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)
Answer:
5 , -5
Step-by-step explanation:
c'mon this is easy. the absolute value of a number is that numbers distance away from zero on the number line. 5 and -5 are the only numbers that are five intergers away from zero.
The answer to this question is 5/25 five out of twenty five will be yellow.
The 2 sides of the right triangle is 25 and 24.
Any multiple of a Pythagorean triple is also considered a Pythagorean triple. Multiplying 3, 4, 5 by 2 gives 6, 8, 10, which is another triple. To see if a set of numbers makes a Pythagorean triple, plug them into the Pythagorean Theorem.
3,4,5 5,12,13 7,24,25 8,15,17 9,12,15
Final Result: 25,24
Answer:
A. b2 – 4ac = –12
Step-by-step explanation:
discriminant must be minus cause the graph doesn't cross the x-axis .in other words it doesn't have an answer.plus , the "a" of the equation is minus too.
Answer:
A)
Step-by-step explanation:
1) To answer this question, we must remember what this discriminant stands for:
2) We must plug in this discriminant, in the square root of Quadratic Formula, to find its roots, i.e.:
The discriminant is a radicand.
3) The graph above shows us no Real roots for this equation then we have roots ∈ Complex Numbers, in other words:
4) So, it's A.