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6*6=36 medals
6 days
1st day=1+35/7=6 remaining 30 medals
2nd day=2+28/7=6 remaining 24 medals
3rd day=3+21/7=6 remaining 18 medals
…
6th day 6 medals
(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
39/50
length of train = 150 m
Answer: 50 kmph.
Explanation:
Let the distances between city A to B and B to C be 2x km and 3x km respectively.
Total time taken to cover from A to C
= (2x)/40 + (3x)/60 = (6x + 6x)/120 = 12x/120 = x/10 Average speed = (2x + 3x)/(x/10) = 50 kmph.
79.2
A parabola is obtained as the intersection of a cone with a
plane
1km = 1000m
Xkm = 300m
X=0.30km
1hr = 3,600s
X = 40.5s
0.01125hr
Therefore; 0.3km/0.1125hr= 26.6666=26.67km/hr
24*4= x(x+4)
solve this
answer is x=8
X+4= 12
share= 5:7
i.e.,
=9600*5/12
= 4000
4