Recent content by Raerin

So how do I convert r=4+4cos(theta) into rectangular form? I know that r^2 = x^2+y^2 and that rcos(theta) = x. Would the start of the solution be: sqrt(x^2 + y^2) = 4+4x If yes, I don't know where to go from there.

I found the cross product between (1,1,0) and (1,0,0). The result was (0,0,1) and I just subbed in (4, 2, -7) to z=0 to find the d value of the equation ax+by+cz+d=0 This method is correct, right? Or was my answer just coincidentally right?

So the second direction vector in the parallel plane just need to have z=0 and be non-colinear with the first vector? For (1,0,0) the equation would be: z + 7 = 0?

What is the equation of a plane that passes through the point (4,2,-7) and is parallel to the xy plane? I know that on the xy plane z = 0, so the direction vector is (1,1,0). Would a direction vector of the parallel line be (2,2,0)? If so the final equation is p=(4,2,-7)+s(1,1,0) + t(2,2,0)?

------ So it's not possible for both parameters to be s? Then there's a mistake in the question. I guess I don't need help anymore. Thanks!

How to find the intersection point between two lines? line 1: r = (3,1,-1) + s(1,2,3) line 2: r = (2,5,0) + s(1,-1,1)

Yep, I did! I just didn't understand how they got to that conclusion from z^2=4i

z = sqrt(2) + sqrt(2)i?

oh so the angle is 45?

720? - - - Updated - - - Sorry, I still don't fully understand how z=2√(1+i)

180 degrees?

r=2

(-16)^2 = sqrt(256) = 16?

I don't understand this part. Is Serena's way easier? Though, I don't really know how to convert it into polar form.

How do I solve z^4+16=0 in a+bi form. I don't see how it can be factored so what do I do?