toandhsp

2 years, 208 days


These are answers submitted by toandhsp

You can try

October 12 2013 toandhsp 190
0 0

> restart;

L:=[]:

N:=10:

for x from -N to N  do

for y from -N to N  do

for z from -N to N  do 

for t from -N to N  do

for s from -N to N  do

if  6*x+10*y+15*z=7  and 24*s+33*t=9 then

L:=[op(L), [x,y,z,s,t]] fi;

od: od: od: od:od:

nops(L);  L;

 

Another way

October 11 2013 toandhsp 190
0 0

Note that, the coordinates centre of pencil of line is solution of the system of equations 3*x+4*y-10=0 and 3*x-y-5=0.

solve([3*x+4*y-10=0,3*x-y-5=0],[x,y]);

We have [[x = 2, y = 1]]

Now, put Delta is the line passing the point A(2, 1) and having the slope k. The equation of the line Delta has the form y=k*(x-2)+1.

And then, you find the distance from the center T to the line Delta.

You can try

> restart:

with(geometry):

circle(C,x^2+y^2+2*x-4*y=0,[x,y],'centername'=T):

R:=radius(C):

line(Delta,y=k*(x-2)+1,[x,y]):

sol:=solve(distance(T,Delta)=R,{k}):

for i to 2 do subs(sol[i],Equation(Delta)) end do;


Equation of two lines are - x/2 + y = 0 and 2 x + y - 5 = 0.



 

You can try

October 11 2013 toandhsp 190
1 0

You can try

with(geometry):

circle(C,x^2+y^2+2*x-4*y=0,[x,y],'centername'=T):

R:=radius(C):

assume(3*a+3*b <> 0):

assume(4*a-b <> 0): 

line(d,a*(3*x+4*y-10)+b*(3*x-y-5)=0,[x,y]):

solve(distance(T,d)=R,[a,b]);

The solutions of the solve are a = b and a = -b/2.

With a = b, you can choice a = b = 1, and using

subs(a=1,b=1,Equation(d));

You get the equation 6*x+3*y-15 = 0.

With a = -b/2, choice b = -2, we have a = 1,

subs(a=1,b=-2,Equation(d));

and we get 

-3*x+6*y = 0

Another way

October 10 2013 toandhsp 190
0 1

solve({50811.89143+x^2+y^2-61.91881776*x-446.5619114*y = 0,1621.451347-0.1838809e-1*x-7.2593834*y = 0},{x,y});

Question 1

May 08 2013 toandhsp 190
0 0

> restart:

with(geom3d):

point(A,1, 0, 0):

point(B,0, 1, 0):

point(C,0,0,1):

plane(P, x+y+z=6,[x,y,z]):

point(T,x,y,z):

sol:=solve([distance(T,A)=distance(T,B),

distance(T,A)=distance(T,C),Equation(P)],[x,y,z]);point(T1,rhs(op(1,sol[1])),rhs(op(2,sol[1])),rhs(op(3,sol[1])) ):

Equation(sphere(C1,[T1,distance(T1,A)],[x,y,z]));

Eq1:=Student[Precalculus][CompleteSquare](Equation(C1));

 

 

 

I tried

April 22 2013 toandhsp 190
0 0

restart;

L:=[]:

k:=10:

for x1 from -k to k  do

for y1 from -k to k  do

for x2 from -k to k  do

for y2 from -k to k  do

for x3 from -k to k  do

for y3 from -k to k do

G:=[1,1]:

H:=[3,3]:

if x1+x2+x3=3*G[1] and y1+y2+y3=3*G[2] and (H[1]-x1)*(x3-x2) +(H[2]-y1)*(y3-y2) = 0 and (H[1]-x2)*(x2-x1) +(H[2]-y2)*(y2-y1)=0 and (x2-x1)*(y3-y1) +(x3-x1)*(y2-y1) <>0 then

L:=[op(L), [[x1,y1], [x2,y2], [x3,y3]]] fi;

od: od: od: od: od: od:

nops(L);  L;

 

Thank you very much.

January 25 2013 toandhsp 190
0 0

Thanks for all helps.

With Mathematica

January 05 2013 toandhsp 190
0 0
I tried solve it with Mathematica
Reduce[Sqrt[x + 3 - 4*Sqrt[x - 1]] + Sqrt[x + 8 - 6*Sqrt[x - 1]] == 
1, x, Reals]

Change codes

December 26 2012 toandhsp 190
0 0

First code

>restart:

f:=x->a/x^2-x+2:

g:=x->a*sin(2*x):

ans:=allvalues(solve(f(f(-1))=sqrt(2)+g(f(2)),a)):

ans2:=[evalf(ans)];

We get

ans2 := [-2.881759230-1.051335058*I, -.7437844448-1.866833976*I, -2.881759230+1.051335058*I, -.7437844448+1.866833976*I, -5.471111560]

Second code

>restart:

f:=x->a/x^2-x+2:

g:=x->a*sin(2*x):

ans:=fsolve(f(f(-1))=sqrt(2)+g(f(2)),a);

We get

ans := -14.51059004

The solutions -5.471111560 and -14.51059004 are different. 



Thank you very much.

December 26 2012 toandhsp 190
0 0

Thank you very much.

Thank you.

December 22 2012 toandhsp 190
1 0

Thank you.

Thank you very much.

December 22 2012 toandhsp 190
1 0

Thank you very much.

Thank you very much.

November 27 2012 toandhsp 190
0 0

Thank you very much.

Another way

November 21 2012 toandhsp 190
0 0

Put the equation of the line passing the point A(4,1) has the form a*(x-4)+b*(y-1)=0. We always assume a^2 + b^2 = 1. And then, you solve the system of equations

solve([-4*a-b=2,a^2+b^2=1],[a,b]);

But I don't know how to get the exactly solutions [a = -8/17-1/17*13^(1/2),b = -2/17+4/17*13^(1/2)] and [[a = -8/17+1/17*13^(1/2), b = -2/17-4/17*13^(1/2)]]

I must solve in two cases 

solve([f(0,0)=2,a^2+b^2=1,b<0],[a,b]);

solve([f(0,0)=2,a^2+b^2=1,b>0],[a,b]);

Note that, when the line passing the point A(2,1), the line x = 2 has not slope, but it is also a tangent of the circle. You try

f:=(x,y)->a*(x-2)+b*(y-1):

solve([f(0,0)=2,a^2+b^2=1],[a,b]);

 

One situation

November 12 2012 toandhsp 190
0 0

How about this code? Maple can not run. Please help me. 

> restart:

a:=2:

eq1:=(a^2+b^2-c^2)/(2*a*b):

eq2:=(b^2+c^2-a^2)/(2*b*c):

eq3:=(c^2+a^2-b^2)/(2*a*c):

sol:=solve([eq1=convert(cos(30*Pi/180), radical), eq2=cos(30*Pi/180), eq3=cos(120*Pi/180)], [b,c]);

with(geom3d):

point(o,0,0,0):

point(A,a,0,0):

sys:=solve([x^2+y^2=(rhs(op(1,sol[1])))^2,(x-a)^2+y^2=(rhs(op(2,sol[1])))^2],[x,y]);

point(B,rhs(op(1,op(1,sys))),rhs(op(2,op(1,sys))),0):

triangle(T,[o,A,B]):

map(coordinates,DefinedAs(T));

simplify(FindAngle(o,T));

simplify(FindAngle(A,T));

simplify(FindAngle(B,T));

line(d,[t+1,-t,-t+2],t):

rotation(W,T,-2*Pi/3,d):

map(coordinates,DefinedAs(W));

 

 

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