Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. x = 8 cos(t), y = 8 sin(t), z = 6 cos(2t); (4 3 , 4, 3)

Step-by-step explanation:

We need to find first th tangent vector. And we know that th tangent vector is the derivate of the vectorial function r(t). If we define r(t) like this:

We can find the derivate like this:

We can see that at the point given (4 sqrt (3),3,4) the value for t= pi/6. Because from the equation for x we have this:

Now since we have the value for t we can find the derivate at the point t=pi/6

And from the definition of an equation of a line passing thorugh a point with position vector a and parallel to th bector b:

Then the equation of th tangent line at (4 sqrt (3),3,4) is:

And then the parametric equations are given by

he used the algebra tiles correctly.

step-by-step explanation:

given : 3x(2x-1)

solution :

since he used algebra tiles .

armando's solution is the same in the solution shown in figure

as he distributed 2x into x and x and wrote them vertically

and also write   -1 vertically

and 3x into x and x and x and wrote it horizontally

performing the multiplication in the algebra tiles we get

and solving the given expression :

3x(2x-1)

since solving the expression using tiles   and solving the expression algebraically we get the same answer

thus   we can choose the first statement as the answer to this problem which is "he used the algebra tiles correctly".