Answer:
4. 54- 8.5x>20
Step-by-step explanation:
Catherine only has $54, so she cannot spend more than that.
The canvas will cost at least $20, but we don't know how much exactly.
The tubes cost $8.50 each.
So, she starts with a total budget of $54, out of which she will buy paints (8.5x) and she wants to have at least $20 left for canvas.
So, we transpose those facts into the inequity:
54 - 8.5x > 20
Answer:
The 185th digit following that pattern of '12345678910111213141516' would be the number '5'
Step-by-step explanation:
The reason the 185th digit is '5' is because the pattern increases it's standard number by the following number (ex, 1,2,3,4,5, etc), and eventually it reached 185 as one of it's numbers, '5' would be the digit because the '18' part of 185 would only be the 184th digit, hence you are needed to include the '5' to complete the sequence of that number to reach the 185th digit.
Let’s look at the permutations of the letters “ABC.” We can write the letters in any of the following ways:
ABC
ACB
BAC
BCA
CBA
CAB
Since there are 3 choices for the first spot, two for the next and 1 for the last we end up with (3)(2)(1) = 6 permutations. Using the symbolism of permutations we have:

. Note that the first 3 should also be small and low like the second one but I couldn’t get that to look right.
Now let’s see how this changes if the letters are AAB. Since the two As are identical, we end up with fewer permutations.
AAB
ABA
BAA
To make the point a bit better let’s think of one A are regular and one as bold
A.
ABA and AB
A look different now because we used bold for one of the As but if we don’t do this we see that these are actual the same. If they represented a word they would be the same exact word.
So in this case the formula would be

. We use 2! In the denominator because there are 2 repeating letters. If there were three we would use 3!
Hopefully, this is enough to let you see that the answer is A. The number of permutations is limited by the number of items that are identical.
Answer:
![x_3 = \left[\begin{array}{c}4&3&1\\0\end{array}\right]](https://tex.z-dn.net/?f=x_3%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D4%263%261%5C%5C0%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
According to the given situation, The computation of all x in a set of a real number is shown below:
First we have to determine the
so that 
![\left[\begin{array}{cccc}1&-3&5&-5\\0&1&-3&5\\2&-4&4&-4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%26-3%265%26-5%5C%5C0%261%26-3%265%5C%5C2%26-4%264%26-4%5Cend%7Barray%7D%5Cright%5D)
Now the augmented matrix is
![\left[\begin{array}{cccc}1&-3&5&-5\ |\ 0\\0&1&-3&5\ |\ 0\\2&-4&4&-4\ |\ 0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%26-3%265%26-5%5C%20%7C%5C%200%5C%5C0%261%26-3%265%5C%20%7C%5C%200%5C%5C2%26-4%264%26-4%5C%20%7C%5C%200%5Cend%7Barray%7D%5Cright%5D)
After this, we decrease this to reduce the formation of the row echelon
![R_3 = R_3 -2R_1 \rightarrow \left[\begin{array}{cccc}1&-3&5&-5\ |\ 0\\0&1&-3&5\ |\ 0\\0&2&-6&6\ |\ 0\end{array}\right]](https://tex.z-dn.net/?f=R_3%20%3D%20R_3%20-2R_1%20%5Crightarrow%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%26-3%265%26-5%5C%20%7C%5C%200%5C%5C0%261%26-3%265%5C%20%7C%5C%200%5C%5C0%262%26-6%266%5C%20%7C%5C%200%5Cend%7Barray%7D%5Cright%5D)
![R_3 = R_3 -2R_2 \rightarrow \left[\begin{array}{cccc}1&-3&5&-5\ |\ 0\\0&1&-3&5\ |\ 0\\0&0&0&-4\ |\ 0\end{array}\right]](https://tex.z-dn.net/?f=R_3%20%3D%20R_3%20-2R_2%20%5Crightarrow%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%26-3%265%26-5%5C%20%7C%5C%200%5C%5C0%261%26-3%265%5C%20%7C%5C%200%5C%5C0%260%260%26-4%5C%20%7C%5C%200%5Cend%7Barray%7D%5Cright%5D)
![R_2 = 4R_2 +5R_3 \rightarrow \left[\begin{array}{cccc}1&-3&5&-5\ |\ 0\\0&4&-12&0\ |\ 0\\0&0&0&-4\ |\ 0\end{array}\right]](https://tex.z-dn.net/?f=R_2%20%3D%204R_2%20%2B5R_3%20%5Crightarrow%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%26-3%265%26-5%5C%20%7C%5C%200%5C%5C0%264%26-12%260%5C%20%7C%5C%200%5C%5C0%260%260%26-4%5C%20%7C%5C%200%5Cend%7Barray%7D%5Cright%5D)
![R_2 = \frac{R_2}{4}, R_3 = \frac{R_3}{-4} \rightarrow \left[\begin{array}{cccc}1&-3&5&-5\ |\ 0\\0&1&-3&0\ |\ 0\\0&0&0&1\ |\ 0\end{array}\right]](https://tex.z-dn.net/?f=R_2%20%3D%20%5Cfrac%7BR_2%7D%7B4%7D%2C%20%20R_3%20%3D%20%5Cfrac%7BR_3%7D%7B-4%7D%20%20%5Crightarrow%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%26-3%265%26-5%5C%20%7C%5C%200%5C%5C0%261%26-3%260%5C%20%7C%5C%200%5C%5C0%260%260%261%5C%20%7C%5C%200%5Cend%7Barray%7D%5Cright%5D)
![R_1 = R_1 +3 R_2 \rightarrow \left[\begin{array}{cccc}1&0&-4&-5\ |\ 0\\0&1&-3&0\ |\ 0\\0&0&0&-1\ |\ 0\end{array}\right]](https://tex.z-dn.net/?f=R_1%20%3D%20R_1%20%2B3%20R_2%20%5Crightarrow%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%260%26-4%26-5%5C%20%7C%5C%200%5C%5C0%261%26-3%260%5C%20%7C%5C%200%5C%5C0%260%260%26-1%5C%20%7C%5C%200%5Cend%7Barray%7D%5Cright%5D)
![R_1 = R_1 +5 R_3 \rightarrow \left[\begin{array}{cccc}1&0&-4&0\ |\ 0\\0&1&-3&0\ |\ 0\\0&0&0&-1\ |\ 0\end{array}\right]](https://tex.z-dn.net/?f=R_1%20%3D%20R_1%20%2B5%20R_3%20%5Crightarrow%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%260%26-4%260%5C%20%7C%5C%200%5C%5C0%261%26-3%260%5C%20%7C%5C%200%5C%5C0%260%260%26-1%5C%20%7C%5C%200%5Cend%7Barray%7D%5Cright%5D)

![x = \left[\begin{array}{c}4x_3&3x_3&x_3\\0\end{array}\right] \\\\ x_3 = \left[\begin{array}{c}4&3&1\\0\end{array}\right]](https://tex.z-dn.net/?f=x%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D4x_3%263x_3%26x_3%5C%5C0%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%5C%5C%20x_3%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D4%263%261%5C%5C0%5Cend%7Barray%7D%5Cright%5D)
By applying the above matrix, we can easily reach an answer
For this on you need to set up two equations
x + y = 155 total number of packages
3x+8y=815 total weight
Then multiply both sides of the first equation by 3: 3x +3y = 465
Then combine the two equations by subtracting the first one from the second one which gets you 5y=350 so y = 70 so we have 70 packages weighting 8 pounds.