the student read 35 pages in class to start the book because the days are like the pages assigned and read for homework
Answer:
20a³
Step-by-step explanation:
Simply multiply:
-5(-4) = 20
-a(-a)(a) = a²(a) = a³
20a³
You would multiply each kilogram amount by 2.2 to get the point equivalent.
Answer:
r - 5 = 2c
r = 75
Step-by-step explanation:
To write an equation for the problem, we first need do declare the value of the number of apps cora has.
Let c = Cora's apps
r - 5 = 2c
r - 5 is used to indicate that Rita deleted 5 apps.
2c is used to represent the twice the number of apps Cora has.
Now you said that Cora had 35 apps.
Let's plug that into the equation.
r - 5 = 2c
r - 5 = 2(35)
r - 5 = 70
Now we transpose the -5 to the other side to leave r.
r = 70 + 5
r = 75
So if Cora has 35 apps, then Rita will have 75 apps.
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!
