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13km
perimeter of rectangle = 2(l+b)
given l abd b as 18 cm and 26 cm
therefore perimeter of rectangle = 2(44)
perimeter of circle = 2*pi*r
given 2*pi*r=2(l+B)
i.e, 2*(22/7)*r= 2(44)
(22/7)*r=44
r=14 cms
area of the circle = pi*r*r
=(22/7)*14*14
=616 sqcm is the answer
(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
240 m
(x-2)^3==x^3-6x^2+12x-8
(x-2)^3==(2^(2/3)+2^(1/3))^3
therefore,by solving
Ans: x^3-6x^2+x== -8-5*2^(2/3)-5*2^(1/3)
Let the money borrowed by Nitin=Rs.P
According to the question,
(P×6×3)/100+(P×9×5)/100+(P×13×3)/100=Rs.8160
ans=8000
2:3
121 I. e it is Prime numbers square
( c ) R or T
when know that 13 is factor of 39 so it will also not be able to divide it differently. so the remainder will be same 20 but as we know that 20 is greater then 13 so we can divide it once again and in the casE of 13 the remainder will be 20-13=7