pipes a and b can fill tank in 5 and 6 hours respectively pipe C can empty in 12 hours the tank is half full of the three pipes are in Operation after how much time will the tank will be full
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f(x) = (2x3 – 5x2 + 9x – 8)
Now, x – 3 = 0 ⇒ x = 3
By the remainder theorem, we know that when f(x) is divided by (x – 3) the remainder is f(3).
Now, f(3) = 2 × 33 – 5 × 32 + 9 × 3 – 8
= 54 – 45 + 27 – 8
= 81 – 53 = 28
∴ The required remainder is 28.
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