Question 1

Select the graph of the polar equation using symmetry, zeros, maximum r-values, and any other additional points.
  r=5cos⁡2θr = 5 \cos 2 	hetar=5cos2θ

Accepted Answer

A)  Symmetric with respect to the polar axis  ∣r∣=5 when θ=0,π2,π,3π2r=0 when θ=π4,3π4,5π4,7π4\begin{array} { c } | r | = 5 	ext { when } 	heta = 0 , \frac { \pi } { 2 } , \pi , \frac { 3 \pi } { 2 } \r = 0 	ext { when } 	heta = \frac { \pi } { 4 } , \frac { 3 \pi } { 4 } , \frac { 5 \pi } { 4 } , \frac { 7 \pi } { 4 }\end{array}∣r∣=5 when θ=0,2π​,π,23π​r=0 when θ=4π​,43π​,45π​,47π​​  
B)  Symmetric with respect to the polar axis  ∣r∣=5 when θ=0,π2,π,3π2r=0 when θ=π4,3π4,5π4,7π4\begin{array} { c } | r | = 5 	ext { when } 	heta = 0 , \frac { \pi } { 2 } , \pi , \frac { 3 \pi } { 2 } \r = 0 	ext { when } 	heta = \frac { \pi } { 4 } , \frac { 3 \pi } { 4 } , \frac { 5 \pi } { 4 } , \frac { 7 \pi } { 4 }\end{array}∣r∣=5 when θ=0,2π​,π,23π​r=0 when θ=4π​,43π​,45π​,47π​​  
C)  Symmetric with respect to the polar axis  ∣r∣=5 when θ=0,π2,π,3π2r=0 when θ=π4,3π4,5π4,7π4\begin{array} { c } | r | = 5 	ext { when } 	heta = 0 , \frac { \pi } { 2 } , \pi , \frac { 3 \pi } { 2 } \r = 0 	ext { when } 	heta = \frac { \pi } { 4 } , \frac { 3 \pi } { 4 } , \frac { 5 \pi } { 4 } , \frac { 7 \pi } { 4 }\end{array}∣r∣=5 when θ=0,2π​,π,23π​r=0 when θ=4π​,43π​,45π​,47π​​   
D)  Symmetric with respect to the polar axis  ∣r∣=5 when θ=0,π2,π,3π2r=0 when θ=π4,3π4,5π4,7π4\begin{array} { c } | r | = 5 	ext { when } 	heta = 0 , \frac { \pi } { 2 } , \pi , \frac { 3 \pi } { 2 } \r = 0 	ext { when } 	heta = \frac { \pi } { 4 } , \frac { 3 \pi } { 4 } , \frac { 5 \pi } { 4 } , \frac { 7 \pi } { 4 }\end{array}∣r∣=5 when θ=0,2π​,π,23π​r=0 when θ=4π​,43π​,45π​,47π​​   
E)  Symmetric with respect to the polar axis  ∣r∣=5 when θ=0,π2,π,3π2r=0 when θ=π4,3π4,5π4,7π4\begin{array} { c } | r | = 5 	ext { when } 	heta = 0 , \frac { \pi } { 2 } , \pi , \frac { 3 \pi } { 2 } \r = 0 	ext { when } 	heta = \frac { \pi } { 4 } , \frac { 3 \pi } { 4 } , \frac { 5 \pi } { 4 } , \frac { 7 \pi } { 4 }\end{array}∣r∣=5 when θ=0,2π​,π,23π​r=0 when θ=4π​,43π​,45π​,47π​​

Question 2

Find the standard form of the equation of the hyperbola with the given characteristics and center at the origin. 
Vertices: (0,±2) ; asymptotes: y = ±  32\frac { 3 } { 2 }23​  x

