For the first digit of the required 20-digit number, we can have 9 choices for the number, those are the digits 1 to 9. In the second digit of the number, we will only have 8 choices. That is because, we don't want to write a number in the second digit that is similar to the first digit and the third digit.
Similarly, in the third digit's place, we will only have 8 choices because we do not want the digit to be similar to the second digit and the fourth digit. This will go on up until the 19th digit. In the 20th digit, we will also have 9 choices (from 0-9) but we do not want the digit to be similar to the 19th digit.
From the explanation given above, we have the number of ways,
n = 9 x 8^(19 - 1) x 9 = 1.459 x 10^18
<em>ANSWER: 1.459 x 10^18</em>
6 rows are empty.
112/7 = 16 rows
70/7 = 10 rows used.
Answer:
For #14, (around) 108 times.
#15 = 53
Step-by-step explanation:
14)
2.69*10^5 = 269000
2.5*10^3 = 2500
269000/2500 = 107.6
Rounding up, we get 108
15)
√(53 - 8)2 + (31 - 3)2
= √(45)2 + (28)2
= √2809
= 53
Hope this helped! I would love a Brainliest if this helped you!
In this question, we are given a matrix, and we have to perform the given operation.
The matrix is:
![\left[\begin{array}{ccc}6&-1&|5\\1&-5&|0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D6%26-1%26%7C5%5C%5C1%26-5%26%7C0%5Cend%7Barray%7D%5Cright%5D)
The following operation is given:

In which
is the element at the first line and
is the element at the second line.
Updating the first line:



Thus, the filled matrix will be given by:
![\left[\begin{array}{ccc}0&29&|5\\1&-5&|0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%2629%26%7C5%5C%5C1%26-5%26%7C0%5Cend%7Barray%7D%5Cright%5D)
For another example where row operations are applied on a matrix, you can check brainly.com/question/18546657
Polynomial :



The rest are not polynomials.
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
I hope that helps you out!!
Any more questions, please feel free to ask me and I will gladly help you out !
~Zoey