Derivative, instantaneous rate of change When water is discharged from the water tank, the remaining water in the water tank after t minutes is Q (T) = 200 (30-t) ^ 2 (unit: liter). How fast does the water flow out just after 10 minutes? What is the average water flow out rate in the first 10 minutes?

Derivative, instantaneous rate of change When water is discharged from the water tank, the remaining water in the water tank after t minutes is Q (T) = 200 (30-t) ^ 2 (unit: liter). How fast does the water flow out just after 10 minutes? What is the average water flow out rate in the first 10 minutes?

Q(t)=200(30-t)^2
1, q '(T) = - 400 (30-t), q' (10) = - 8000 A. just after 10 minutes, the water flows out at 8000 liters per minute
2, q (10) = 80000, q (0) = 180000, so the average outflow rate is [Q (0) - Q (10)] / 10 = 10000 L / min

Root 1 + x minus root 1 - the limit of X divided by X, X → 1

A: X tends to zero
lim(x→0) [√(1+x)-√(1-x) ] / x
=lim(x→0) [√(1+x)-√(1-x) ] * [√(1+x)-√(1-x) ] /{ x [√(1+x)+√(1-x) ]}
=lim(x→0) [1+x-(1-x) ] / { x [√(1+x)+√(1-x) ]}
=lim(x→0) 2x / { x [√(1+x)+√(1-x) ]}
=lim(x→0) 2 / [√(1+x)+√(1-x) ]
=2/(1+1)
=1

Given that x equals root 6 minus root 5 divided by root 6 plus root 5, y equals root 6 plus root 5 divided by root 6 minus root 5, find the value of Y / x-x / y

X equals root 6 minus root 5 divided by root 6 plus root 5x = √ 6 - √ 5 / √ 6 + √ 5x = (√ 6 - √ 5) ^ 2 / (√ 6 + √ 5) (√ 6 - √ 5) x = (√ 6 - √ 5) ^ 2Y equals root 6 plus root 5 divided by root 6 minus root 5Y = √ 6 + √ 5 / √ 6 - √ 5Y = (√ 6 + √ 5) ^ 2 / (√ 6 + √ 5) (√ 6 - √ 5) y = (√ 6 + √ 5) ^ 2Y /

Root 32 minus root 8 and then divided by root 2 equals?

(√32-√8)/√2=(√64-√16)/2=(8-4)/2=2