43543546601331 is the answer
The matrix equation that represents this situation is
![\left[\begin{array}{ccc}3&2&0\\1&0&4\\3&1&1\end{array}\right]*\left[\begin{array}{ccc}x\\y\\z\end{array}\right]=\left[\begin{array}{ccc}13.50\\16.50\\14.00\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%262%260%5C%5C1%260%264%5C%5C3%261%261%5Cend%7Barray%7D%5Cright%5D%2A%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D13.50%5C%5C16.50%5C%5C14.00%5Cend%7Barray%7D%5Cright%5D%20)
Use technology to find the inverse of matrix A:
![A^{-1}= \left[\begin{array}{ccc}-\frac{2}{5}&-\frac{1}{5}&\frac{4}{5}\\ \frac{11}{10}&\frac{3}{10}&-\frac{6}{5}\\\frac{1}{10}&\frac{3}{10}&-\frac{1}{5}\end{array}\right]](https://tex.z-dn.net/?f=A%5E%7B-1%7D%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-%5Cfrac%7B2%7D%7B5%7D%26-%5Cfrac%7B1%7D%7B5%7D%26%5Cfrac%7B4%7D%7B5%7D%5C%5C%0A%5Cfrac%7B11%7D%7B10%7D%26%5Cfrac%7B3%7D%7B10%7D%26-%5Cfrac%7B6%7D%7B5%7D%5C%5C%5Cfrac%7B1%7D%7B10%7D%26%5Cfrac%7B3%7D%7B10%7D%26-%5Cfrac%7B1%7D%7B5%7D%5Cend%7Barray%7D%5Cright%5D%20)
Multiplying A inverse by B, we get the solution matrix
[tex]\left[\begin{array}{ccc}2.50\\3\\3.50\end{array}\right][\tex]
Answer:
I believe it's enlargement
Step-by-step explanation:
because you have a positive scale factor and it goes to the right 3 and up 4
Answer:
he can get 12 bowls
Step-by-step explanation:
You have to divide 8% by 100% to get the answer but I rounded down
Answer:
B. 600,000 (1.15)^{n-1}
Step-by-step explanation:
The <em>n-th</em> term of a geometric sequence with initial value a and common ratio r can be determined by multiplying the first term of the sequence (i.e. initial value a) by r^{n-1}.
The first term (i.e. initial value a) is 600,000.
The common ratio r can be calculated by dividing any two consecutive terms in the sequence:
r = 690,000/600,000 = 1.15 <em>or</em> r = 793,500/690,000 = 1.15
Thus, we get the answer:
the explicit rule that can be used to determine the value of the art collection n years after that is 600,000 (1.15)^{n-1}
