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:- If there are 9 yellow balls, 3 red balls and 2 green balls. What is the probability that the second ball picked is yellow given the first ball is yellow a. 8/13 b. 9/13 c. 8/14 d. 9/14
Recent Answer : Added by G MOHANAKANNAN On 2022-05-17 08:55:23:
Mm
Ques:- A 5 litre jug contains 4 litres of a salt water solution that is 15 percent salt. If 1.5 litres of the solution spills out of the jug, and the jug is then filled to capacity with water, approximately what percent of the resulting solution in the jug is salt?
A. 7.5%
B. 9.5%
C. 10.5%
D. none
Recent Answer : Added by Admin On 2020-05-17 12:05:46:
5 litre jug contains 4 litres of salt water solution
with 15% of salt, that means it has 4000*.15 = 600 ml salt
in it.
If 1.5 litres solution is spills out, remaining solution
is 2.5 litres, then the salf content is 2500*.15 = 375ml.
Now, the jug is filled to full capacity with water i.e.,
now the jug has 5 litre solution in it.
Now, the salt content is 375/5000 = 7.5%
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Ques:- The H.C.F and L.C.M of two numbers are 44 and 264 respectively. If the first number is divided by 2, the quotient is 44. The other number is
A. 66
B. 132
C. 264
D. 528
Recent Answer : Added by Shiva Akhil On 2023-06-09 12:28:38:
let,first num be x and second num be y
hcf =264
lcm=44
given equation
x/2=44 i.e x=88
product of 2 numbers is equal to lcm*hcf
88*y=44*264
y=132
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1972
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Recent Answer : Added by MGK On 2021-08-06 04:58:03:
Servlets are the Java programs that run on the Java-enabled web server or application server. They are used to handle the request obtained from the webserver, process the request, produce the response, then send the response back to the webserver.
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Ques:- if a man byes 1 lt of milk for 12 Rupees and mixes with 20% of water and sells it for Rs. 15 then what is percentage of gain
Recent Answer : Added by Admin On 2020-05-17 12:00:07:
After the mixing, 1lt of the mixture has 0.8lt of milk and 0.2lt of water. 1lt of milk costs Rs.12 so 0.8lt of milk costs 0.8 x 12 = 9.6
He sells the mixture at Rs.15,
so profit%=(15-9.6)/9.6 x 100 = 56.25%
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Recent Answer : Added by Mousumi On 2022-10-09 06:56:21:
1
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find out what is the problem and taking the bad out always works.
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Ques:- The H.C.F. and L.C.M. of two numbers are 12 and 336 respectively. If one of the numbers is 84, the other is
A. 24
B. 36
C. 48
D. 72
Recent Answer : Added by anonymous On 2022-03-09 16:38:02:
12*336/84=48
Ques:- 343 : 729 :: 125 : __
Recent Answer : Added by Sunil Kumar On 2022-08-14 16:25:38:
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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