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Ques:- When x is real what is the least value of (x**2-6*x+5)/(x**2+2*x+1)
Recent Answer : Added by Admin On 2020-05-17 12:01:23:
(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
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Ques:- Find the missing number of the series 1, 2, 4, 7, 11, ..
Recent Answer : Added by Vishnu Das On 2021-10-13 14:55:31:
Ques:- X, Y and Z can do a piece of work in 24, 30 and 40 days respectively. They start the work together but Z leaves 4 days before the completion of the work. In how many days is the work done?
Recent Answer : Added by Admin On 2022-09-29 17:23:19:
One day’s work of A, B and C = (1/24 + 1/30 + 1/40) = 1/10.
C leaves 4 days before completion of the work, which means only A and B work during the last 4 days.
Work done by A and B together in the last 4 days = 4 (1/24 + 1/30) = 3/10.
Remaining Work = 7/10, which was done by A,B and C in the initial number of days.
Number of days required for this initial work = 7 days.
Thus, the total numbers of days required = 4 + 7 = 11 days.
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Ques:- A can finish a piece of work in 5 days. B can do it in 10 days. They work together for two days and then A goes away. In how many days will B finish the work?
Recent Answer : Added by Roopesh K p On 2021-06-15 17:06:48:
4 days
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Recent Answer : Added by Admin On 2020-05-17 12:02:10:
(1/2)*x*y
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Ques:- What is the angle between the two hands of a clock when time is 8:30
Recent Answer : Added by Mustafa On 2022-08-14 16:32:02:
180*2/3 = 60degree
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Ques:- In 1978, a kg of paper was sold at Rs25/-. I f the paper rate increases at 1.5% more than inflation rate which is of 6.5% a year , then what wil be the cost of a kg of paper after 2 years? a)29.12 (b) 29.72 (c) 30.12 (d) 32.65 (e) none of these
Recent Answer : Added by Admin On 2020-05-17 12:00:04:
(a) 29.12 (actual)
first year rate of paper (inflation rate) : 25 + 25 *
6.5/100 = 26.625
paper rate in first year is 1.5 % more . ..thr4
inflated cost in 1 yr = 26.625 + 1.5 / 100 * 26.625 =
27.024.
second year : 27.024 + 27.024 * 6.5 / 100 = 28.78
paper cost : 1.5 /100 * 28.78 + 28.78 = 29.12 ( approx..)
Ques:- what will be reminder after (16)^15 is devided by the 254…
A. 2
B. 0
C. 1
D. 128
Recent Answer : Added by Damodharan M On 2023-05-22 03:48:14:
I need a answers from hakuna matata questions
Ques:- A work which could be finished in 9 days was finished 3 days earlier after 10 more men joined. The number of men employed was?
Recent Answer : Added by Admin On 2022-09-28 17:11:47:
No of men employed = N
No of days to finish the work = 9 days
No of men after increase = (N + 10)
No of days to finish the work = 6 days
Equating mandays
9N = (N+10)*6
9N — 6N = 60
3N = 60
N = 20
No of men employed = 20
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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