Answer:
It's Impossible.
Matrix can only be raised to a power if it has same number of rows and columns
Step-by-step explanation:
![P= \begin{pmatrix}1&3&2\\ 3&2&1\end{pmatrix}\\\\P^2 = \begin{pmatrix}1&3&2\\ 3&2&1\end{pmatrix}^2\\\\\mathrm{Dimensions\:of\:matrix}\:\\\begin{pmatrix}1&3&2\\ 3&2&1\end{pmatrix}:\:2x3\\\mathrm{Matrix\:can\:only\:be\:raised\:to\:a\:power,\\\:if\:it\:has\:same\:number\:of\:rows\:and\:columns}](https://tex.z-dn.net/?f=P%3D%20%5Cbegin%7Bpmatrix%7D1%263%262%5C%5C%203%262%261%5Cend%7Bpmatrix%7D%5C%5C%5C%5CP%5E2%20%3D%20%5Cbegin%7Bpmatrix%7D1%263%262%5C%5C%203%262%261%5Cend%7Bpmatrix%7D%5E2%5C%5C%5C%5C%5Cmathrm%7BDimensions%5C%3Aof%5C%3Amatrix%7D%5C%3A%5C%5C%5Cbegin%7Bpmatrix%7D1%263%262%5C%5C%203%262%261%5Cend%7Bpmatrix%7D%3A%5C%3A2x3%5C%5C%5Cmathrm%7BMatrix%5C%3Acan%5C%3Aonly%5C%3Abe%5C%3Araised%5C%3Ato%5C%3Aa%5C%3Apower%2C%5C%5C%5C%3Aif%5C%3Ait%5C%3Ahas%5C%3Asame%5C%3Anumber%5C%3Aof%5C%3Arows%5C%3Aand%5C%3Acolumns%7D)
Everyone should do this derivation, because otherwise the "quadratic formula" is some sort of "magic" LOL...
ax^2+bx+c=0
x^2+bx/a+c/a=0
x^2+bx/a=-c/a
x^2+bx/a+b^2x/(4a^2)=b^2/(4a^2)-c/a
(x+b/(2a))^2=(b^2-4ac)/(4a^2)
x+b/(2a)=±√(b^2-4ac)/(2a)
x=-b/(2a)±√(b^2-4ac)/(2a)
x=(-b±√b^2-4ac)/(2a)
Answer:
The answer is 11,337,408
Step-by-step explanation:
6^2 = 36, 3^2 = 9, 3^3 = 27, and 6^4 = 1296.
when you multiply them all together, you get 11,337,408.
Your answer is 27.31 because u multiply 25.95 by 5.25 then divide it by 100 then add 25.95 to it
I would say that the option you have selected in the image is the only one that is correct because it is the only one that uses a similar sign, ~, rather that a congruent symbol.