 divided by 4 is .5 .777 is greater than .5, and so 7/9ths is the greater of the two equations
That all depends upon the first term of the sequence: As long as the first number is less than, or equal to 777 and as long as (first_number MOD 7) ≡ (777 MOD 7) will 777 be in the sequence. 777 MOD 7 ≡ 0 ⇒ if, and only if, first_number ≤ 777 and first_number MOD 7 ≡ 0 (ie 7 divides the first number) will 777 be in the sequence.
77 / 7 - 777 / 7 - 777 / 7 - 777 / 7 - 7
.777 is greater than the rest.
-7 is not greater than 7.
Yes, 777 is in the 7 times tables. To determine this, you would divide 777 by 7. If the result is a whole number, then 777 is in the 7 times tables. In this case, 777 divided by 7 equals 111, which is a whole number, confirming that 777 is indeed in the 7 times tables.
14
0.009
The GCF is 7.
1
777*7 = 5439
There are: 777/7 = 111