Forwarded from Deleted Account
15 20 25..... Triplate
CD=15*20/25=12
AD=โ(15^2 - 12^2)=9
BD=โ(20^2 - 12^2)=16
โACD.... S=18 and โ=54
R1=54/18=3
โBCD.... S=24 and โ=96
RE=96/24=4
So PQ= โ(7^2+1)=โ50
CD=15*20/25=12
AD=โ(15^2 - 12^2)=9
BD=โ(20^2 - 12^2)=16
โACD.... S=18 and โ=54
R1=54/18=3
โBCD.... S=24 and โ=96
RE=96/24=4
So PQ= โ(7^2+1)=โ50
Forwarded from Jagriti Singh
R (small)= 9+12-15/2=3
R (bigg)==16+12-20/2=4
Pq=_/(3+4)ยฒ+(4-3)ยฒ
. =_/49+1=_/50
R (bigg)==16+12-20/2=4
Pq=_/(3+4)ยฒ+(4-3)ยฒ
. =_/49+1=_/50
Forwarded from Deleted Account
xยณ + 4x = 8
Squaring both sides
(xยณ + 4x)ยฒ = 8ยฒ
x^6 + 16xยฒ + 8xโด = 64
Multiplying by x both sides
x^7+ 16xยณ + 8x^5= 64x
Adding 16xยณ both sides
x^7 + 32xยณ + 8x^5 = 64x + 16xยณ
x^7+ 8xยฒ(4x + xยณ) = 16(4x + xยณ)
using 4x + xยณ = 8
x^7+ 8xยฒ * 8 = 16 * 8
x^7+ 64xยฒ = 128
Squaring both sides
(xยณ + 4x)ยฒ = 8ยฒ
x^6 + 16xยฒ + 8xโด = 64
Multiplying by x both sides
x^7+ 16xยณ + 8x^5= 64x
Adding 16xยณ both sides
x^7 + 32xยณ + 8x^5 = 64x + 16xยณ
x^7+ 8xยฒ(4x + xยณ) = 16(4x + xยณ)
using 4x + xยณ = 8
x^7+ 8xยฒ * 8 = 16 * 8
x^7+ 64xยฒ = 128
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Here you will find only ๐Favourite ๐ผsongs.... ๐ง
.
With Full of Quality Sounds
.
๐บ๐ฅ๐น๐ฃ๐ฏ๐กโ๏ธ
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Advertise or cross with us: @jaihind_bot
Forwarded from Deleted Account
sin20sin40sin80=
1/2(cos20-cos60)โขsin80=
1/4(sin60+sin100)-1/4sin80=
1/4โขsin60=sqrt(3)/8
1/2(cos20-cos60)โขsin80=
1/4(sin60+sin100)-1/4sin80=
1/4โขsin60=sqrt(3)/8
Forwarded from Deleted Account
x-1/x=1
x^2-1=x
1/(x-1)-1/(x+1)+1/(x^2+1)-1/(x^2-1)=
2/(x^2-1)-2/(x^2-1)(x^2+1)=
(2x^2+2-2)/(x^2-1)(x^2+1)=
2x^2/x(x^2+1)=
2x/(x^2+1)=2x/(x^2+1)
x-1/x=1 -> (x+1/x)^2-4=1, x+1/x=sqrt(5)
=> +-2/sqrt(5)
x^2-1=x
1/(x-1)-1/(x+1)+1/(x^2+1)-1/(x^2-1)=
2/(x^2-1)-2/(x^2-1)(x^2+1)=
(2x^2+2-2)/(x^2-1)(x^2+1)=
2x^2/x(x^2+1)=
2x/(x^2+1)=2x/(x^2+1)
x-1/x=1 -> (x+1/x)^2-4=1, x+1/x=sqrt(5)
=> +-2/sqrt(5)
Forwarded from Deleted Account
x^2-2x+4=0
x1+x2=2, x1โขx2=4
ax^2+bx+c=0
X1=a^3/b^2, X2=b^3/a^2
