Messages posted by magiclko

Let the respective speeds be a, b and c.

then when a, nd c meets, 88 (a+c) = 2X11Xpi => a+c = pi/4 .................(1)

when b and c meets, 110(b+c)= 2X11Xpi => b+c=pi/5 .................(2)

now, since a and b are moving in same direction, when they meet after time 't'

t (a-c) = 2X11Xpi .................subtracting (2) from (1)

=> t [(pi/4) - (pi/4) ] = 2X11Xpi

=> t = 440 sec

A friend function is a function that is not a member of a class but has access to the class's private and protected members. Friend functions are not considered class members; they are normal external functions that are given special access privileges. Friends are not in the class's scope, and they are not called using the member-selection operators (. and –>) unless they are members of another class

hemang has solved the question already...thanx :)

whats the question dear :?

the question has been already solved by prahlad !!

thank u :)

The catalyst increases rate reaction by providing a different reaction mechanism to occur with a lower activation energy.
It does not affect the position of the equilibria, as the catalyst speeds up the backward and forward reactions equally.

what do u mean by the 'absolute meaning' of any theorem?

the question is already solved :)

If one vector is equal to the sum of scalar multiples of other vectors, it is said to be a linear combination of the other vectors.

For example, suppose a = 2b + 3c, as shown below.

	11
	16

=

2

	1
	2

+

3

	3
	4

=

	21 + 33
	22 + 34

a

b

c

Note that 2b is a scalar multiple and 3c is a scalar multiple. Thus, a is a linear combination of b and c.

what is the next term??

a⁴ +a³b+a²b²+ ab³+b⁴

or,

a⁴ +a³b+2a²b²+ ab³+b⁴

Pls specify!!

domain of sin^-1(x-x²)

-1 =< x-x²< = 1

x² - x - 1 =< 0

=> ( x-1/2 )² - (5/4) =< 0

=> [ x - {(1-sqrt5)/2} ] [ x - {(1+sqrt5)/2} ] =< 0

=> [ x - (- 0.6 ) ] [ x - 1.6 ] =< 0

=> x

[ -0.6, 1.6] .....................(A)

domain of (1-(1/ IXI))^1/2

firstly, IxI should not be 0 => x (not =) 0 .................(i)

1-(1/ IxI) > = 0

=> 1 > = (1/ IxI)

=> |x| > = 1

=> x =< -1 or x >= 1 ..........................(ii)

combining (i) and (ii), domain : x (-infinity, -1] [1, infinity) .............................(B)

domain of 1/[x²-1]

[x²-1] (not =) 0

=> (x²-1)<0 or, (x²-1) >=1

(x²-1) < 0

=> x

(-1,1) .............................................(iii)

(x²-1) > = 1

=> x

(-infinity , -sqrt2 ]

[sqrt2, infinity ) ..........................(iv)

combining (iii) and (iv), x (-infinity , -sqrt2 ] (-1,1) [sqrt2, infinity ) .....................(C)

as all the three conditions given by (A), (B) and (C), should be fullfilled, the domain of the entire expression can be found out by taking the intersection of the three :)

cos # = 2.4 / 3

= 4/5

=> # = 37degrees

now tan (180-$) = slope of the line = -3/4

=> $ = 37degrees

therefore, the slope of the required line = tan @ = tan (90- $ - # )

= tan (90- 74)

= cot 74

now u have the slope, and the point.. u can write the equation of the line :)

two species are isolobal, if the number, symmetry properties, approximate energy and shape of the frontier orbitals and the number of electrons in them are similar – not identical

STRUCTURED QUERY LANGUAGE... It is a standard language for accessing and manipulating databases

in the bigger triangle sin @ = R / (L²+ R²)^1/2 ....................(A)

cos @ = L / (L²+ R²)^1/2....................(B)

Now the area of the required triangle (the shaded area) = b X h

in smaller triangle, with angle @ .... h = L cos @ and b = L sin @

therefore, the area = L² sin @ cos @ substituting values from (A) and (B)

= RL³ / (L²+ R²)

=>

hemang has already solved the question!

r d sharma isnt that gud buk for preperation...its gud fr practising the basic prblms...but for jee standard...go for TMH or arihant!

sinx cosx dx/ [a²cos²x + b²sin²x]

=> sin2x dx/2 [a²cos²x + b²sin²x] .............(A)

Let [a²cos²x + b²sin²x] = t

Differentiating both the sides, we have

sin 2x [b²-a²] dx = dt

=> sin 2x dx = dt / [b²-a²]

Substituting in equation (A)

integral of dt /{2t [b²-a²] }

=> ln (t) / {2 [b²-a²]}

substitute the value of 't' and get the answer :)

jagdish has already solved the question...thank u for reducing the wrk :D

well done :)