minhthien2016

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3 years, 15 days

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These are questions asked by minhthien2016

How to get the only solution x = (1/6)*sqrt(114)*a of this equation?
restart;
with(Student:-MultivariateCalculus);
u := `<,>`(-(1/4)*a, -(1/12)*sqrt(3)*a, -x);
v := `<,>`(-(1/2)*a, (1/6)*sqrt(3)*a, (1/2)*x);
`assuming`([solve(Angle(u, v) = arccos(2*sqrt(26)*(1/13)), x)], [And(a > 0, x > 0)])

I have the sequence f(n) satify the conditions

f(n) = 4*f(n-1)-2*f(n-2), f(1) = 1, f(2) = 28.

I find the formula of f(n)

rsolve({f(1) = 1, f(2) = 28, f(n) = 4*f(n-1)-2*f(n-2)}, {f})


I got

f(n) = (-25*sqrt(2)*(1/4)-6)*(2-sqrt(2))^n+(25*sqrt(2)*(1/4)-6)*(2+sqrt(2))^n

Is there an integeral number k satisfy the equation f(n) = k^2, (k <> 0) ?

I have problem: Let  A(2,0,0), B(0,3,0), C(0,0,6) and D(1,1,1) be four points and Delta is the line passing through the point D so that sum of distances from the points A, B, C to Delta maximum. Find a direction vector of Delta.

I tried. Let v(a,b,c) where a^2 + b^2 + c^2 = 1 and my code
 

with(Student:-MultivariateCalculus):
with(Optimization):
A := [2, 0, 0]: 
B := [0, 3, 0]: 
C := [0, 0, 6]: 
DD := [1, 1, 1]: 
v := <a, b, c>: 
d := Line(DD, v): 
d1 := Distance(A, d): 
d2 := Distance(B, d): 
d3 := Distance(C, d): 
S := d1+d2+d3: 
Maximize(S, {a^2+b^2+c^2 = 1});

I didn't get the result. How to get the numbers a, b, c?

I want to get solutions (approximate solutions) of the inequality g(x) > 0. I tried. 
 

restart; 
fprime := x-> (x-1)*(x-2)^2*(x-3)*(x-4); 
f := unapply(simplify(int(fprime(x), x)), x); 
g := unapply(expand(f(3*x+1)-x^3+3*x), x); 
solve(g(x) > 0, x);

I got

I have triangle TAD. Now I want to take any point inside the triagle TAD. How can I take it? I tried
 

restart; 
with(Student:-MultivariateCalculus); 
A := [1, 2, 3]; 
DD := [-2, 1, 0]; 
T := [1/2, 1/2, 3/2]; 
P := Plane(T, A, DD); 
GetRepresentation(P)

I got the quation of the plane TAD is
 

-3*x-3*y+4*z = 3

With the point P(x,y,z). What is the condition of x, y, z?

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