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secondary 4 | E Maths
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unsure of the working for part C
Date Posted: 5 years ago
Views: 581
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Here. Let me know if you do not understand what I wrote. Unfortunately, it’s hard to pen down such explanations on paper.
Date Posted:
5 years ago
I need explanation please for throwing one 2
784 = 2^4 times 3^2
There are two things we can do to the factor “2” to make 784 a perfect cube.
One, include two “2”s.
Two, throw away a “2”.
Both are possible. However, there is this extra condition that for the m/n which we must introduce, both m and n must be prime numbers.
If you include 2^2, then the number to introduce, m = 2^2 = 4, is clearly not prime.
So, we cannot include two “2”s.
This leaves us with throwing away one “2”, so this goes into the expression for n.
There are two things we can do to the factor “2” to make 784 a perfect cube.
One, include two “2”s.
Two, throw away a “2”.
Both are possible. However, there is this extra condition that for the m/n which we must introduce, both m and n must be prime numbers.
If you include 2^2, then the number to introduce, m = 2^2 = 4, is clearly not prime.
So, we cannot include two “2”s.
This leaves us with throwing away one “2”, so this goes into the expression for n.
Without the “m and n are prime numbers” stipulation, you could introduce two “2”s as well.
Remember that you need the powers of each prime factor to be multiples of 3 for the number to be a perfect cube.
So, 2^a times 3^b can only be a perfect cube if a and b are multiples of 3.
So, 2^a times 3^b can only be a perfect cube if a and b are multiples of 3.
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