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This page contains the most recently asked technical questions and answers in the Gupshup Technology India.
All of the questions listed below were collected by students recently placed at Gupshup Technology India.
but how?
20&130
(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
x+y+z
—– = 6800
3
x>=6400
6400 + y + z = 20400
y + z = 14000
to get the greatest of y and z, lets assume y = 6400
so, z = 7600
so ANS is 7600
c=a/b
c=a-1
=>b=a/c
=>b=a/a-1
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Answer:
11 days.
Step-by-step explanation:
In the question,
Time taken by Ramesh to finish a piece of work = 20 days
Time taken by Sushil to finish a work = 25 days
Time for which they worked together = 5 days
Sushil left after = 5 days
So,
One day work of Ramesh is,
One day work of Sushil is,
So,
Work done in 5 days is given by,
Therefore, Remaining work is given by,
Now, as the Sushil left the remaining work was done by Ramesh,
Time taken by Ramesh for the remaining work is,
Therefore, the remaining work will be completed in 11 days by Ramesh.
24kmph
time taken = x/40 + 2x/20
=> x=8
so, 3x = 24
516 tones
Product of two numbers = 1320
HCF = 6
LCM = x
Formula:
Product of two numbers =(HCF *LCM)
1320=(6*x)
x=1320/6
x=220
LCM Of the numbers is220