1. The given rectangular equation is
.
We substitute
.
![r\cos \theta=2](https://tex.z-dn.net/?f=r%5Ccos%20%5Ctheta%3D2)
Divide through by ![\cos \theta](https://tex.z-dn.net/?f=%5Ccos%20%5Ctheta)
![r=\frac{2}{\cos \theta}](https://tex.z-dn.net/?f=r%3D%5Cfrac%7B2%7D%7B%5Ccos%20%5Ctheta%7D)
![r=2}\sec \theta](https://tex.z-dn.net/?f=r%3D2%7D%5Csec%20%5Ctheta)
![\boxed{x=2\to r=2\sec \theta}](https://tex.z-dn.net/?f=%5Cboxed%7Bx%3D2%5Cto%20r%3D2%5Csec%20%5Ctheta%7D)
2. The given rectangular equation is:
![x^2+y^2=36](https://tex.z-dn.net/?f=x%5E2%2By%5E2%3D36)
This is the same as:
![x^2+y^2=6^2](https://tex.z-dn.net/?f=x%5E2%2By%5E2%3D6%5E2)
We use the relation ![r^2=x^2+y^2](https://tex.z-dn.net/?f=r%5E2%3Dx%5E2%2By%5E2)
This implies that:
![r^2=6^2](https://tex.z-dn.net/?f=r%5E2%3D6%5E2)
![\therefore r=6](https://tex.z-dn.net/?f=%5Ctherefore%20r%3D6)
![\boxed{x^2+y^2=36\to r=6}](https://tex.z-dn.net/?f=%5Cboxed%7Bx%5E2%2By%5E2%3D36%5Cto%20r%3D6%7D)
3. The given rectangular equation is:
![x^2+y^2=2y](https://tex.z-dn.net/?f=x%5E2%2By%5E2%3D2y)
This is the same as:
We use the relation
and ![y=r\sin \theta](https://tex.z-dn.net/?f=y%3Dr%5Csin%20%5Ctheta)
This implies that:
![r^2=2r\sin \theta](https://tex.z-dn.net/?f=r%5E2%3D2r%5Csin%20%5Ctheta)
Divide through by r
![r=2\sin \theta](https://tex.z-dn.net/?f=r%3D2%5Csin%20%5Ctheta)
![\boxed{x^2+y^2=2y\to r=2\sin \theta}](https://tex.z-dn.net/?f=%5Cboxed%7Bx%5E2%2By%5E2%3D2y%5Cto%20r%3D2%5Csin%20%5Ctheta%7D)
4. We have ![x=\sqrt{3}y](https://tex.z-dn.net/?f=x%3D%5Csqrt%7B3%7Dy)
We substitute
and ![x=r\cos \theta](https://tex.z-dn.net/?f=x%3Dr%5Ccos%20%5Ctheta)
![r\cos \theta=r\sin \theta\sqrt{3}](https://tex.z-dn.net/?f=r%5Ccos%20%5Ctheta%3Dr%5Csin%20%5Ctheta%5Csqrt%7B3%7D)
This implies that;
![\tan \theta=\frac{\sqrt{3}}{3}](https://tex.z-dn.net/?f=%5Ctan%20%5Ctheta%3D%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B3%7D)
![\theta=\frac{\pi}{6}](https://tex.z-dn.net/?f=%5Ctheta%3D%5Cfrac%7B%5Cpi%7D%7B6%7D)
![\boxed{x=\sqrt{3}y\to \theta=\frac{\pi}{6}}](https://tex.z-dn.net/?f=%5Cboxed%7Bx%3D%5Csqrt%7B3%7Dy%5Cto%20%5Ctheta%3D%5Cfrac%7B%5Cpi%7D%7B6%7D%7D)
5. We have ![x=y](https://tex.z-dn.net/?f=x%3Dy)
We substitute
and ![x=r\cos \theta](https://tex.z-dn.net/?f=x%3Dr%5Ccos%20%5Ctheta)
![r\cos \theta=r\sin \theta](https://tex.z-dn.net/?f=r%5Ccos%20%5Ctheta%3Dr%5Csin%20%5Ctheta)
This implies that;
![\tan \theta=1](https://tex.z-dn.net/?f=%5Ctan%20%5Ctheta%3D1)
![\theta=\frac{\pi}{4}](https://tex.z-dn.net/?f=%5Ctheta%3D%5Cfrac%7B%5Cpi%7D%7B4%7D)
![\boxed{x=y\to \theta=\frac{\pi}{4}}](https://tex.z-dn.net/?f=%5Cboxed%7Bx%3Dy%5Cto%20%5Ctheta%3D%5Cfrac%7B%5Cpi%7D%7B4%7D%7D)
Ninty three point six five I think
Hi There!
Answer:
25x² - 4
Step-by-step explanation:
Solve: (5x - 2)²
Step 1: 25x² - 2²
Step 2: 25x² - 4
Hope This Helps :)
Answer:
The first answer is 3, and the second answer is groups of 3 allow 4 rows of 9 students.
Step-by-step explanation:
Not sure I guessed on the question and got it right, just trying to help people out.
Solve for x or y or 1
solving for y
divide both sides by x
y=25/x
not a line because x has t ohave exponent of 1, this one is y=25x^-1
ok, solve for something else
solve for 1
xy/25=1
this doesn't fit into the standard forms of any of the conic sections
if we were to subsitute points we would see it is a hyperbola that is diagonal with the x and y axises as assemtots
it is a hyperbola