obvious
e^(i*pi) + 1 = 0
12
35 km/h = 10 m/s
Speed = Distance/Time
Distance= (100+150)m = 250m
Speed = 10 m/s
Time= 250/10 = 25s
Assume Q can do the work alone in Q days:
work of a day by both P and Q =>
(1/15) + (1/Q ) = 1/6
solving this Q = 10.
The greatest number that will divide 187, 233 and 279 leaving the same remainder in each case.
To find : The number ?
Solution :
First we find the difference between these numbers.
The required numbers are
233-187=46
279-233=46
279-187=92
Now, We find the HCF of 46 and 92.
Therefore, The required largest number is 46.
EMNXDS
Let’s say, length of train = x metres
Speed at pole = Speed at platform
x/15 = 100+x / 25
5x = 300 + 3x
2x = 300
x = 150m
So, train is 150 metres long.
23. 15s squre is 225and 8 squre is 64 addition is 289 product also 120 addition 15+8=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
name
As fast as possible. For example, I will suitable with the working environment within one month.
32000