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Secondary 1 | Maths
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Pls help with 3 questions
Date Posted: 5 years ago
Views: 372
I can't see the whole of question 6, but learn from Q4 and Q5, try solving it on your own. If you've learnt the concept, it should be easy. Enjoy doing maths man :)
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a) is self explanatory. For b, as the lowest common multiple of the 2 numbers is 600, 24 only has 2 and 3, thus, m must have a factor of 5 in it. The highest common factor between m and 24 is 4, thus m must have the factor 2, but it must be 2^2, anything lower makes the HCF less than 4 and anything more makes it 8 or more.
Thus, m=2^2 × 5^2 (from 600, must have the 5^2 factor)=100
Little bit complicated, just practice. Hope you'll get better :)
Thus, m=2^2 × 5^2 (from 600, must have the 5^2 factor)=100
Little bit complicated, just practice. Hope you'll get better :)
Date Posted:
5 years ago
Thanks
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To make something a perfect square, all of its factors must have an even number power, 0 included. Anything to the power of zero is 1, so anything works, less your term is zero.
In this question, they asked for the lowest value of k, so try not to change anything that doesn't need to be changed (like 5), so by dividing 1350 by 2, you change 2's power from 1 to 0, making it even. Good job. Move on. Dividing 1350 by 3 makes 3's power reduce from 3 to 2. Thus, K= 2×3.
In this question, they asked for the lowest value of k, so try not to change anything that doesn't need to be changed (like 5), so by dividing 1350 by 2, you change 2's power from 1 to 0, making it even. Good job. Move on. Dividing 1350 by 3 makes 3's power reduce from 3 to 2. Thus, K= 2×3.
Date Posted:
5 years ago
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