BCS MATH PRACTICAL

REGULA FALSI METHODFormula:

C=(a*f(b)-b*f(a)/f(b)-f(a)
Program:
deff(‘d=f(x)’,’d=x^3-100′)

a=input(“Enter the value of a:”)

b=input(“Enter the value of b:”)

n=input(“Enter the number of iterations n:”)

for i=1:n

    c=(a*f(b)-b*f(a))/(f(b)-f(a))

    disp([i,c])

if f(a)*f(c)<0 then

    b=c

end

if f(b)*f(c)<0 then

    a=c

end

c1=(a*f(b)-b*f(a))/(f(b)-f(a))

if abs(c1-c)<0.00001 then

    disp(“We get accurate roots”)

    break;

    end

end

/*PROGRAM FOR NEWTON RAPHSON METHOD*/

deff(‘y=f(x)’,’y=x^2-2*x-1′)

deff(‘dy=df(x)’,’dy=2*x-2′)

a=0,n=5;

for i=1:n

    c=(a-f(a)/df(a))

    disp([a,f(a),df(a)])

a=c

end

printf(“c=%f”,c)

1] Newton’s Forward Gregory Interpolation:

Program:

x=[1 2 3 4 5]

y=[1 4 9 16 25]

d1=mtlb_diff(y)

d2=mtlb_diff(d1)

d3=mtlb_diff(d2)

d4=mtlb_diff(d3)

disp([d1d2d3d4])

d=[d1(1)d2(1)d3(1)d4(1)]

disp(d)

x0=1.5;

h=1;

u=(x0-x(1))/h;

printf(“u=%f”,u)

ans=0;

for i=1:4

q=1

for j=1:i

p=4-(j-1);

q=q*p

end

t=d(i)*q/factorial(i)

ans=ans+t

end

y=y(1)+ans

printf(“\n The value of function at %f:%f”,x0,y)

Program:- (Sort a set of points w.r.t. REACTANGLE-2D)
#include<stdio.h>

#include<conio.h>

#include<math.h>

void main()

{

int i,n;

float x[10],y[10],Xmin,Ymin,Xmax,Ymax;

clrscr();

printf(“\n\n Enter the values of Xmin & Ymin =\t”);

scanf(“%f%f”,&Xmin,&Ymin);

printf(“\n\n Enter the values of Xmax & Ymax =\t”);

scanf(“%f%f”,&Xmax,&Ymax);

printf(“\n\n Enter the number of points to be sort =\t”);

scanf(“%d”,&n);

for(i=1;i<=n;i++)

{

printf(“\n\n Enter the cordinates of point %d =\t”,i);

scanf(“%f%f”,&x[i],&y[i]);

if(((x[i]>Xmin) && (Xmax>x[i])) && ((y[i]>Ymin) && (Ymax>y[i])))

printf(“\n\n The input point is INSIDE the rectangle.”);

else

if((x[i]<Xmin) || (Xmax<x[i]) || (y[i]<Ymin) || (Ymax<x[i]))

printf(“\n\n The input point is OUTSIDE the rectangle.”);

else

printf(“\n\n The input point is ON the rectangle.\n\n”);

}

getch();

}

/*
 OUTPUT:-
 Enter the values of Xmin & Ymin = 1 3

 Enter the values of Xmax & Ymax = 7 8

 Enter the number of points to be sort = 3

 Enter the cordinates of point 1 = 1 7

 The input point is ON the rectangle.

 Enter the cordinates of point 2 = 4 6

 The input point is INSIDE the rectangle.
 

 Enter the cordinates of point 3 = 1 2

 The input point is OUTSIDE the rectangle.
*/


Program:- (Sort a set of points w.r.t. LINE)

#include<stdio.h>

#include<conio.h>

#include<math.h>
void main()

{

int i,n;

float x[20],y[20],z[20],x1,x2,y1,y2;

clrscr();

printf(“\n Enter cordinates of points of line (x1,y1) & (x2,y2) =\t”);

scanf(“%f%f%f%f”,&x1,&x2,&y1,&y2);

printf(“\n Enter the number of points to be sort =t”);

scanf(“%d”,&n);

for(i=1;i<=n;i++)

