91
91 ÷ 12 = 7 remainder 7
To find the number of tens in 209 and 710, we first need to break down each number. In 209, there are 2 tens. In 710, there are 1 ten. Adding these together, we have a total of 2 + 1 = 3 tens in the numbers 209 and 710 altogether.
Every ten there is a nine in the ones place, therefore there are 100 nines in the ones place.Every hundred there are 10 nines in the tens place, therefore add 10 * 10 nines. eg 90, 91, 92, 93, ... in the tens placeEvery thousand there are 100 nines in the hundreds place, therefore add 100 nines.Total is 300.
i can tel one digit ie... 1 tens digit i dont knoww sorry will tel u later
That is $922.(if you're talking about dollar bills)
91
The number that is 2 tens (or 20) more than 71 is 91.
91 since the ones digit is 8 less than 9 and the two digits, 9+1 = 10, a two-digit number.
Assuming that the question refers to 2-digit numbers, the units [ones] digit must be 1 3 7 or 9 otherwise it cannot be prime. Since the two digits add to ten, the number must belong to the set {91, 73, 37, 19} The ten digit is the larger of the two so the number cannot be 37 or 19. So it must belong to the set {91, 73) 91 = 7*13 so is not a prime, whereas 73 is. So the answer is 73.
Let's assume the number is represented by 'x'. The given statement "Twice a number is added to seventy nine is the result of ninety one" can be translated into the equation 2x + 79 = 91. Simplifying the equation, we get 2x = 12. Therefore, the number x is equal to 6.
91, and 97 are the only ones.
10*10 - 10 + 10/10 = 100 - 10 + 1 = 91
12 x 91 = 1092
12 divided 91 = 0.13186813186813187
90.763 rounded to the ones place is 91
91 ÷ 12 = 7 remainder 7