Ques:- If the sum of five consecutive nos. 35? how many prime nos are there?
Recent Answer : Added by DHKrish On 2021-02-24 17:31:30:
Ans is 2(Prime numbers)
Sum of 5 consecutive nos is 35
so X + (X+1) + (X+2) + (X+3)+ (X+4) = 35
5X + 10 = 35
X = 5
So the 5 consecutive numbers are : 5, 6, 7, 8, 9
The Prime numbers are 5 and 7
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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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Ans is 2(Prime numbers)
Sum of 5 consecutive nos is 35
so X + (X+1) + (X+2) + (X+3)+ (X+4) = 35
5X + 10 = 35
X = 5
So the 5 consecutive numbers are : 5, 6, 7, 8, 9
The Prime numbers are 5 and 7