{

printf(“\n Enter the %d co-ordinates of point(x[i],y[i]) =”,i);

scanf(“%f%f”, &x[i],&y[i]);

z[i]=((y[i]-y1)*(x2-x1)) – ((x[i]-x1)*(y2-y1));

printf(“\n z[i]=%f”,z[i]);

if(z[i]>0)

{

printf(“\n z[i] is greater than zero, therefore point(%f,%f) is ABOVE the line.”,x[i],y[i]);

}

else

if(z[i]<0)

{

printf(“\n z[i] is less than zero, therefore point(%f,%f) is BELOW the line.”,x[i],y[i]);

}

else

{

printf(“\n z[i] is equal to zero, therefore point(%f,%f) is ON the line.”,x[i],y[i]);

}

}

getch();

}

/*
OUTPUT:-
Enter cordinates of points of line (x1,y1) & (x2,y2) = 1 3 2 1
Enter the number of points to be sort =3
Enter the 1 co-ordinates of point(x[i],y[i]) = 1 2

z[i]=0.000000

z[i] is equal to zero, therefore point(1.000000,2.000000) is ON the line.
Enter the 2 co-ordinates of point(x[i],y[i]) = -1 2

z[i]=-2.000000

z[i] is less than zero, therefore point(-1.000000,2.000000) is BELOW the line.
Enter the 3 co-ordinates of point(x[i],y[i]) = 4 5

z[i]=9.000000

z[i] is greater than zero, therefore point(4.000000,5.000000) is ABOVE the line*/

Ques. Evaluate ?_0^6¦x2 dx by using Simpson’s 3/8 rule & value of n=6.

Solu.:
Program:

deff(‘y=f(x)’,’y=x^2′)

a=0;b=6;n=6;

h=(b-a)/n

printf(“h=%f”,h)

fori=0:n

x=a+i*h

y=f(x)

disp([xy])

end

sum1=0;sum2=0;

fori=1:n-1

x=a+i*h

ifmodulo(i,3)==0then

sum2=sum2+f(x)

else

sum1=sum1+f(x)

end

end

t=(3*h/8)*[(f(a)+f(b))+3*sum1+2*sum2]

printf(“t=%f”,t)

Output:

exec(‘D:\poojardbms\scilab\prat12_1.sce’, -1)

h=1.000000 

    0. 0.  

    1. 1.  

    2. 4.  

    3. 9.  

    4. 16.  

    5. 25.  

    6. 36.  

t=72.000000

/*PROGRAM TO FIND SIMPSONS 1/3RD RULE/*

deff(‘y=f(x)’,’y=x^2′)

a=1,b=6,n=5;

h=(b-a)/n

printf(“h=%f”,h)

for i=0:n

    x=a+i*h

    y=f(x)

    disp([x y])

end

sum1=0,sum2=0

for i=1:n-1

    x=a+i*h

    if modulo(i,2)==0 then

        sum1=sum1+f(x)

    else

        sum2=sum2+f(x)

    end

  th=(h/3)*(f(a)+f(b)+(4*(sum1))+(2*(sum2)))

end

printf(“th=%f”,th)
OUTPUT:-
h=1.000000 

    1. 1.  

 

    2. 4.  

 

    3. 9.  

 

    4. 16.  

 

    5. 25.  

 

    6. 36.  

th=71.000000
Ques. Eulars methode

Program:

deff(‘y=f(x,y)’,’y=(x^2+y)’)

x0=0;y0=1;h=0.02;

next=y0+h*f(x0,y0)

x0=x0+h;

y0=next;

disp([x0y0])

end

Program:- (sort a set of points w.r.t. RECTANGULAR BOX-3D)

#include<stdio.h>

#include<conio.h>

#include<math.h>

void main()

{

int i,n;

float x[10],y[10],z[10],Xmin,Xmax,Ymin,Ymax,Zmin,Zmax;

clrscr();

printf(“\n Enter the minium & maximum values for line X=\t”);

scanf(“%f%f”,&Xmin,&Xmax);

printf(“\n Enter the minium & maximum values for line Y=\t”);

scanf(“%f%f”,&Ymin,&Ymax);

printf(“\n Enter the minium & maximum values for line Z=\t”);

scanf(“%f%f”,&Zmin,&Zmax);

printf(“\n Enter how many points to be sorts=\t”);

scanf(“%d”,&n);

for(i=1;i<=n;i++)

