Exam (elaborations)
MIP2602 Assignment 3 2024 - DUE 23 July 2024
- Course
- MIP2602 (MIP2602)
- Institution
- University Of South Africa (Unisa)
MIP2602 Assignment 3 2024 QUESTIONS AND ANSWERS
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OPM1501 Assignment3 2024 (839387) - DUE
8 July 2024
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, OPM1501 Assignment 3 2024 (839387) - DUE 8 July 2024
Question 1
1.1. Complete the following function machines so that the inputs give the correct outputs:
a) 32 224 (3)
b) 1080 − 960 (3)
c) 525 25 (3)
d) 510 270 (3) (12)
1.2. If you were asked to complete the following number line by filling in the missing numbers, explain
why it would not be possible to do so: 19 31 42 (4)
1.3. Provide two ways you can use to teach the students how to generate the rule for each of the following:
a) 32 224 (3) ….. − ….. …. .. …. …..
b) 1080 − 960 (3)
c) 525 25 (3)
d) 510 270 (3) (12)
1.4. Fill out the table below to show how many matches you would need for each diagram number: (5)
a) Write the rule for this pattern. (2) b) What formula could you use for this pattern? (2)
c) How many matchsticks would diagram number 37 have? (3) (12) [40]
1.1. Function Machines
For each pair of inputs and outputs, we need to determine the function that transforms the input to the output.
a) 32 → 224 We need to find a function f(x)f(x)f(x) such that f(32)=224f(32) = 224f(32)=224.
One possible function is: f(x)=7xf(x) = 7xf(x)=7x f(32)=7×32=224f(32) = 7 \times 32 = 224f(32)=7×32=224
b) 1080 → −960 We need to find a function f(x)f(x)f(x) such that f(1080)=−960f(1080) = -960f(1080)=−960.
One possible function is: f(x)=x−2040f(x) = x - 2040f(x)=x−2040 f(1080)=1080−2040=−960f(1080) = 1080 -
2040 = -960f(1080)=1080−2040=−960
c) 525 → 25 We need to find a function f(x)f(x)f(x) such that f(525)=25f(525) = 25f(525)=25.
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