Accepted Answer

A)    y29+x24=1\frac { y ^ { 2 } } { 9 } + \frac { x ^ { 2 } } { 4 } = 19y2​+4x2​=1 
B)    y24−x29=1\frac { y ^ { 2 } } { 4 } - \frac { x ^ { 2 } } { 9 } = 14y2​−9x2​=1 
C)    y29−x24=−1\frac { y ^ { 2 } } { 9 } - \frac { x ^ { 2 } } { 4 } = - 19y2​−4x2​=−1 
D)    x24+y29=1\frac { x ^ { 2 } } { 4 } + \frac { y ^ { 2 } } { 9 } = 14x2​+9y2​=1 
E)    y29−x24=1\frac { y ^ { 2 } } { 9 } - \frac { x ^ { 2 } } { 4 } = 19y2​−4x2​=1

Question 3

Find the standard form of the equation of the hyperbola with the given characteristics and center at the origin. 
Vertices:  (±4,0) ( \pm 4,0 ) (±4,0)   ; foci:  (±8,0) ( \pm 8,0 ) (±8,0)

Accepted Answer

A)    y248−x216=1\frac { y ^ { 2 } } { 48 } - \frac { x ^ { 2 } } { 16 } = 148y2​−16x2​=1 
B)    y216−x248=1\frac { y ^ { 2 } } { 16 } - \frac { x ^ { 2 } } { 48 } = 116y2​−48x2​=1 
C)    x216−y248=1\frac { x ^ { 2 } } { 16 } - \frac { y ^ { 2 } } { 48 } = 116x2​−48y2​=1 
D)    y216+x248=1\frac { y ^ { 2 } } { 16 } + \frac { x ^ { 2 } } { 48 } = 116y2​+48x2​=1 
E)    x216+y248=1\frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 48 } = 116x2​+48y2​=1

Question 4

Find the center and vertices of the hyperbola.  12x2−4y2−48x+32y−64=012 x ^ { 2 } - 4 y ^ { 2 } - 48 x + 32 y - 64 = 012x2−4y2−48x+32y−64=0

Accepted Answer

A)  center: (-2, -4) , vertices: (-2, -6) , (-2, -2) 
B)  center: (2, 4) , vertices: (2, 2) , (2, 6) 
C)  center: (2, 4) , vertices: (0, 4) , (4, 4) 
D)  center: (-2, -4) , vertices: (-4, -4) , (0, -4) 
E)  center: (-4, -2) , vertices: (-6, -2) , (-2, -2)

Question 5

Find the vertex and directrix of the parabola.  x2−14x−8y+73=0x ^ { 2 } - 14 x - 8 y + 73 = 0x2−14x−8y+73=0

Accepted Answer

A)  vertex:  (7,3) ( 7,3 ) (7,3)   directrix:  y=5y = 5y=5 
B)  vertex:  (−7,−3) ( - 7 , - 3 ) (−7,−3)   directrix:  y=5y = 5y=5 
C)  vertex:  (7,3) ( 7,3 ) (7,3)   directrix:  y=1y = 1y=1 
D)  vertex:  (−7,−3) ( - 7 , - 3 ) (−7,−3)   directrix:  y=−5y = - 5y=−5 
E)  vertex:  (−7,−3) ( - 7 , - 3 ) (−7,−3)   directrix:  y=9y = 9y=9

Question 6

Find the vertex and focus of the parabola.  y2=−17xy ^ { 2 } = - \frac { 1 } { 7 } xy2=−71​x

Accepted Answer

A)  vertex:  (0,−54) \left( 0 , - \frac { 5 } { 4 } \right) (0,−45​)   focus:  (0,−128) \left( 0 , - \frac { 1 } { 28 } \right) (0,−281​)  
B)  vertex:  (0,54) \left( 0 , \frac { 5 } { 4 } \right) (0,45​)   focus:  (−17,−17) \left( - \frac { 1 } { 7 } , - \frac { 1 } { 7 } \right) (−71​,−71​)  
C)  vertex: (0, 0)  focus:  (−128,0) \left( - \frac { 1 } { 28 } , 0 \right) (−281​,0)  
D)  vertex: (0, 0)  focus:  (−17,0) \left( - \frac { 1 } { 7 } , 0 \right) (−71​,0)  
E)  vertex: (0, 0)  focus:  (0,−17) \left( 0 , - \frac { 1 } { 7 } \right) (0,−71​)

Question 7

Sketch the graph of the ellipse, using the lateral recta.   x24+y216=1\frac { x ^ { 2 } } { 4 } + \frac { y ^ { 2 } } { 16 } = 14x2​+16y2​=1

