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Secondary 1 | Maths
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Did i factorise correctly? I would be grateful if u could help me.
Date Posted: 5 years ago
Views: 259
25/15 = 5/3
(a - b)⁻ⁿ⁺² / (a - b)⁻ⁿ⁻¹
= (a - b)⁻ⁿ⁺² / [ (a - b)⁻ⁿ/(a - b) ]
= (a - b)² / [ 1/(a - b) ]
= (a - b)² x (a - b)
= (a - b)³
(1 - c)³/(1 - c)
= (1 - c)²
hence,
[25(a-b)⁻ⁿ⁺²(1 - c)³ / 15(a-b)⁻ⁿ⁻¹(1 - c) ]
= [5(a - b)³ (1 - c)²]/3
(a - b)⁻ⁿ⁺² / (a - b)⁻ⁿ⁻¹
= (a - b)⁻ⁿ⁺² / [ (a - b)⁻ⁿ/(a - b) ]
= (a - b)² / [ 1/(a - b) ]
= (a - b)² x (a - b)
= (a - b)³
(1 - c)³/(1 - c)
= (1 - c)²
hence,
[25(a-b)⁻ⁿ⁺²(1 - c)³ / 15(a-b)⁻ⁿ⁻¹(1 - c) ]
= [5(a - b)³ (1 - c)²]/3
Thanku i understand where i went wrong
Full treatment of the concept is seen in Sec 3 A/E Maths under the chapter 'Indices'.
I doubt powers -n + 2 and -n - 1 should be seen in Sec 1 though.
I doubt powers -n + 2 and -n - 1 should be seen in Sec 1 though.
See 1 Answer
done
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clear
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You will need to use laws of indices to solve this, typically taught in Secondary 3.
Date Posted:
5 years ago
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