Exam (elaborations)
MIP1502 ASSIGNMENT 4 2024
- Course
- MIP1502
- Institution
- University Of South Africa (Unisa)
MIP1502 ASSIGNMENT 4 2024
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MIP1502ASSIGNMENT 4
ANSWERS 2024
MIP1502 ASSIGNMENT 4 ANSWERS 2024
Due date: 16 August 2024
,MIP1502
Assignment 4: Compulsory
Contributes 25% to the final pass mark
Unique number: 397869
Due date: 16 August 2024
Question 1
1.1 A tiling pattern is made by arranging black and red squares, as shown below:\
1.1.1 Complete the table below for tile numbers 5 and 6.
complete the table below for tile numbers 5 and 6 (6 points):
To find the missing values for tile numbers 5 and 6, let's first check the patterns in
the given data:
- The number of red squares \( R \) follows an arithmetic sequence with a common
difference of 2:
- 4, 6, 8, 10 ...
,- It appears to be \( R = 2n + 2 \).
- The number of black squares \( B \):
- 5, 10, 17, 26...
- The difference progresses as follows:
- Difference: 5, 7, 9 (which increases by 2 each time)
- Therefore, it appears to follow a quadratic sequence. We can express this as a
polynomial equation.
To calculate the total squares \( S \) for tile numbers 5 and 6:
- The total number of squares seems to be the square of the tile length \( l \).
- The relationship \( S = l^2 \); where \( l \) is 3 (for tile number 1), so it is likely that
each tile length is \( n + 2 \).
Assuming tile number \( n \):
- For \( n = 5 \), \( l \) would equal 5, and for tile \( n = 6 \), \( l \) would equal 6.
Hence, squares will be:
- \( 5^2 = 25 \)
- \( 6^2 = 36 \)
, Using these observations, we can complete the table:
Tile no 1 2 3 4 5 6 27
(𝑛)
Tile 3 4 5 6 7 8 29
length
(𝑙)
Number 4 6 8 10 12 14 58
of red
squares
(𝑅)
Number 5 10 17 26 37 50 89
of black
squares
(𝐵)
Total 9 16 25 36 49 64 841
number
of
squares
(𝑆)
1.1.2 Given that the length of tile number 1 is 3, what is the length of tile number 7?
Explain your thinking (4 points):
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