Answer:
680
Step-by-step explanation:
Use the binomial coefficient where you choose
numbers out of
possible numbers and find the total amount of combinations since order does not matter:

Thus, you can make 680 three-non-repeating-number codes
Answer:
around 1.3
Step-by-step explanation:
Answer: 3 stickers
Step-by-step explanation:
From the question, we know that Pippa's 8 friends have an equal amount of stickers, meaning that the number of stickers that Pippa gave out is a multiple of 8.
Also, we are able to know that Pippa gave as much as she can, meaning that she gave out the stickers until the number is the maximum multiple of 8.
First Five Multiple of 8 = 8, 16, 24, 32, 40
As we can see from the list, 8, 16, and 24 are all multiples of 8, but they are not the maximum number that could fit under 35 stickers. Similarly, 40 exceeds the number of stickers Pippa has. Thus, we are left with 32.
This means, Pippa gave out 32 stickers in total, and each friend got 4 stickers.
Also, this means Pippa would keep <u>3 stickers</u> for herself.
Hope this helps!! :)
Please let me know if you have any questions
Step-by-step explanation:
The angle between vectors is given by

The magnitudes for the vectors are as follows:




The dot product between the vectors is

Therefore, the angle between the two vectors is

or

Answer: Hello your question is poorly written attached below is the complete question
answer:
![y = \left[\begin{array}{ccc}-4\\-11\\5\end{array}\right]](https://tex.z-dn.net/?f=y%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-4%5C%5C-11%5C%5C5%5Cend%7Barray%7D%5Cright%5D)
![x = \left[\begin{array}{ccc}16\\12\\-40\end{array}\right]](https://tex.z-dn.net/?f=x%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D16%5C%5C12%5C%5C-40%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
![y = \left[\begin{array}{ccc}-4\\-11\\5\end{array}\right]](https://tex.z-dn.net/?f=y%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-4%5C%5C-11%5C%5C5%5Cend%7Barray%7D%5Cright%5D)
![x = \left[\begin{array}{ccc}16\\12\\-40\end{array}\right]](https://tex.z-dn.net/?f=x%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D16%5C%5C12%5C%5C-40%5Cend%7Barray%7D%5Cright%5D)
attached below is the detailed solution using LU factorization