Accepted Answer

A)   
B)   
C)   
D)   
E)

Question 8

Eliminate the parameter and write the corresponding rectangular equation whose graph represents the curve.   x=4cos⁡θy=2sin⁡θ\begin{array} { l } x = 4 \cos 	heta \y = 2 \sin 	heta\end{array}x=4cosθy=2sinθ​

Accepted Answer

A)    x24+y216=1\frac { x ^ { 2 } } { 4 } + \frac { y ^ { 2 } } { 16 } = 14x2​+16y2​=1 
B)    x216+y24=1\frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 4 } = 116x2​+4y2​=1 
C)    y=x4y = \frac { x } { 4 }y=4x​ 
D)    y=x2y = \frac { x } { 2 }y=2x​ 
E)    x216−y24=1\frac { x ^ { 2 } } { 16 } - \frac { y ^ { 2 } } { 4 } = 116x2​−4y2​=1

Question 9

Find the angle  θ	hetaθ  (in radians and degrees)  between the lines.Round your answer to four decimal places for radians and round your answer to one decimal places for degree.   x−y=05x−4y=−3\begin{array} { l } x - y = 0 \5 x - 4 y = - 3\end{array}x−y=05x−4y=−3​

Accepted Answer

A)    Θ≈0.1107 radian ≈7.3∘\Theta \approx 0.1107 	ext { radian } \approx 7.3 ^ { \circ }Θ≈0.1107 radian ≈7.3∘ 
B)    Θ≈0.1107 radian ≈5.3∘\Theta \approx 0.1107 	ext { radian } \approx 5.3 ^ { \circ }Θ≈0.1107 radian ≈5.3∘ 
C)    Θ≈0.1107 radian ≈6.3∘\Theta \approx 0.1107 	ext { radian } \approx 6.3 ^ { \circ }Θ≈0.1107 radian ≈6.3∘ 
D)    Θ≈0.1107 radian ≈8.3∘\Theta \approx 0.1107 	ext { radian } \approx 8.3 ^ { \circ }Θ≈0.1107 radian ≈8.3∘ 
E)    Θ≈0.1107 radian ≈4.3∘\Theta \approx 0.1107 	ext { radian } \approx 4.3 ^ { \circ }Θ≈0.1107 radian ≈4.3∘

Question 10

Select the graph of the polar equation using symmetry, zeros, maximum r-values, and any other additional points.
  r=3π7r = \frac { 3 \pi } { 7 }r=73π​

Accepted Answer

A)  Symmetric with respect to  θ=π2	heta = \frac { \pi } { 2 }θ=2π​  , polar axis, poleCircle with radius  3π7\frac { 3 \pi } { 7 }73π​   
B)  Symmetric with respect to  θ=π2	heta = \frac { \pi } { 2 }θ=2π​  , polar axis, poleCircle with radius  3π7\frac { 3 \pi } { 7 }73π​   
C)  Symmetric with respect to  θ=π2	heta = \frac { \pi } { 2 }θ=2π​  , polar axis, poleCircle with radius  3π7\frac { 3 \pi } { 7 }73π​   
D)  Symmetric with respect to  θ=π2	heta = \frac { \pi } { 2 }θ=2π​  , polar axis, poleCircle with radius  3π7\frac { 3 \pi } { 7 }73π​   
E)  Symmetric with respect to  θ=π2	heta = \frac { \pi } { 2 }θ=2π​  , polar axis, poleCircle with radius  3π7\frac { 3 \pi } { 7 }73π​

Question 11

The  x′y′x ^ { \prime } y ^ { \prime }x′y′  -coordinate system has been rotated  θ	hetaθ  degrees from the  xyx yxy  -coordinate system.The coordinates of a point in the  xyx yxy  -coordinate system are given.Find the coordinates of the point in the rotated coordinate system.   θ=30∘	heta = 30 ^ { \circ }θ=30∘  ,  (2,6) ( 2,6 ) (2,6)

