toandhsp

2 years, 349 days


These are answers submitted by toandhsp

One way

October 14 2014 toandhsp 215

f:=x->a* x^2+b*x+c:
sol:=solve([f(2)=0,f(6)=1,f(8)=0]);
subs(sol,f(x));

You can try

October 12 2013 toandhsp 215

> 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 215

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 215

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 215

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 215

> 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 215

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 215

Thanks for all helps.

With Mathematica

January 05 2013 toandhsp 215
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 215

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 215

Thank you very much.

Thank you.

December 22 2012 toandhsp 215

Thank you.

Thank you very much.

December 22 2012 toandhsp 215

Thank you very much.

Thank you very much.

November 27 2012 toandhsp 215

Thank you very much.

Another way

November 21 2012 toandhsp 215

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]);

 

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