If A and B are equal:
Matrix A must be a diagonal matrix: FALSE.
We only know that A and B are equal, so they can both be non-diagonal matrices. Here's a counterexample:
![A=B=\left[\begin{array}{cc}1&2\\4&5\\7&8\end{array}\right]](https://tex.z-dn.net/?f=A%3DB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%262%5C%5C4%265%5C%5C7%268%5Cend%7Barray%7D%5Cright%5D)
Both matrices must be square: FALSE.
We only know that A and B are equal, so they can both be non-square matrices. The previous counterexample still works
Both matrices must be the same size: TRUE
If A and B are equal, they are literally the same matrix. So, in particular, they also share the size.
For any value of i, j; aij = bij: TRUE
Assuming that there was a small typo in the question, this is also true: two matrices are equal if the correspondent entries are the same.
Answer:
15/4
Step-by-step explanation:
3/4:1/5= 3/4*5= 15/4
It’s option B. Because that’s the try statement.
Answer:
See a solution process below:
Explanation:
Let's call the number of miles driven we are looking for
m
.
The the total cost of ownership for the first car model is:
12000
+
0.1
m
The the total cost of ownership for the second car model is:
14000
+
0.08
m
We can equate these two expressions and solve for
m
to find after how many miles the total cost of ownership is the same:
12000
+
0.1
m
=
14000
+
0.08
m
Next, we can subtract
12000
and
0.08
m
from each side of the equation to isolate the
m
term while keeping the equation balanced:
−
12000
+
12000
+
0.1
m
−
0.08
m
=
−
12000
+
14000
+
0.08
m
−
0.08
m
0
+
(
0.1
−
0.08
)
m
=
2000
+
0
0.02
m
=
2000
Now, we can divide each side of the equation by
0.02
to solve for
m
while keeping the equation balanced:
0.02
m
0.02
=
2000
0.02
0.02
m
0.02
=
100000
After 100,000 miles the total cost of ownership of the two cars would be the same.