Accepted Answer

A)    (3+3,33+1) ( \sqrt { 3 } + 3,3 \sqrt { 3 } + 1 ) (3c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​+3,33c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​+1)  
B)    (3+3,33−1) ( \sqrt { 3 } + 3,3 \sqrt { 3 } - 1 ) (3c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​+3,33c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​−1)  
C)    (3−32,33+12) \left( \frac { 3 - \sqrt { 3 } } { 2 } , \frac { 3 \sqrt { 3 } + 1 } { 2 } \right) (23−3c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​​,233c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​+1​)  
D)    (3−3,33−1) ( \sqrt { 3 } - 3,3 \sqrt { 3 } - 1 ) (3c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​−3,33c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​−1)  
E)    (3−32,33+32) \left( \frac { 3 - \sqrt { 3 } } { 2 } , \frac { 3 \sqrt { 3 } + 3 } { 2 } \right) (23−3c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​​,233c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​+3​)

Question 12

A point in rectangular coordinates is given.Convert the point to polar coordinates, r > 0.   (−7,7) ( - 7,7 ) (−7,7)

Accepted Answer

A)  Polar coordinates:  (−7,π4) \left( - \sqrt { 7 } , \frac { \pi } { 4 } \right) (−7c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​,4π​)  
B)  Polar coordinates:  (98,3π4) \left( \sqrt { 98 } , \frac { 3 \pi } { 4 } \right) (98c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​,43π​)  
C)  Polar coordinates:  (98,−3π4) \left( \sqrt { 98 } , - \frac { 3 \pi } { 4 } \right) (98c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​,−43π​)  
D)  Polar coordinates:  (−98,3π4) \left( - \sqrt { 98 } , \frac { 3 \pi } { 4 } \right) (−98c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​,43π​)  
E)  Polar coordinates:  (7,π4) \left( \sqrt { 7 } , \frac { \pi } { 4 } \right) (7c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​,4π​)

Question 13

Find the angle  θ	hetaθ  (in radians and degrees)  between the lines.Round your answer to four decimal places for radians and round your answer to one decimal places for degree.  12x+6y=183x−2y=−1\begin{array} { l } 12 x + 6 y = 18 \3 x - 2 y = - 1\end{array}12x+6y=183x−2y=−1​

Accepted Answer

A)    Θ≈1.1071 radians ≈64.4∘\Theta \approx 1.1071 	ext { radians } \approx 64.4 ^ { \circ }Θ≈1.1071 radians ≈64.4∘ 
B)    θ≈1.1071 radians ≈62.4∘	heta \approx 1.1071 	ext { radians } \approx 62.4 ^ { \circ }θ≈1.1071 radians ≈62.4∘ 
C)    θ≈1.1071 radians ≈65.4∘	heta \approx 1.1071 	ext { radians } \approx 65.4 ^ { \circ }θ≈1.1071 radians ≈65.4∘ 
D)    θ≈1.1071 radians ≈63.4∘	heta \approx 1.1071 	ext { radians } \approx 63.4 ^ { \circ }θ≈1.1071 radians ≈63.4∘ 
E)    θ≈1.1071 radians ≈61.4∘	heta \approx 1.1071 	ext { radians } \approx 61.4 ^ { \circ }θ≈1.1071 radians ≈61.4∘

Question 14

Select correct graph to graph rotated conic.   r=66+sin⁡(θ−π/3) r = \frac { 6 } { 6 + \sin ( 	heta - \pi / 3 )  }r=6+sin(θ−π/3) 6​

Accepted Answer

A)  
B)   
C)  
D)   
E)

Question 15

Select the graph of the equation as a circle, a parabola, an ellipse, or a hyperbola.   22x2−13x−200y−112=022 x ^ { 2 } - 13 x - 200 y - 112 = 022x2−13x−200y−112=0

Accepted Answer

A)  Hyperbola
B)  Circle
C)  Parabola
D)  Ellipse
E)  None of the above

Question 16

Use the Quadratic Formula to solve for  yyy  .   x2−12xy−10y2−22=0x ^ { 2 } - 12 x y - 10 y ^ { 2 } - 22 = 0x2−12xy−10y2−22=0

