Answer:
Step-by-step explanation:
- First, lets get rid of the fraction. I did this by multiplying both sides by
.
- We want to isolate
on one side of this equation. Let's put any values with on one side of the equation, and normal integers on the other.
- Divide both sides by
.
- If your teacher wants you to leave your final answer as an improper fraction, your final answer is this. If they want it to be a mixed number or decimal, your final answer will be:
2x - y = 7
y = 2x - 7 this line has slope = 2
Perpendicular lines, slope is opposite and reciprocal so slope = -1/2
Answer
A . -1/2
Cos(x) = BC/AB = 10.5/20 = 0.525 ⇒ x = 58.3°
Answer:
![\left[\begin{array}{cc}2&8\\5&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%268%5C%5C5%261%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
The <em>transpose of a matrix </em>
is one where you swap the column and row index for every entry of some original matrix 
. Let's go through our first matrix row by row and swap the indices to construct this new matrix. Note that entries with the same index for row and column will stay fixed. Here I'll use the notation 
and 
to refer to the entry in the i-th row and the j-th column of the matrices 
and 
respectively:
Constructing the matrix from those entries gives us
![P^T=\left[\begin{array}{cc}2&8\\5&1\end{array}\right]](https://tex.z-dn.net/?f=P%5ET%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%268%5C%5C5%261%5Cend%7Barray%7D%5Cright%5D)
which is option a. from the list.
Another interesting quality of the transpose is that we can geometrically represent it as a reflection over the line traced out by all of the entries where the row and column index are equal. In this example, reflecting over the line traced from 2 to 1 gives us our transpose. For another example of this, see the attached image!
Step-by-step explanation:
using the quadratic formula......
