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secondary 4 | A Maths
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Date Posted: 7 years ago
Views: 294
y = ax^n
substituting x=2 and y=40 into the equation
40=a(2^n)
a = 40/(2^n) ------ (1)
substituting x=3 and y=135 into the equation
135 = a(3^n)
a = 135/(3^n) ------ (2)
(1) = (2)
40/(2^n) = 135/(3^n)
2×2×2×5/(2^n) = 3×3×3×5/(3^n)
(2^3)/(2^n) = (3^3)/(3^n)
2^(3-n) = 3^(3-n)
(2/3)^(3-n) = 1
3-n = 0
n = 3
substitute n=3 into (1)
a = 40/(2^3)
= 5
thus the equation is y = 5(x^3)
so when x=4,
y = 320
thus k = 320
substituting x=2 and y=40 into the equation
40=a(2^n)
a = 40/(2^n) ------ (1)
substituting x=3 and y=135 into the equation
135 = a(3^n)
a = 135/(3^n) ------ (2)
(1) = (2)
40/(2^n) = 135/(3^n)
2×2×2×5/(2^n) = 3×3×3×5/(3^n)
(2^3)/(2^n) = (3^3)/(3^n)
2^(3-n) = 3^(3-n)
(2/3)^(3-n) = 1
3-n = 0
n = 3
substitute n=3 into (1)
a = 40/(2^3)
= 5
thus the equation is y = 5(x^3)
so when x=4,
y = 320
thus k = 320
See 1 Answer
done
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is this ok?
Date Posted:
7 years ago
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