(x+2)^2 -x^2 = 84
X=20
So (20,22)
Sum= 42
(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
(1/2)*x*y
A = (B*H)/2
H = 2B
formula becomes A = (B*2*B)/2
this can be rewritten as A = (2*B^2)/2
the 2 in the numerator and denominator cancel out and you get:
A = B^2
41, 43, 47, 53, 61, 71, 73, 81
81
FUHAZIBA
Speed of train = 54km/hr
= 54 x (5/18) m/s
= 15 m/s
Length of train = 165m
Time required to cross a bridge of 660m in length = (660+165) / 15
= 55 seconds
625
B
Had is correct ! Anil went in right direction but ended up
with wrong answers
Soln: B lied ( see anil’s comment ) which implies B has not
stolen mule. So B could steal camel or sheep.
1) Lets assume B stole camel.
In that case, A lies ( refer – A says “B had stolen sheep”
), which implies A had stolen sheep ( becoz a lier cant
steal mule ). So A->sheep B->camel C->mule. But this cant
happen because C cant lie ( refer – C says ” B had stolen
mule” ).
2) Lets assume B stole sheep.
In that case, A is true ( refer – A says “B had stolen
sheep” ), which implies A had stolen mule. So A->Mule
B->Sheep C->camel. Here C lies ( refer – C says ” B had
stolen mule” )
SO ANS: answer A- mule, B-sheep, C camel
500