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Q: How do you use three 9s to make 11?

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9s + 7 = -11 9s = -18 s = -18/9 s = -2

(99)/(99) +9.

The next year after 1999 that will have three 9s in its title will be 2999.

how should I know....(Use google bruh)

Write the repeating digits over the same number of 9s and simplify: 0.81... has two repeating digits ⇒ the denominator has two 9s, ie 99. ⇒ 0.81... = 81/99 = 9/11

11

Only 1 exists, and it is "999"

11 with 1 remaining 100 - 1 = 99 = 9 x 11

The GCF of 9s and 63s^3 is 9s.

There are 27 of them.

999 + (999/999) = 1000

99 + (9 / 9) = 100

9 + (9/9) = 10

The GCF is 9s.

The numeral value 9s (nines) is the plural of 9 (nine).

(9*9)+(9+9)+(9/9) = 100

[(√9)! X (√9)!] + 9 + (√9)! = [6 X 6] + 9 + 6 = 51 Note: It can also be done with just three 9s. [(√9)! x 9] - √9 = [6 x 9] - 3 = 51

wait not the discription srry

Since 9s is a factor of 36s, it is automatically the GCF.

Take two 9s and make them a fraction like this 9/9. Then put the other two 9s together to make 99 and you get 9/9+99=100. Because 9/9=1 so it becomes 1+99=100.

9*9/9 = 9 or 9+9-9 = 9

99 9/9 ninety nine plus nine over nine equals 100

Three, master gunny, sgt major and sgt major of the marine corps

The largest number you can get before rounding into 45,000 would be 44,999. The only way to get something larger than that would to be use decimal places after the final three 9s. So 44,999 is the largest whole number.

L = S + 16; 3 x (S + 16) = 9S ie 3S + 48 = 9S ie 48 = 6S so S = 8 and L = 24

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