99/9 + sqrt(9) = 11 + 3 = 14
(99/9) + sqrt(9) = 14
Sixes and Nines - 1913 was released on: USA: 14 March 1913
14 nines, with a remainder of 3 !
Given these kind of questions, there are often several answers, if we are allowed to freely use the four normal ways of calulating. We will focus on one answer here, however. We have four cars; two nines and two aces. Remember that the ace card is allowed two values: 1 and 14. We don't need to complicate matters further, though. We can give one possible answer by simply adding the two aces with the highest value (14) and subtract the two 9's: (14 + 14)-( 9 + 9) = 10 Written out: 28 - 18 = 10
There are infinitely many numbers between 14 and 15. Just type "14." then continue typing any arbitrary digits (except all zeroes, or all nines), and you have a number between 14 and 15.There are infinitely many numbers between 14 and 15. Just type "14." then continue typing any arbitrary digits (except all zeroes, or all nines), and you have a number between 14 and 15.There are infinitely many numbers between 14 and 15. Just type "14." then continue typing any arbitrary digits (except all zeroes, or all nines), and you have a number between 14 and 15.There are infinitely many numbers between 14 and 15. Just type "14." then continue typing any arbitrary digits (except all zeroes, or all nines), and you have a number between 14 and 15.
(33 ÷ 3) + 3 = 14
4+4+4+(4/4)+(4/4) = 14 :) 4+4+4=12 4/4 = 1 4/4 = 1 12+1+1=14
Use two of the 3s to make 33. Then: 33 - 3 = 11 11 + 3 = 14
One possibility is 4 + 4 + 4 + √4 = 4 + 4 + 4 + 2 = 14
This answer contains a non mathematical part:4/4 =1; then write a four14Or... 4*4 - sqrt(4) = 14
No, the only even prime number is 2. 14 is composite. All whole numbers ending in four are composite.
If you're allowed to use any symbols, then 4 + 4 + 4 + sqrt(4) will do it.