{

printf(“\n Enter the co-ordinates of a point (x,y,z) =\t”,i);

scanf(“%f%f%f”,&x[i],&y[i],&z[i]);

if(((x[i]>Xmin) && (Xmax>x[i])) && ((y[i]>Ymin) && (Ymax>y[i])) && ((z[i]>Zmin) && (Zmax>z[i])))

printf(“\n\n The input point is INSIDE the rectangular box.\n\n”);

else if((x[i]<Xmin) || (Xmax<x[i]) || (y[i]<Ymin) || (Ymax<y[i]) || (z[i]<Zmin) || (Zmax<z[i]))

printf(“\n\n The input point is OUTSIDE the rectangular box.\n\n”);

else

printf(“\n\n The input point is ON the rectangular box.\n\n”);

}

getch();

}

/*
 OUTPUT:-
 Enter the minium & maximum values for line X= 1 7
 Enter the minium & maximum values for line Y= 2 8
 Enter the minium & maximum values for line Z= 5 9
 Enter how many points to be sorts= 3
 Enter the co-ordinates of a point (x,y,z) = 4 4 6

 The input point is INSIDE the rectangular box.

 Enter the co-ordinates of a point (x,y,z) = 9 8 7

 The input point is OUTSIDE the rectangular box.

 Enter the co-ordinates of a point (x,y,z) = 3 2 5

 The input point is ON the rectangular box.

*/

Output:

exec(‘D:\poojardbms\scilab\prat10_1.sce’, -1)
Warning : redefining function: f . Use funcprot(0) to avoid this message

    0.02 1.020101

Ques. ?_0^6¦x2 dx using Trapezoid Rule into 6 parts.

Program:

deff(‘y=f(x)’,’y=x^2′)

a=0;b=1;h=0.5;

n=(b-a)/h

printf(“n=%f”,n)

fori=0:n

x=a+i*h

y=f(x)

disp([xy])

end

sum=0

fori=1:n-1

x=a+i*h

sum=sum+f(x)

end

printf(“sum=%f”,sum)

t=h/2*[(f(a)+f(b))+2*sum]

printf(“t=%f”,t)
Output:

exec(‘D:\poojardbms\scilab\prat9_1.sce’, -1)

n=2.000000 

    0. 0.   

    0.5 0.25  

    1. 1.  

Warning : redefining function: sum . Use funcprot(0) to avoid this message

sum=0.250000t=0.375000

/*PROGRAM TO FIND SIMPSONS 1/3RD RULE/*

deff(‘y=f(x)’,’y=x^2′)

a=1,b=6,n=5;

h=(b-a)/n

printf(“h=%f”,h)

for i=0:n

    x=a+i*h

    y=f(x)

    disp([x y])

end

sum1=0,sum2=0

for i=1:n-1

    x=a+i*h

    if modulo(i,2)==0 then

        sum1=sum1+f(x)

    else

        sum2=sum2+f(x)

    end

    th=(h/3)*(f(a)+f(b)+(4*(sum1))+(2*(sum2)))

end

printf(“th=%f”,th)
OUTPUT:-
h=1.000000 

    1. 1.  

 

    2. 4.  

 

    3. 9.  

 

    4. 16.  

 

    5. 25.  

 

    6. 36.  

th=71.000000

Ques. Use Rangakutta method of order second to solve the following using step size using h=1,

dy/dx=(5×2-y)/(ex+y)

Program:

deff(‘y=f(x,y)’,’y=(5*(x^2)-y)/(%e^(x+y))’)

x0=0;y0=1;h=0.1;n=2;
fori=1:n

k1=h*f(x0,y0)

k2=h*f(x0+h,y0+k1)

disp([k1k2])

printf(“\n k1=%f, k2=%f”,k1,k2)

t=y0+(1/2)*[k1+k2]

disp(t)

printf(“\n t=%f”,t)

x0=x0+h

y0=t

end
Output:

