🍁A∂vαทcє Maths🍁 @advance_collection Channel on Telegram

🍁A∂vαทcє Maths🍁

07 Jan, 09:05

Put p=q=r=1
1=y+z
1=z+x
1=x+y

Find =(x/1+x)+(y/1+y)+(z/1+z)
==X+y+z/x+y+z==1

533

🍁A∂vαทcє Maths🍁

27 Dec, 15:45

Put A=60,B=30,C=0
Ans=D

572

🍁A∂vαทcє Maths🍁

05 Oct, 01:39

A=2
satisfied
2^2-4/√2=0

760

🍁A∂vαทcє Maths🍁

05 Oct, 01:38

(√a -1)(a+1)=3
Then find
(a²-4)/√a

765

🍁A∂vαทcє Maths🍁

28 Sep, 13:12

CosA=3/√10
Cos2A=9/5-1=4/5
Cos4A=7/25
Sin4A=24/25
Sin8A=2*24*7/625==336/625
Credit = @L_earner

844

🍁A∂vαทcє Maths🍁

28 Sep, 13:06

Tana=1/3
Sina=1/√10
Sin2a =2sinacosa
2*1/√10*3/√10
3/5
Sin4a=2sin2acos2a
2*3/5*4/5=24/25
Sin8a=2*24/25*7/25=
336/625

802

🍁A∂vαทcє Maths🍁

06 Sep, 09:24

Ladder theorm
1/x= 1/70+1/60-1/40
1/x= 5/840
X = 168

826

🍁A∂vαทcє Maths🍁

03 Aug, 03:22

n²x² + 1/x² = n
n(nx² + 1/nx²) = n
(nx² + 1/nx²) = 1
So
(nx²)³ = - 1
n³x⁶ = - 1
n³x³ = - 1/x³
(nx)³ = - 1/x³
1/(nx)³ = - x³
Now
x³ + 1/(nx)³
x³ - x³
0

864

🍁A∂vαทcє Maths🍁

17 Jul, 05:34

Credit - Vivek Gupta😍

925

🍁A∂vαทcє Maths🍁

17 Jul, 05:34

Ye kriye

774

🍁A∂vαทcє Maths🍁

23 Jun, 18:07

CD
+AB
1CE

It’s given that
C+A=C(mod10)
A=0(mod10)
So A=0 but this is not possible because then 1CE will not be possible

So it means B+D is a two digit number
Which means C+A+1=C(mod10)
A+1=0(mod10)
A=-1(mod10)
A=9(mod10)

So A=9

768

🍁A∂vαทcє Maths🍁

23 Jun, 17:51

A,B,C,D,E can take max value of 9 as they are digits
10A+B+10C+D= 100+10C+E
10A+B+D=100+E
At Max value of D=B=9 and E=1
10A=83
A= 8.3
So min value of A should be 8.3 in worst case so A has to take 9
90+B+D= 100+E
B+D= 10+E
B,D,E should be distinct and none can be 9
8+7= 10+5 or any other combination
C can take any number from 1 to 4
So numbers will be
For example
98 and 17
115
Many such numbers are possible
But value of A is fixed at 9
Credit - Vivek Gupta
@viki600

645