Accepted Answer

A)    y=12x±144x2+40(x2+22) −20y = \frac { 12 x \pm \sqrt { 144 x ^ { 2 } + 40 \left( x ^ { 2 } + 22 \right)  } } { - 20 }y=−2012x±144x2+40(x2+22) c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​​ 
B)    y=12x±144x2+40(x2−22) −20y = \frac { 12 x \pm \sqrt { 144 x ^ { 2 } + 40 \left( x ^ { 2 } - 22 \right)  } } { - 20 }y=−2012x±144x2+40(x2−22) c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​​ 
C)    y=12x±144x2+40(x2−22) 20y = \frac { 12 x \pm \sqrt { 144 x ^ { 2 } + 40 \left( x ^ { 2 } - 22 \right)  } } { 20 }y=2012x±144x2+40(x2−22) c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​​ 
D)    y=12x±144x+40(x2−22) −20y = \frac { 12 x \pm \sqrt { 144 x + 40 \left( x ^ { 2 } - 22 \right)  } } { - 20 }y=−2012x±144x+40(x2−22) c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​​ 
E)    y=12x±144x2+40(x2+22) 20y = \frac { 12 x \pm \sqrt { 144 x ^ { 2 } + 40 \left( x ^ { 2 } + 22 \right)  } } { 20 }y=2012x±144x2+40(x2+22) c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​​

Question 17

By using a graphing utility select the correct graph of the polar equation.Identify the graph.   −32+8sin⁡θ\frac { - 3 } { 2 + 8 \sin 	heta }2+8sinθ−3​

Accepted Answer

A)    1 \Rightarrow  Hyperbola 
B)      e = 4 > 1 \Rightarrow  Hyperbola 
C)      e = 4 > 1 \Rightarrow  Hyperbola 
D)     e = 4 > 1 \Rightarrow  Hyperbola 
E)      e = 4 > 1 \Rightarrow  Hyperbola " class="answers-bank-image d-inline" rel="preload" >   1 \Rightarrow">e=4>1⇒e = 4 > 1 \Rightarrowe=4>1⇒  Hyperbola
B)    1 \Rightarrow  Hyperbola 
B)      e = 4 > 1 \Rightarrow  Hyperbola 
C)      e = 4 > 1 \Rightarrow  Hyperbola 
D)     e = 4 > 1 \Rightarrow  Hyperbola 
E)      e = 4 > 1 \Rightarrow  Hyperbola " class="answers-bank-image d-inline" rel="preload" >   1 \Rightarrow">e=4>1⇒e = 4 > 1 \Rightarrowe=4>1⇒  Hyperbola
C)    1 \Rightarrow  Hyperbola 
B)      e = 4 > 1 \Rightarrow  Hyperbola 
C)      e = 4 > 1 \Rightarrow  Hyperbola 
D)     e = 4 > 1 \Rightarrow  Hyperbola 
E)      e = 4 > 1 \Rightarrow  Hyperbola " class="answers-bank-image d-inline" rel="preload" >   1 \Rightarrow">e=4>1⇒e = 4 > 1 \Rightarrowe=4>1⇒  Hyperbola
D)   1 \Rightarrow  Hyperbola 
B)      e = 4 > 1 \Rightarrow  Hyperbola 
C)      e = 4 > 1 \Rightarrow  Hyperbola 
D)     e = 4 > 1 \Rightarrow  Hyperbola 
E)      e = 4 > 1 \Rightarrow  Hyperbola " class="answers-bank-image d-inline" rel="preload" >   1 \Rightarrow">e=4>1⇒e = 4 > 1 \Rightarrowe=4>1⇒  Hyperbola
E)    1 \Rightarrow  Hyperbola 
B)      e = 4 > 1 \Rightarrow  Hyperbola 
C)      e = 4 > 1 \Rightarrow  Hyperbola 
D)     e = 4 > 1 \Rightarrow  Hyperbola 
E)      e = 4 > 1 \Rightarrow  Hyperbola " class="answers-bank-image d-inline" rel="preload" >   1 \Rightarrow">e=4>1⇒e = 4 > 1 \Rightarrowe=4>1⇒  Hyperbola