  – 0.0367879 – 0.0315373  

 k1=-0.036788, k2=-0.031537 

    0.9658374  

 t=0.965837 

  – 0.0315450 – 0.0236185  

 k1=-0.031545, k2=-0.023618 

    0.9382556  

 t=0.938256

Ques. Runge kutta fourth order

 dy/dx=y-x, where y(0)=2, obtain y(0.1) & y(0.2), take h=0.1 

Program:

deff(‘y=f(x,y)’,’y=y-x’)

x0=0;y0=2;h=0.1;n=2;
fori=1:n

k1=h*f(x0,y0)

k2=h*f(x0+h/2,y0+k1/2)

k3=h*f(x0+h/2,y0+k2/2)

k4=h*f(x0+h,y0+k3)

disp([k1k2k3k4])

printf(“\n k1=%f, k2=%f, k3=%f, k4=%f”,k1,k2,k3,k4)

t=y0+(1/6)*[k1+2*k2+2*k3+k4]

disp(t)

printf(“\n t=%f”,t)

x0=x0+h

y0=t

end
Output:

    0.2 0.205 0.20525 0.2105 

 k1=0.200000, k2=0.205000, k3=0.205250, k4=0.210525 

    2.2051708  

 t=2.205171 

    0.2105171 0.2160429 0.2163192 0.2221490  

 k1=0.210517, k2=0.216043, k3=0.216319, k4=0.222149 

    2.4214026  

 t=2.421403

Program:- (To find least MUTUAL DISTANCE between the points)
*/

#include<stdio.h>

#include<conio.h>

#include<math.h>

#include<stdlib.h>

void main()

{

int i,j,n,a;

float x[10],y[10],d[10][10],Dmax;

clrscr();

printf(“\n\n Enter the number of points to be processing :\t”);

scanf(“%d”,&n);

for(i=1;i<=n;i++)

{

printf(“\n Enter the cordinates of point %d =\t”,i);

scanf(“%f%f”,&x[i],&y[i]);

}

for(i=1;i<=n;i++)

{

Dmax=0;

for(j=1;j<=n;j++)

{

if(i!=j)

{

d[i][j]=sqrt(((x[i]-x[j] ) * (x[i]-x[j] )) + ((y[i]-y[j] ) * (y[i]-y[j])));

if(d[i][j]>Dmax)

{

Dmax=d[i][j];

a=j;

}

}

}

printf(“\n Maximum distance between the points p(%d) & p(%d)=%f”,i,a,Dmax);

printf(“\n\n”);

}

getch();

}

/*
 OUTPUT:-

 

 Enter the number of points to be processing : 4
 Enter the cordinates of point 1 = 1 3
 Enter the cordinates of point 2 = 6 0
 Enter the cordinates of point 3 = 4 2
 Enter the cordinates of point 4 = 3 1
 Maximum distance between the points p(1) & p(2)=5.830952

 Maximum distance between the points p(2) & p(1)=5.830952

 Maximum distance between the points p(3) & p(1)=3.162278

 Maximum distance between the points p(4) & p(2)=3.162278

*/

Program:- ( To find the FARTHEST DISTANCE between points)

#include<stdio.h>

#include<conio.h>

#include<math.h>

#include<stdlib.h>

void main()

{

int i,j,n,a;

float x[10],y[10],d[10][10],Dmax;

clrscr();

printf(“\n\n Enter the number of points to be processing :\t”);

scanf(“%d”,&n);

for(i=1;i<=n;i++)

{

printf(“\n Enter the cordinates of point %d =\t”,i);

scanf(“%f%f”,&x[i],&y[i]);

}

for(i=1;i<=n;i++)

{

Dmax=0;

for(j=1;j<=n;j++)

{

if(i!=j)

{

d[i][j]=sqrt(((x[i]-x[j] ) * (x[i]-x[j] )) + ((y[i]-y[j] ) * (y[i]-y[j])));

if(d[i][j]>Dmax)

{

Dmax=d[i][j];

a=j;

}

}

}

printf(“\n Maximum distance between the points p(%d) & p(%d)=%f”,i,a,Dmax);

printf(“\n\n”);

}

getch();

}

/*
 OUTPUT:-

 

