Given that y=2/3 x^2:Realtime－線上遊戲排行榜2013/2014,進擊的巨人線上看,candy crush saga外掛,正妹寫真三圍

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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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