Exam (elaborations)
Mip1502 Assignment 4 due 16 Aug 2024
- Course
- Mip1502 (MIP1502)
- Institution
- University Of South Africa (Unisa)
Mip1502 Assignment 4 2024 2024 Assignment 4 Date 16 August 2024
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MIP1502ASSIGNMENT 4
UNIQUE CODE:
397869
DUE DATE: 16 AUGUST 2024
Page 1 of 6
, QUESTION 1
2.1.
1.1.1. Complete the table below for tile numbers 5 and 6.
Tile no. (𝑛) 5 6 27
Tile length (𝑙) 7 8 29
Number of red squares 12 14 56
Number of black squares 37 50 785
Total number of squares 49 64 841
1.1.2. To determine the length of Tile number 7, we can use the established pattern
that the length of the of the tile increases by 1 unit for each subsequent tile
number.
Given that tile number 1 has length of 3. The formula to find the length of any
tile number 𝑛 can be expressed as 𝑙 = 3 + (𝑛 − 1). So, tile number 7 length will
be : = 3 + (7 − 1)
:=3+6=9
Another thought, the length of tile increases by 1 for each subsequent tile
number, so as tile number 1 has length and tile number 2 has a length of 4 =
(3+1). And tile number 3 has length of 5 = (4+1). Which means Tile number 7
has length of 9 = (8+1 = 9).
So, the length of tile number 7 is 9
1.1.3. Formulas
1.1.3.1. 𝑅 = 2𝑙 − 2
1.1.3.2. 𝐵 = 𝑙 2 − 2𝑙 + 2
1.1.4. Column 27
Tile no. (𝑛) 27
Tile length (𝑙) 29
Number of red squares 56
Number of black squares 785
Page 2 of 6
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