 Enter the number of points to be processing : 4
 Enter the cordinates of point 1 = 1 3
 Enter the cordinates of point 2 = 6 0
 Enter the cordinates of point 3 = 4 2
 Enter the cordinates of point 4 = 3 1
 Maximum distance between the points p(1) & p(2)=5.830952

 Maximum distance between the points p(2) & p(1)=5.830952

 Maximum distance between the points p(3) & p(1)=3.162278

 Maximum distance between the points p(4) & p(2)=3.162278

*/

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बेंजामिन फ्रांकलीन

image

Benjamin Franklin Drawing Electricity from the Sky c. 1816 at the Philadelphia Museum of Art, by Benjamin West

image

Franklin and Electricity vignette engraved by the BEP (c. 1860).

बेंजामिन फ्रांकलीन(Benjamin Franklin)
जन्म1706-मृत्यु1790
बोस्टन

1747.पंतगाला रेशमी दोरा लाऊन खेळत असतांना दोन ढगांच्या धडकेने आकाशातील विज त्या दोरीच्या सहाय्याने त्याच्या हातापर्यंत वाहत आली व त्या दोरीला बांधलेल्या खीळ्यापासुन जानवली हा विजेचा सबंध ह्या अचानक झालेल्या सबंधातुन सीध्द झाला,आणी वेजेच्या शोधाची चाहुल लागली,त्याने या सबंधाला “lighting is an electric current ” असे म्हटले,आणि विजेच्या शोधाची क्रांतिच झाली.

हा एक मुद्रक,शास्त्रज्ञ ,अविष्कारक,राजनेता,दर्शनिक,संगीतकार,अर्थत््ज म्हनुन ओळखला जातो,

त्याने आविष्कार केलेले शोध,
Bifocal glass,Franklin stove,glass harmonica…etc

Lighting Is an electric Current

When rain has wet the kite twine so that it can conduct the electric fire freely, you will find it streams out plentifully from the key at the approach of your knuckle, and with this key a phial, or Leyden jar, may be charged: and from electric fire thus obtained spirits may be kindled, and all other electric experiments [may be] performed which are usually done by the help of a rubber glass globe or tube; and therefore the sameness of the electrical matter with that of lightening completely demonstrated.”

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Chenge boot audio

Application list
1) Root Explorer
2)key root master

Available in ” play store ”

“Root Explorer” ये अॉप युज करके बुट ऑडीओ यानी की स्विच ऑफ ऑडीओ चेंज कर सकते है!

“Key root master” ईस अॉप्लीकेशन द्वारा आप root explorer को rw से ro मे बदल सकते हो यानी की आपका मोबाईल टेम्पोररी root हो जायेगा.

अगर आप root explorer use भी करते हो तब भी आप key root master के सीवा फाईल चेंज यानी ऑडीओ बदल नही सकते .

ऑडीओ चेंज करने का तरीका:

फस्ट ऊपर दीये गये दोनो अॉपलीकेशन ईंस्टाल करे.

आप जो भी ” सॉंग ऑडीओ ” बूट ऑडीओ के लीए रखना चाहते हो ओ एक फाईल मे निचे दीये गये नाम से Rename करे
“Bootaudio.Mp3 ”
“Shutaudio. Mp3 ”

ओर सेव करे .

अब key root master ओपन करे
और आगेकी दीखाई गई निर्दशो का पालन करे

Open>>Mulai root>>root>>red option

अब Root explorer खोले
और नीचे दीये गये स्टेप फोलो करे
Open>> system file>>media file>>

अब सामने Bootaudio.Mp3 Shutaudio. Mp3  दो फाईल्स दीखाई देगी .

ये फाईल्स आपने पहीले जीस फाईल में आपकी बुट ऑडीओ वाली फाईले सेव की है वहा पर मुव करे . या डीलीट करे

पर पहले आपकी विंडो मे ऊपर के साईट पर जो RW OR RO नामक बटन है उसपर क्लीक करे वो rw से ro हो जायेगा.

बाद मे आपकी सीलेक्ट फाईल
यहा पर मुव या कॉफी करे.
अब मोबाईल स्विच ऑफ करे और ऐंजोय करे

Android mobaile users only
Whats app feedback
No  8888575548

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