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hey guys let d eqn b mx-y+2m-3=0 hence by solvng dis eqn wid d oder 2 u wud get B & C hence as per gven data dat AB X AC =20 find m & substitute m in it itsidea wich comes in our minds try dis

1. A parachutist after billing out falls 50m without friction . when parachute opens, it decelerates at 2m/s^2 . He reaches d ground with a speed of 3m/s^2 . At what height , did he ball out?

2. A boy playing on d roof a 10m high building throws a ball with a speed of 10m/s at an angle 30 with horizontal . How far from d throwing point will d ball b at d height of 10m from d ground [g=10m/s^2 ]

3. An inclined track ends in a circular loop of diameter D. From wat hiit on d track a particle shud b released so dat it completes dat loop in d vertical plane , assuming dat der is no friction - [ans- 5D/4 ]

A nail is located at a certain distance vertically below d point of suspension of a simple pendulum of length 1m . D pedulum is released from horizontal position . If it rotates in a vertical circle wid nail as centr , mim distance of d nail from point of suspension

its very urgent 2maro is exam

excuse me hs Bhatt iam unable to catch the third step of yours
can u explain me

if d diagonals of a trapezium r equal den d ratio of lengths of non parallel sides is

actually it is integral lowr alpa to higher lt beta

f[x-beta]
e / e^f[x-alpa] +e^f[x-beta]

=let it b I
hence 2I = S dx
I =[ alpa-beta] /2
=rt[b^2 -4ac] /2a

aree yaaron wads dis can i say a simplest soln its actually
f(x^2) -f(x)*(x-1)
________
x(x-1)
__________
f(x) -f(0)
____________
x
hence d whole ding becoms Lt x->0 x-1 hence its -1

is der any reln btw x & y

d range is (-infinty, infinty )

oyy wats en here

i dont understand ur quesn

let me present mine its simpler point out d mistakes if ive committed

1stlycos^2 [2x] +cos^2[3x] =sin^2x

hence by trans4masions cos^2x + cos4x + cos2x =0

so cos2x[cos2x +cos4x] =0
hence d soln r 2x=(2n+1)/2 =>x=(2n+1)pie/4 & by d above equation

x=(2k+1)pie/6 & (2m+1)pie/2

a short cut 4 dis is 1st as it is mentioned ven |z| =1 lets take z=i

hence p(z)=c-1+ib hence |p(z)| =1

so b^2 +c^2 -2c =0 hence b^2+c(c-2) =0

since b^2 is non -ve so b=0 c=0 c=2 r d so.lution der

d ordocentre of a triangl is (2,1) & d circumcentr is [7/2,5/2] den d centroid is [wanted shortcut]

yuhoo hey guys its a tough quesn i got up2 a lt n den due 2 hevi calculasn i cudnt continue help me plz actually quesn is a^2 b^2

--------------------- + ----------------

rt[1+a^3][1+b^3] rt[1+b^3][1+c^3]

+ c^2

----------------- <=4/3

rt[1+a^3][1+c^3]

a small circular coil of radius r n no.of turns n is placed at d centre of anoder big circular coil of radius R n no.of turns N. initially d 2 coils r coplanar n d same current i flows thru both coils den d amount of work done in rotating d small coil about any of its diameter by an angle pie vil b ?

3 long wires of resistances in d ratio 3:4:5 r connected in parallel 2 each oder as shown if net force on middl wire is 0 den d1/d2 = ?