How many different 2-digit numbers are there with the following property: the tenth digit is greater than the units digit?

There is 45 different 2-digit numbers.

Step-by-step explanation:

My way is kinda dumb, but it still works. So, 2 digit numbers is from 10-99. We can start from the 10-19 first.

10, 11,12,13,14,15,16,17,18,19

We need to find the numbers that the tenth digit is greater than the units digit.

10, 11,12,13,14,15,16,17,18,19

Since 1 is the tenth digit, all the ones digits are all going to be bigger. Same goes with 20-29. Then you will have 2 numbers that the tenth digit is greater. My way applies when the tenth digit is only less than the ones digit, not greater. If you do this way...

10-19: 1 number

20-29: 2 numbers

30-39: 3 numbers

40-49: 4 numbers

50-59: 5 numbers

60-69: 6 numbers

70-79: 7 numbers

80-89: 8 numbers

90-99: 9 numbers

So, if you add 1+2+3+4++9(which all of you probably memorized by now) it would be 45. The answer is 45. Hope this helped!

x² + 11x + 18 < 0  

[-11 ± sqrt(121 - 72)]/2 = (-11 ± 7)/2  

solution: (-9, -2)

step-by-step explanation:

x= 3^y

step-by-step explanation: