Opm1501 assignment 2022 Study guides, Study notes & Summaries
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OPM1501 Assignment 4 Semester 2 2022
- Exam (elaborations) • 9 pages • 2022
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OPM1501 Assignment 4 Semester 2 2022. 
OPM1501 - Orientation To Intermediate Phase Mathematics 
Which of the following best describe the difference between length and area? 
 (2) 
 
a) Length is the size of a line segment; area is the size of a closed region in a plane 
 
b) Distance and space 
c) Distance formula and volume 
d) Area is the size of a closed region; length is the height of an object 
2. Which of the following best describe the difference between standard units and 
non-standar...
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OPM1501 Assignment 4 Semester 2 2022
- Exam (elaborations) • 9 pages • 2022
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- R56,65
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OPM1501 Assignment 4 Semester 2 2022. 
Orientation To Intermediate Phase Mathematics. 
Which of the following best describe the difference between length and area? 
 (2) 
 
a) Length is the size of a line segment; area is the size of a closed region in a plane 
 
b) Distance and space 
c) Distance formula and volume 
d) Area is the size of a closed region; length is the height of an object 
2. Which of the following best describe the difference between standard units and 
non-standard units? ...
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OPM1501 ASSIGNMENT 03 2022
- Other • 13 pages • 2022
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OPM1501 ASSIGNMENT 03 2022
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OPM1501 Assignment 3 2022
- Exam (elaborations) • 10 pages • 2022
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- R141,61
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OPM1501 
STUDENT NUMBER: 
 
M.UYS 
Assignment 4 
UNIQUE NUMBER: 
OPM1501 – Assignment 4 M. UYS () Unique no: 
2 
Table of Contents 
Question 1 .............................................................................................................................................. 3 
Question 2 .............................................................................................................................................. 3 
Question 3 ..........................................
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OPM1501 Assignment 3 Due 15 July 2022 (Answers)
- Exam (elaborations) • 15 pages • 2022
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- R47,20
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PART A 
1. 
 0 0 
 1 ×3 -4 1 
 2 2 
k 3 
K = 3 
Therefore 3×3 = 9 
 9-4 = 5 
2. 
Input value 0 1 2 k 
Output value 0 -1 2 5 
3. 
0, 3 , 8 , 15 , 24 , 35 , 48 , 63 , 80 , 99, 120 
 3 5 7 9 11 13 15 17 19 20 
 2 2 2 2 2 2 2 2 2 
Therefore, the eleventh for this sequence is 12 
4. 
(i) 5 6 7 
 12 13 14 
 19 20 21 
 1 
(ii) 7 8 10 
 14 15 16 
 2 2 
 1 2 3 
(iii) 
 8 9 10 
 15 16 17 
 
(iv) 2 3 4 
 9 10 11 
 16 17 18 
5. Yes 
6 + 14 + 12 + 20 = 52 
6. 
YES 
Of 19 
Same at the right-hand side 13 + 7...