Question 18

Find the center, vertices and foci of the hyperbola.   x216−y225=1\frac { x ^ { 2 } } { 16 } - \frac { y ^ { 2 } } { 25 } = 116x2​−25y2​=1

Accepted Answer

A)  Center: (0, 0) Vertices:  (4,0) ( 4,0 ) (4,0)   Foci:  (±41,0) ( \pm \sqrt { 41 } , 0 ) (±41c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​,0)  
B)  Center: (0, 0) Vertices:  (−4,0) ( - 4,0 ) (−4,0)   Foci:  (−41,0) ( - \sqrt { 41 } , 0 ) (−41c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​,0)  
C)  Center: (0, 0) Vertices:  (4,0) ( 4,0 ) (4,0)   Foci:  (−41,0) ( - \sqrt { 41 } , 0 ) (−41c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​,0)  
D)  Center: (0, 0) Vertices:  (4,0) ( 4,0 ) (4,0)   Foci:  (41,0) ( \sqrt { 41 } , 0 ) (41c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​,0)  
E)  Center: (0, 0) Vertices:  (±4,0) ( \pm 4,0 ) (±4,0)   Foci:  (±41,0) ( \pm \sqrt { 41 } , 0 ) (±41c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​,0)

Question 19

Graph the hyperbola.   9y2−16x2+36y+64x=1729 y ^ { 2 } - 16 x ^ { 2 } + 36 y + 64 x = 1729y2−16x2+36y+64x=172

Accepted Answer

A)   
B)   
C)   
D)

Question 20

Find the center and vertices of the hyperbola and sketch its graph, using asymptotes as sketching aids.   x29−y225=1\frac { x ^ { 2 } } { 9 } - \frac { y ^ { 2 } } { 25 } = 19x2​−25y2​=1

Accepted Answer

A)  Center: (0,0) Vertices: (-5,0)  
B)  Center: (0,0) Vertices: (-3,0)  
C)  Center: (0,0) Vertices: (3,0)  
D)  Center: (0,0) Vertices: (±5,0)  
E)  Center: (0,0) Vertices: (±3,0)

Exam 10: Topics In Analytic Geometry

A point in rectangular coordinates is given.Convert the point to polar coordinates, r > 0. $( - 7,7 )$

Correct Answer
verified

Find the standard form of the equation of the hyperbola with the given characteristics and center at the origin. Vertices: (0,±2) ; asymptotes: y = ± $\frac { 3 } { 2 }$ x

Correct Answer
verified

Select the graph of the equation as a circle, a parabola, an ellipse, or a hyperbola. $22 x ^ { 2 } - 13 x - 200 y - 112 = 0$

Correct Answer
verified

Find the center and vertices of the hyperbola and sketch its graph, using asymptotes as sketching aids. $\frac { x ^ { 2 } } { 9 } - \frac { y ^ { 2 } } { 25 } = 1$

Correct Answer
verified

Select the graph of the polar equation using symmetry, zeros, maximum r-values, and any other additional points. $r = \frac { 3 \pi } { 7 }$

Correct Answer
verified

Find the vertex and directrix of the parabola. $x ^ { 2 } - 14 x - 8 y + 73 = 0$

Correct Answer
verified

Find the center and vertices of the hyperbola. $12 x ^ { 2 } - 4 y ^ { 2 } - 48 x + 32 y - 64 = 0$

Correct Answer
verified

Select correct graph to graph rotated conic. $r = \frac { 6 } { 6 + \sin ( \theta - \pi / 3 ) }$

Correct Answer
verified

The $x ^ { \prime } y ^ { \prime }$ -coordinate system has been rotated $\theta$ degrees from the $x y$ -coordinate system.The coordinates of a point in the $x y$ -coordinate system are given.Find the coordinates of the point in the rotated coordinate system. $\theta = 30 ^ { \circ }$ , $( 2,6 )$

Correct Answer
verified

Use the Quadratic Formula to solve for $y$ . $x ^ { 2 } - 12 x y - 10 y ^ { 2 } - 22 = 0$

Correct Answer
verified

Find the vertex and focus of the parabola. $y ^ { 2 } = - \frac { 1 } { 7 } x$

Correct Answer
verified

Eliminate the parameter and write the corresponding rectangular equation whose graph represents the curve. $\begin{array} { l } x = 4 \cos \theta \\y = 2 \sin \theta\end{array}$

