Gautam Gurijala

In her majesty's diaper service, Financial Analyst , Deep Learning Engineer.

[Gautam Gurijala @Gauguri]

@sonukg4india (3-m)/(1+3m)=1 => m=1/2, passing thru (1,5). y=1/2x+9/2.
Intersection of y=3x+8
is (-7/5,19/5)
The four edges are : E(-8/3,0) O(0,0) M(0,9/2) D(-7/5,19/5)
formula: 1/2((x1y2-y1x2)+(x2y3-y2x3)+(x3y4-y3x4)+(x4y1-y4x1))
=> 1/2(9/2*7/5+ 19/5*8/3)
=> 1/2(16.432)
~ 8.216

[Gautam Gurijala @Gauguri]

@sonukg4india a = log6/log5, b= log7/log6
c = log8/log7, d = log 625/log 8
Sqrt(abcd) = log6/log5 * log7/log6 * log8/log7* log 625/log 8
=> sqrt(4)
= 2

[Gautam Gurijala @Gauguri]

@ybgoi Let 1st circle radius = r1
r1/(p+r1) = sin(30) => 2r1 = p+ r1
=>r1= p=1
r2/(p+2r1+r2)= 1/2=>3
r3/ (p+2r1+2r2+r3)= 1/2=>r3 =9
r4/(1+2r1+2r2+2r3+r4)= r4=> 27
So the series is 3^0,3^1… 3^(n-1)
=>3^9
=19683

[Gautam Gurijala @Gauguri]

@ZahlenRMD Dividing by x
=> 8 / ( x -8 + 1/x )
=> 8 / ( 10 - 8 )
=> 4

[Gautam Gurijala @Gauguri]

@ybgoi Tan(BDC) = a/2a = 1/2
=> 5b = 2a.
=> Shaded fraction =2b^2/2a^2
=> (2/5)^2a^2/a^2
=> 4/25

[Gautam Gurijala @Gauguri]

@ZahlenRMD f(2025) = f(405)+f(5)
=>f(135•3)= f(135)+f(3)+f(5)
=>f(45•3)= f(45)+f(3)+f(3)+f(5)
=>f(15•3)= f(15)+f(3)+f(3)+f(3)+f(5)
=> f(5)+4•f(3)+f(5)
3+8+3
= 14.

[Gautam Gurijala @Gauguri]

@ybgoi

[Gautam Gurijala @Gauguri]

@gaashuOfficial @UniCanberra @PerHenrikChris1 @MantillaIgnacio @MsLenaYmaths @profmath @alshareef1289 @Mathinity_ @Nikki_mathsol Int(e^x-1/x) [1,2]
=> (e^2-e)-(ln(2)-ln(1))

[Gautam Gurijala @Gauguri]

@sonukg4india Int(-x^2+4x)[0,2]-Int(x^2)[0,2]
=> (-x^3/3+2x^2)[0,2]-(x^3/3)[0,2]
=>(-8/3+8)-(8/3)
=>8/3

[Gautam Gurijala @Gauguri]

@ybgoi

[Gautam Gurijala @Gauguri]

@ybgoi

[Gautam Gurijala @Gauguri]

@sonukg4india A(a,0), B(4,0), C(4,Sqrt(a)),D(a,Sqrt(a))
Area = (4-a)(Sqrt(a)
Diff’ing => (4-a)(1/2a^(-1/2))+ (-1)a^(1/2)=0
(4-a)(1/2a^(-1/2))= a^(1/2)
=>4-a = 2a => a=4/3
Max Area = (4-4/3)* 2/(Sqrt(3))
=> 16/(3*Sqrt(3))
Approx = 3.079

[Gautam Gurijala @Gauguri]

@ybgoi Gautam Gurijala
@Gauguri
·
5m
The coordinates of Sq are A(0,0), B( pi/2,0), C (pi/2,pi/2), D(0,pi/2)
Area Sq ABCD = pi^2/4.
Area shaded = Int[pi/4,pi/2](sinx-cosx)dx
=> -(cosx-sinx)[pi/4,pi/2]
=> -[(1-Sqrt(2))]
=> Sqrt(2)-1
Fraction = (Sqrt(2)-1)/pi^2/4) = 4(Sqrt(2)-1)/pi^2

[Gautam Gurijala @Gauguri]

@ybgoi

[Heather Endresen @EndresenHeather]

Want to understand how SBA views their loan programs?

Please join us this Thursday where I will be moderating a discussion with the Associate Administrator for the Office of Capital Access, Kathryn Frost, organized by Harvard Alumni Entrepreneurs.

Capital Access for Entrepreneurs — HARVARD ALUMNI ENTREPRENEURS (https://t.co/onQZuRXdKl)

[Gautam Gurijala @Gauguri]

@mamekebi

[Gautam Gurijala @Gauguri]

@Maths_India @sonukg4india @azure_keke @MagicPi2 @ZahlenRMD @math_rehash @ChahalJaireet The center of the circle is (0,0) with radius 5.
The slope of
3x + y -5 = 0 -> (1)
is -3. Eqn of Perp to it passing thru (0,0) is
y= 1/3x -> (2)
Solving 3x + 1/3x = 5 or x = 3/2=> y= 1/2
(3/2,1/2) -> (b)

[Gautam Gurijala @Gauguri]

@CoreMath80 The diagonal of the Square is = r=> side of Square = r/Sqrt(2) =>
ArcTan ( 1 / Sqrt(2) )

[Gautam Gurijala @Gauguri]

@CoreMath80 Dividing by x =>
x + 1/x = 4 -> (1)
Let Sqrt(x)-1/Sqrt(x) = p =>
p^2 = x + 1/x -2
=> x + 1/x = p^2 + 2 -> (2)
Since (1) = (2)
p^2 = 2 => p= Sqrt(2)

[Gautam Gurijala @Gauguri]

@ChahalJaireet Let x^2 + 1 = t => x dx = 1/2dt
Integral(t^1/2) 1/2dt => 2/3t^3/2 + C => 2/3 (x^2+1)^(3/2) + C