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)
Answer:
the movie is 4 dollars and games are 5
Step-by-step explanation:
The sum of -3 and a number equal to -91 is -88
Answer:
20
Step-by-step explanation:
pilot flew a 400-mile flight in 2.5 hours flying into the wind . Flying the same rate and with the same wind speed, the return trip took only 2hours with a tailwind
The speed can be defined as the ratio distance over time, but in this case the pilot is flying at the speed of the plane as well of speed of the wind. We can denote speed of plane as "s" and speed of the wind as "k"
Then the speed of the plane=(s-k) this is because it is moving in the opposite direction of the wind.
But From the question, the plane distance=400mile and the corresponding time =2.5 h
Then the speed of the plane=(s-k) =400mile/2.5
=160mph
By the time it return from trip, the speed the plane flew with is (s+k) because the wind is at the rear.
From the question, the plane distance=400mile and the corresponding time =2 h
Then (s+k)= (400/2)
(s+k)=200mph