find out what is the problem and taking the bad out always works.
57 and 3
12 x 57 = 684
-8 x 3 = -24
684 – 24 = 660
660/ 60 = 11
None of these
Neice
28 Sec
c
Distance =speed ×time
Speed =90kmph
90km take 60min time to reach
In 10min the distance covered =?
D=90×10/60
900/60
D=15
Total=60reems
Utilized=40reems
Un utilized= 60-40=20
Percentage of remainder 20/60=.3333
0.33333*100=33.333%
(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
Distance covered by B to meet A=Total distance – Distance covered by A hrs
[using : distance = speed x time]
Putting value of from equation (1),
hrs
Therefore, time at which both A and B will meet is = 7 a.m. + 3 hrs =10 am