X1+X2=(a^5+b^5)/(ab)^2=
(a+b)^5=a^5+b^5+5ab(a^4+b^4)+10a^2b^2(a+b)
->
a^5+b^5=(a+b)^5-5ab((a+b)^2-2ab))^2-2a^2b^2)-10โข(ab)^2(a+b)
a^5+b^5=2^5-5โข4((4-8)^2-2โข16)-10โข16โข2=>
a^5+b^5=32+320-320=32
=> X1+X2=(a^5+b^5)/(ab)^2=32/16=2
=> X1โขX2=a^3/b^2โขb^3/a^2=ab=4
=> ax^2+bx+c=0, x^2-2x+4=0
x1+x2=2, x1โขx2=4
ax^2+bx+c=0
X1=a^3/b^2, X2=b^3/a^2
X1+X2=(a^5+b^5)/(ab)^2=
(a+b)^5=a^5+b^5+5ab(a^4+b^4)+10a^2b^2(a+b)
->
a^5+b^5=(a+b)^5-5ab((a+b)^2-2ab))^2-2a^2b^2)-10โข(ab)^2(a+b)
a^5+b^5=2^5-5โข4((4-8)^2-2โข16)-10โข16โข2=>
a^5+b^5=32+320-320=32
=> X1+X2=(a^5+b^5)/(ab)^2=32/16=2
=> X1โขX2=a^3/b^2โขb^3/a^2=ab=4
=> ax^2+bx+c=0, x^2-2x+4=0
Forwarded from Deleted Account
x^2+1=3x
x+1/x=3
x^2+2+1/x^2=9 -> 1/x^2=7-x^2
(x^4+x^-2)/(x^2+5x+1)=
(x^4+7-x^2)/(3x+5x)=
(x^2(x^2-1)+7)/8x=
((3x-1)(3x-2)+7)/8x=
(9x^2-9x+9)/8x=
9/8โข(x^2-x+1)/x=9/8โข2x/x=
9/4
C
x+1/x=3
x^2+2+1/x^2=9 -> 1/x^2=7-x^2
(x^4+x^-2)/(x^2+5x+1)=
(x^4+7-x^2)/(3x+5x)=
(x^2(x^2-1)+7)/8x=
((3x-1)(3x-2)+7)/8x=
(9x^2-9x+9)/8x=
9/8โข(x^2-x+1)/x=9/8โข2x/x=
9/4
C
Forwarded from Deleted Account
x^2+2x+4=0,
x^4=(-2x-4)^2=4x^2+16x+16
-> 4(-2x-4)+16x+16=8x
x+4/x=-2 -> 8/x=-4-2x
x^4/4+8/x-3=
x^4/4-4-2x-3=
8x/4-4-2x-3=-7
c
x^4=(-2x-4)^2=4x^2+16x+16
-> 4(-2x-4)+16x+16=8x
x+4/x=-2 -> 8/x=-4-2x
x^4/4+8/x-3=
x^4/4-4-2x-3=
8x/4-4-2x-3=-7
c
Forwarded from Deleted Account
x+1/(16x)=3
xโข4+1/(16x)โข4=12
4x+1/(4x)=12
16x^2+1/16x^2=144-2=142
(16x^2+1/16x^2)(x+1/16x)=3โข142
16x^3+x+1/16x+1/256x^3=426
=>
16x^3+1/(256x^3)=423
xโข4+1/(16x)โข4=12
4x+1/(4x)=12
16x^2+1/16x^2=144-2=142
(16x^2+1/16x^2)(x+1/16x)=3โข142
16x^3+x+1/16x+1/256x^3=426
=>
16x^3+1/(256x^3)=423
Forwarded from Vicky NARCOTICS SSC CGL 2025 SELECTED
Inke question ke solution bhej do koi please bhut confusion h
Forwarded from Deleted Account
ax^2+bx+c=0
x1=5+3sqrt(3)
-> x2=5-3sqrt(3)
x1+x2=5+5=10
x1โขx2=5^2-9โข3=-2
=> x^2-10x-2=0
a=1, b=-10, c=-2
(a^2+b^2+c^2)/(a+b+c)=
(1+100+4)/(-11)=-105/11
C
x1=5+3sqrt(3)
-> x2=5-3sqrt(3)
x1+x2=5+5=10
x1โขx2=5^2-9โข3=-2
=> x^2-10x-2=0
a=1, b=-10, c=-2
(a^2+b^2+c^2)/(a+b+c)=
(1+100+4)/(-11)=-105/11
C