Answer:
Idk
Step-by-step explanation:
The answer should be £6.70
Given:
Composite figure.
The figure splitted into two shapes.
One is vertical cuboid and other is horizontal cuboid
To find:
Total surface area of the figure
Solution:
<u>Vertical cuboid:</u>
Length = 14 inches
Width = 12 inches
Height = 24 inches
Surface area = 2(lw + wh + lh)
= 2(14 × 12 + 12 × 24 + 14 × 24)
= 2(168 + 288 + 336)
Surface area = 1584 square inches
<u>Horizontal cuboid:</u>
Length = 14 inches
Width = 10 inches
Height = 30 - 12 = 18 inches
Surface area = 2(lw + wh + lh)
= 2(14 × 10 + 10 × 18 + 14 × 18)
= 2(140 + 180 + 252)
Surface area = 1144 square inches
Total surface area = 1584 + 1144
= 2728 square inches
The total surface area of the figure is 2728 square inches.
564/24 equals to 47/2 or 23.5
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.