Answer:
![\left[\begin{array}{cc}2&0\\-2&3\end{array}\right]+\left[\begin{array}{cc}0&0\\0&0\end{array}\right]=\left[\begin{array}{cc}2&0\\-2&3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%260%5C%5C-2%263%5Cend%7Barray%7D%5Cright%5D%2B%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D0%260%5C%5C0%260%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%260%5C%5C-2%263%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
If you have two matrices:
![A=\left[\begin{array}{cc}a&b\\c&d\end{array}\right]\\and\\B=\left[\begin{array}{cc}e&f\\g&h\end{array}\right]\\\\\\A+B=\left[\begin{array}{cc}a+e&b+f\\c+g&d+h\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26b%5C%5Cc%26d%5Cend%7Barray%7D%5Cright%5D%5C%5Cand%5C%5CB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7De%26f%5C%5Cg%26h%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%5C%5CA%2BB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%2Be%26b%2Bf%5C%5Cc%2Bg%26d%2Bh%5Cend%7Barray%7D%5Cright%5D)
We have:
![A=\left[\begin{array}{cc}2&0\\-2&3\end{array}\right]\\ and\\B=\left[\begin{array}{cc}0&0\\0&0\end{array}\right]\\\\\\A+B=\left[\begin{array}{cc}2&0\\-2&3\end{array}\right]+\left[\begin{array}{cc}0&0\\0&0\end{array}\right]\\\\\\A+B=\left[\begin{array}{cc}2+0&0+0\\-2+0&3+0\end{array}\right]\\\\\\A+B=\left[\begin{array}{cc}2&0\\-2&3\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%260%5C%5C-2%263%5Cend%7Barray%7D%5Cright%5D%5C%5C%20and%5C%5CB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D0%260%5C%5C0%260%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%5C%5CA%2BB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%260%5C%5C-2%263%5Cend%7Barray%7D%5Cright%5D%2B%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D0%260%5C%5C0%260%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%5C%5CA%2BB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%2B0%260%2B0%5C%5C-2%2B0%263%2B0%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%5C%5CA%2BB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%260%5C%5C-2%263%5Cend%7Barray%7D%5Cright%5D)
The sum between the two matrices is:
![\left[\begin{array}{cc}2&0\\-2&3\end{array}\right]+\left[\begin{array}{cc}0&0\\0&0\end{array}\right]=\left[\begin{array}{cc}2&0\\-2&3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%260%5C%5C-2%263%5Cend%7Barray%7D%5Cright%5D%2B%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D0%260%5C%5C0%260%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%260%5C%5C-2%263%5Cend%7Barray%7D%5Cright%5D)
If two pentagons are similar, then their sides are proportional. To find PT, we'll need to set up a proportion.
AE / PT = AB / PQ
---There are many ways to set up this proportion, it just depends on what side lengths you have and ensuring that they match up on both shapes.
6 / x = 5 / 12.5
5x = 75
x = 15
Length of PT = 15 cm
Option C
Hope this helps!
Answer: 6 book set = 69.95
Step-by-step explanation:
6 separate books = 14.95 = 6x14.95= 89.70
69.95<89.70, 6 book set is a better deal