Correct Answer
verified

Graph the hyperbola. $9 y ^ { 2 } - 16 x ^ { 2 } + 36 y + 64 x = 172$

Correct Answer
verified

Sketch the graph of the ellipse, using the lateral recta. $\frac { x ^ { 2 } } { 4 } + \frac { y ^ { 2 } } { 16 } = 1$

Correct Answer
verified

Select the graph of the polar equation using symmetry, zeros, maximum r-values, and any other additional points. $r = 5 \cos 2 \theta$

Correct Answer
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Find the angle $\theta$ (in radians and degrees) between the lines.Round your answer to four decimal places for radians and round your answer to one decimal places for degree. $\begin{array} { l } 12 x + 6 y = 18 \\3 x - 2 y = - 1\end{array}$

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Find the center, vertices and foci of the hyperbola. $\frac { x ^ { 2 } } { 16 } - \frac { y ^ { 2 } } { 25 } = 1$

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Find the standard form of the equation of the hyperbola with the given characteristics and center at the origin. Vertices: $( \pm 4,0 )$ ; foci: $( \pm 8,0 )$

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By using a graphing utility select the correct graph of the polar equation.Identify the graph. $\frac { - 3 } { 2 + 8 \sin \theta }$

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Exam 10: Topics In Analytic Geometry

A point in rectangular coordinates is given.Convert the point to polar coordinates, r > 0. (−7,7) ( - 7,7 ) (−7,7)

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Find the standard form of the equation of the hyperbola with the given characteristics and center at the origin. Vertices: (0,±2) ; asymptotes: y = ± 32\frac { 3 } { 2 }23​ x

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Select the graph of the equation as a circle, a parabola, an ellipse, or a hyperbola. 22x2−13x−200y−112=022 x ^ { 2 } - 13 x - 200 y - 112 = 022x2−13x−200y−112=0

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Find the center and vertices of the hyperbola and sketch its graph, using asymptotes as sketching aids. x29−y225=1\frac { x ^ { 2 } } { 9 } - \frac { y ^ { 2 } } { 25 } = 19x2​−25y2​=1

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Select the graph of the polar equation using symmetry, zeros, maximum r-values, and any other additional points. r=3π7r = \frac { 3 \pi } { 7 }r=73π​

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Find the vertex and directrix of the parabola. x2−14x−8y+73=0x ^ { 2 } - 14 x - 8 y + 73 = 0x2−14x−8y+73=0

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Find the center and vertices of the hyperbola. 12x2−4y2−48x+32y−64=012 x ^ { 2 } - 4 y ^ { 2 } - 48 x + 32 y - 64 = 012x2−4y2−48x+32y−64=0

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Select correct graph to graph rotated conic. r=66+sin⁡(θ−π/3) r = \frac { 6 } { 6 + \sin ( \theta - \pi / 3 ) }r=6+sin(θ−π/3) 6​

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Use the Quadratic Formula to solve for yyy . x2−12xy−10y2−22=0x ^ { 2 } - 12 x y - 10 y ^ { 2 } - 22 = 0x2−12xy−10y2−22=0

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Find the vertex and focus of the parabola. y2=−17xy ^ { 2 } = - \frac { 1 } { 7 } xy2=−71​x

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Eliminate the parameter and write the corresponding rectangular equation whose graph represents the curve. x=4cos⁡θy=2sin⁡θ\begin{array} { l } x = 4 \cos \theta \\y = 2 \sin \theta\end{array}x=4cosθy=2sinθ​

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Graph the hyperbola. 9y2−16x2+36y+64x=1729 y ^ { 2 } - 16 x ^ { 2 } + 36 y + 64 x = 1729y2−16x2+36y+64x=172

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Sketch the graph of the ellipse, using the lateral recta. x24+y216=1\frac { x ^ { 2 } } { 4 } + \frac { y ^ { 2 } } { 16 } = 14x2​+16y2​=1

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Select the graph of the polar equation using symmetry, zeros, maximum r-values, and any other additional points. r=5cos⁡2θr = 5 \cos 2 \thetar=5cos2θ

