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Yahoo!奇摩知識+ - 分類問答 - 科學常識 - 已解決
Yahoo!奇摩知識+ - 分類問答 - 科學常識 - 已解決 
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Given that y=2/3 x^2
Apr 29th 2014, 09:10

(Q.a) When x increases by 0.2%, new y is :
(2/3)[(1 + 0.2%)x]^2
= (2/3)(1.002x)^2
= (2/3)(1.004004x^2)
So, y increases :
[(2/3)(1.004004x^2) - (2/3)x^2] / [(2/3)x^2]
= 0.004004
= 0.4004%
therefore, the appr. percentage increases in y is 0.4%.

(Q.b.) Let 9x^4 - 12x^3 + 28x^2 + ax + b ≡ (3x^2 - px + q)^2
As (3x^2 - px + q)^2 = 9x^4 - 6px^3 + (p^2 + 6q)x^2 - 2pqx + q^2
So,
6p = 12
p^2 + 6q = 28
-2pq = a
q^2 = b
After solving the above 4 equations, we get,
p = 2, q = 4, a = -16, b = 16

Therefore, a = -16, b = 16.


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