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the ans will be pi/2 in the given lines
a=1,b=-1, h=0,
putting these values in formula,
angle(x)=tan(inv)[2(h^2-ab)^1/2]/(a+b)
x=tan(inv)[2(1-(-1))^1/2]/1-1
x=tan(Inv)(infinity)
therefore x=pi/2

i think u took b=1 instead of -1
if u find my sol correct plz plz rate me.

if wrong plz tell me

hi
the cord d on giving a sudden jerk will break because the block m will remain at rest due to inertia.however if the force on d were to be slowly increased then the cord c would have broken.
plz plz rate me if u find my solution correct.

did u get the method????

i have posted the question as amaron has understood it.maybe i should have made it more clear.but thanks amaron.

i think i can explain
the formula is
m1v1/n1=m2v2/n2
where n1&n2 are no of moles in equation
and not m1v1=m2v2
as told by magiclko basicity of citric acid is 4
therefore 1/2 mole is required to neutralise imle of Ba(OH)2
31.65x0.1645/1=25xM/0.5
we get 0.1044M
which should be the correct ans
i you find my solution correct plz plz rate me.if not plz point out the mistake

plz plz send them to me 2 if u dont mind .my id is
abhijeet_0201@rediffmail.com

thanks a lot

plz solve this problem as son as possible

in any triangle ABC

[sin^2 A+sinA+1/sinA](the whole expression is upon sinA)

is always greater than

(a) 9 (b)3 (c)27 (d)10

ans (c)

can anyone tell me when do we use Am>Gm property.
for eg in this ques how can we use Am>Gm property.???????

hey nayanmange i think you are forgetting that there need not be all different letters.

q89

sinA+cosA=

7/2

since sinA=2tan(A/2)/1+tansquare(A/2)

also cosA=1-tansquare(A/2)/1+tansquare(A/2)

therefore

2tan(A/2)+1-tansquare(A/2)/(1+tansquare(A/2)=

7/2

⁽

7+2)tansquare(A/2)-4tan(A/2)+(

7-2)=0

solving we get tan(A/2)=2+- 1/(

7+2)

since A<pi/6

therefore tan(A/2)=2-1/

7+2

tan(A/2)=

7-2/3

plz plz rate me if u find my solution right

hi priyesh, the letters are (aa),(ii),(nn),e,x,m,t,o that is we group similar letters together.
the possible combinations are
(a)2alike ,2alike=3C2=3
(b)2alike,2different=3C1*7C2=63
(c)all 4 different=8C4=70
the total no of permutations from (a) to (c) are
3*4!/(2!*2!) +63*4!/2! +70*4! =18+756+1680=2454

plzplz rate me if you find my solution correct.

hey how do i bookmark a page .when i click on the bookmark option,a"page not found" message appears

i got it now.thanks a lot to both.

i cant understand the steps after seprating.plz explain elaborately

i am waiting for an answer