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i am having one idea i.e first obtain locus of above coordinate system

x^2/3 +y^2/3=16 now obtain dy/dx for this curve that comes out to be -tan(t)

now obtain equation of tangent to this curve

but what i understand that you want to apply concept of gradient in above problem but that always gives equation of normal

we know that irrational root occur in conjugate pair

we assume g(x)=(x-y)f(x)

(x-y)(x^2+ax+b)

x^3+ax^2+bx-yx^2-axy-by=x^3+px+q

x^3+(a-y)x^2+(b-ay)x-by=x^3+px+q

a=y

b-ay=p

-by=q

from here no relation will match

please mention that a,b are positive or not

assume a,b>0

is tending to zero at theta=pi/2-(tan^-1b/a)

if any faimly of line f(x)+tF(x)=0 is given then minimum distance friom origin is (constant term)/[coef. of(x^2)+coef. of(y^2)]

since ans of above ques. =t^2 +2t +4

anet={Fnet/2}=(3i+j)........Ans

Vx=V(initial)+ax*t

4+3*3=13

Vy=0+3*1=3

Vnet=Vxi+Vyj

(13i+3j).........Ans

X=X0+v(initial)*t+1/2*ax*t^2

=2+4*3+1/2*3*9

=27.5

Y=Y0+0+1/2*1*9

=-1+4.5

=3.5

="hence requried position vector=(27.5i+3.5j)".........Ans

i think that all plots are correct but in second plot at x=0 the slope will must be{ -infinite}

......it will be 0 if field will be coservative

.......coservative means du/dy=+or-dv/dx but it will be applicable if u is function of x only and v is function of y only

.......if it will function of x &y both then integration over close path will be o if curl of that field vector will be o

hello friend E=mc^2 fails if any material particle travell more faster than {c} "& such particle will having immaginary mass &hence not possible to exist"

MATHEMATICALY

m=m0/squrt(1-v^2/c^2)

IF......v>c then 1-v^2/c^2<0

hence m is immaginary

you are wrong you see my answer i answerd in this forum

a black hole will create or form if

1.......a star will die

2......star will be 4to20 times more massive than sun

FRIST CONDITION WILL BE NECESSARY

so how it can be explode

obiously a blackhole is hot"because any body is black means it absoarbs the tempreature by "steafn's law"....."

there will be no doubt it is dence also FIELD=GM/R^2

if R.....> 0 then Field is tending to infinite "this R is known as critical radius"

because R is much small after destroying the planet so it is dence also as comparison of its larger mass

example sun is also a black hole

hello friend E=mc^2 fails if any material particle travell more faster than {c} "& such particle will having immaginary mass &hence not possible to exist"

MATHEMATICALY

m=m0/squrt(1-v^2/c^2)

IF......v>c then 1-v^2/c^2<0

hence m is immaginary

p(x)=x^2-squrt2 *x+1=0 is minimal polynomial

frist let k is biggest side of triangle

then for formation of triangle {sum of two smallest side will always greater than biggest side}

(13+9)>k

k<22.................(1)

let in above triangle ABC the k is side infront of angle A

then for obtuse angle cosA<0

hence cosA=(9^2 +13^2-K^2)/2*9*13

2nd lowest statement emplies...........9^2 +13^2<k^2

on solving above inequality the accepted valu of k is greater than 15.81

hence integer values are 16 17 18 19 20 21 (total 6 integer value)

"NOW CHANGING CONDITION"

i.e 13 is biggest side

13^2>k^2+9^2

-9.38< k<9.38

but k+9>13

k>4

hence total integer valu 5 6 7 8 9

hence total no of integer value will be 11

Let ->

A=Xi+Yj+Zk

B=3i+2j+5k

A.B=mod(A)mod(B) cos (t) "max if cos(t)=1,min if cos(t)=-1"

3x+2y+5z=squrt(x^2+y^2+z^2)*squrt(38) cos(t)

put valu of x^2+y^2+z^2=45 $get ans

sorry friend in 1/(3-5cosx) you are wrong

CORRECT ANS

the denominator 3-5cosx=0 ->cosx=3/5 which is in the region [-1 1]

hence correct range is(-∞ -1/2]U [1/8 ∞)

hence you can see wht is max$min

what is maximum $minimum valu of 1/(3-5cosx)

then prove that

log(A)+log(4)+log(R)>=3(log(x)*log(y)*log(z))^1/3

hello friend.........

find max,min (3x+2y+5z)

if[x^2+y^2+z^2]=45

what is natural log of squt(-1)

if know then reply please

squt(24) is answer of max(3x-y-3z)

both are true I check your lagrange multiplier method

squt(5) is ans of max(x+2y)