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Find the angle θ\thetaθ (in radians and degrees) between the lines.Round your answer to four decimal places for radians and round your answer to one decimal places for degree. 12x+6y=183x−2y=−1\begin{array} { l } 12 x + 6 y = 18 \\3 x - 2 y = - 1\end{array}12x+6y=183x−2y=−1​

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Find the center, vertices and foci of the hyperbola. x216−y225=1\frac { x ^ { 2 } } { 16 } - \frac { y ^ { 2 } } { 25 } = 116x2​−25y2​=1

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Find the standard form of the equation of the hyperbola with the given characteristics and center at the origin. Vertices: (±4,0) ( \pm 4,0 ) (±4,0) ; foci: (±8,0) ( \pm 8,0 ) (±8,0)

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Find the angle θ\thetaθ (in radians and degrees) between the lines.Round your answer to four decimal places for radians and round your answer to one decimal places for degree. x−y=05x−4y=−3\begin{array} { l } x - y = 0 \\5 x - 4 y = - 3\end{array}x−y=05x−4y=−3​

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By using a graphing utility select the correct graph of the polar equation.Identify the graph. −32+8sin⁡θ\frac { - 3 } { 2 + 8 \sin \theta }2+8sinθ−3​

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A point in rectangular coordinates is given.Convert the point to polar coordinates, r > 0. $( - 7,7 )$

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Find the standard form of the equation of the hyperbola with the given characteristics and center at the origin. Vertices: (0,±2) ; asymptotes: y = ± $\frac { 3 } { 2 }$ x

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Select the graph of the equation as a circle, a parabola, an ellipse, or a hyperbola. $22 x ^ { 2 } - 13 x - 200 y - 112 = 0$

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Find the center and vertices of the hyperbola and sketch its graph, using asymptotes as sketching aids. $\frac { x ^ { 2 } } { 9 } - \frac { y ^ { 2 } } { 25 } = 1$

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Select the graph of the polar equation using symmetry, zeros, maximum r-values, and any other additional points. $r = \frac { 3 \pi } { 7 }$

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Find the vertex and directrix of the parabola. $x ^ { 2 } - 14 x - 8 y + 73 = 0$

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Find the center and vertices of the hyperbola. $12 x ^ { 2 } - 4 y ^ { 2 } - 48 x + 32 y - 64 = 0$

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Select correct graph to graph rotated conic. $r = \frac { 6 } { 6 + \sin ( \theta - \pi / 3 ) }$

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Use the Quadratic Formula to solve for $y$ . $x ^ { 2 } - 12 x y - 10 y ^ { 2 } - 22 = 0$

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Find the vertex and focus of the parabola. $y ^ { 2 } = - \frac { 1 } { 7 } x$

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Eliminate the parameter and write the corresponding rectangular equation whose graph represents the curve. $\begin{array} { l } x = 4 \cos \theta \\y = 2 \sin \theta\end{array}$

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Graph the hyperbola. $9 y ^ { 2 } - 16 x ^ { 2 } + 36 y + 64 x = 172$

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Sketch the graph of the ellipse, using the lateral recta. $\frac { x ^ { 2 } } { 4 } + \frac { y ^ { 2 } } { 16 } = 1$

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Select the graph of the polar equation using symmetry, zeros, maximum r-values, and any other additional points. $r = 5 \cos 2 \theta$

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Find the angle $\theta$ (in radians and degrees) between the lines.Round your answer to four decimal places for radians and round your answer to one decimal places for degree. $\begin{array} { l } 12 x + 6 y = 18 \\3 x - 2 y = - 1\end{array}$

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Find the center, vertices and foci of the hyperbola. $\frac { x ^ { 2 } } { 16 } - \frac { y ^ { 2 } } { 25 } = 1$

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Find the standard form of the equation of the hyperbola with the given characteristics and center at the origin. Vertices: $( \pm 4,0 )$ ; foci: $( \pm 8,0 )$

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By using a graphing utility select the correct graph of the polar equation.Identify the graph. $\frac { - 3 } { 2 + 8 \sin \theta }$

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