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(x**2 – 6* x + 5) = (x-1)*(x-5)
(x**2 + 2 * x + 1) = (x + 1) * (x+1) = (x+1)**2
For what x is (x-1)*(x-5)/( (x+1)**2) a minimum?
One way to answer this question is by using calculus.
Take the derivative, and set to zero.
Since this is a fraction of polynomials, and a fraction is
zero only if it’s numerator is zero, we need calculate only
the numerator of the derivative and set it to zero.
The numerator of the
Derivative of (x-1)*(x-5)/( (x+1)**2) is
( (x-1) + (x-5) ) ( x+1)**2 – (x-1)(x-5)( 2 (x+1) )
= (2 x – 6) (x+1)**2 – (2) (x-1)(x-5) (x+1)
= 0
Divide through by 2 (x+1)
(x-3)(x+1) – (x-1)(x-5) = 0
(x**2 – 2 x – 3 ) – (x**2 – 6 x + 5) = 0
x**2 – x**2 – 2 x + 6 x – 3 – 5 = 0
4 x – 8 = 0
x = 2
Plugging in x = 2 into the original
(x**2-6*x+5)/(x**2+2*x+1)
gives us (2**2 – 6 * 2 + 5)/(2**2 + 2*2 + 1)
= (4 – 12 + 5) / (4 + 4 + 1) = -3/9 = -1/3
Least value is -1/3
99/1020
1/4, 1/8
4*8=32
32-4=28
28-8=20
28*20/(4^2)=35.00
9s
8days
30mins
Let the total time taken to fill the tank be x minutes.
Then, work done by B in half hour + work done by (A+B) in half hour =1.
(
40
1
×
2
x
)+(
60
1
+
40
1
)×
2
x
=1
⇒
80
x
+
48
x
=1
⇒
240
3x+5x
=1
⇒8x=240
⇒x=30 min
1 through 24, divisible by 2 in descending order.
That leaves you with:
24, 22, 20, 18, 16, 14, 12, 10, 8, 6, 4, 2
8th place FROM the bottom means your answer will be 16.