Exam (elaborations)
OPM1501 Assignment 3 2024 (839387) - DUE 8 July 2024
- Course
- OPM1501 (OPM1501)
- Institution
- University Of South Africa
OPM1501 Assignment 3 2024 (839387) - DUE 8 July 2024 QUESTIONS AND ANSWERS
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OPM1501 Assignment3 2024 (839387) -
DUE 8 July 2024
QUESTIONS WITH COMPLETE ANSWERS
[School]
[Course title]
,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.
One possible function is: f(x)=x21f(x) = \frac{x}{21}f(x)=21x
f(525)=52521=25f(525) = \frac{525}{21} = 25f(525)=21525=25
d) 510 → 270 We need to find a function f(x)f(x)f(x) such that f(510)=270f(510) =
270f(510)=270.
One possible function is: f(x)=x1.888f(x) = \frac{x}{1.888}f(x)=1.888x
f(510)=5101.888≈270f(510) = \frac{510}{1.888} \approx 270f(510)=1.888510
≈270
1.2. Number Line Completion
Number line: 19, __, __, 31, __, 42
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