Pretty sure that's right.
So just you need to rationalize the denominator so multiple top and bottom by conjugate so by sqrt6 +sqrt3
8 8(sqrt6 +sqrt3) 8(sqrt6 +sqrt3)
----------------- = ------------------------------------- = ----------------------- =
sqrt6 - sqrt3 (sqrt6 -sqrt3)(sqrt6 +sqrt3) 6 -3
8(sqrt6 +sqrt3)
= -----------------------
3
hope this will help you
The number of integers from 100 to 900 that can be written in base ten without using the digit 7 is 648.
From 100 to 999, all integers are three (3) digit numbers and available digits are (0, 1, 2, 3, 4, 5, 6, 7, 8, 9).
It implies that the total number of digits is 9. The next thing to do will be to arrange the 9 digits in three (3) places in such a way that repetition is allowed.
In the first place;
- the number of ways the number can be positioned since it can't be zero and seven = 8.
Hence, the first chance is 8 out of 10 chance.
In the second and third place;
∴
The number of integers from 100 to 900 that can be written in base ten without using the digit 7 is:
= 8 × 9 × 9
= 648
Therefore, the number of integers from 100 to 900 that can be written in base ten without using the digit 7 is 648.
Learn more about integers here:
brainly.com/question/15276410?referrer=searchResults
180 - 78 = 102 hope this help
