Solved

Select the Correct Graph of the Polar Equation r=9π2r = \frac { 9 \pi } { 2 }

Question 92

Multiple Choice

Select the correct graph of the polar equation.Describe your viewing window. r=9π2r = \frac { 9 \pi } { 2 }


A) Select the correct graph of the polar equation.Describe your viewing window. r = \frac { 9 \pi } { 2 } A) \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 18 \\ X _ { \max } = 18 \\ X _ { s c l } = 3 \\ Y _ { \min } = - 18 \\ Y _ { \max } = 18 \\ Y _ { s c l } = 3 \end{array} B) \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 18 \\ X _ { \max } = 18 \\ X _ { s c l } = 3 \\ Y _ { \min } = - 18 \\ Y _ { \max } = 18 \\ Y _ { s c l } = 3 \end{array} C) \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 18 \\ X _ { \max } = 18 \\ X _ { s c l } = 3 \\ Y _ { \min } = - 18 \\ Y _ { \max } = 18 \\ Y _ { s c l } = 3 \end{array} D) \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 18 \\ X _ { \max } = 18 \\ X _ { s c l } = 3 \\ Y _ { \min } = - 18 \\ Y _ { \max } = 18 \\ Y _ { s c l } = 3 \end{array} E) \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 18 \\ X _ { \max } = 18 \\ X _ { s c l } = 3 \\ Y _ { \min } = - 18 \\ Y _ { \max } = 18 \\ Y _ { s c l } = 3 \end{array} θmin⁡=0θmax⁡=2πθstep=π/24Xmin⁡=−18Xmax⁡=18Xscl=3Ymin⁡=−18Ymax⁡=18Yscl=3\begin{array} { l } \theta _ { \min } = 0 \\\theta _ { \max } = 2 \pi \\\theta _ { s t e p } = \pi / 24 \\X _ { \min } = - 18 \\X _ { \max } = 18 \\X _ { s c l } = 3 \\Y _ { \min } = - 18 \\Y _ { \max } = 18 \\Y _ { s c l } = 3\end{array}
B) Select the correct graph of the polar equation.Describe your viewing window. r = \frac { 9 \pi } { 2 } A) \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 18 \\ X _ { \max } = 18 \\ X _ { s c l } = 3 \\ Y _ { \min } = - 18 \\ Y _ { \max } = 18 \\ Y _ { s c l } = 3 \end{array} B) \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 18 \\ X _ { \max } = 18 \\ X _ { s c l } = 3 \\ Y _ { \min } = - 18 \\ Y _ { \max } = 18 \\ Y _ { s c l } = 3 \end{array} C) \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 18 \\ X _ { \max } = 18 \\ X _ { s c l } = 3 \\ Y _ { \min } = - 18 \\ Y _ { \max } = 18 \\ Y _ { s c l } = 3 \end{array} D) \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 18 \\ X _ { \max } = 18 \\ X _ { s c l } = 3 \\ Y _ { \min } = - 18 \\ Y _ { \max } = 18 \\ Y _ { s c l } = 3 \end{array} E) \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 18 \\ X _ { \max } = 18 \\ X _ { s c l } = 3 \\ Y _ { \min } = - 18 \\ Y _ { \max } = 18 \\ Y _ { s c l } = 3 \end{array} θmin⁡=0θmax⁡=2πθstep=π/24Xmin⁡=−18Xmax⁡=18Xscl=3Ymin⁡=−18Ymax⁡=18Yscl=3\begin{array} { l } \theta _ { \min } = 0 \\\theta _ { \max } = 2 \pi \\\theta _ { s t e p } = \pi / 24 \\X _ { \min } = - 18 \\X _ { \max } = 18 \\X _ { s c l } = 3 \\Y _ { \min } = - 18 \\Y _ { \max } = 18 \\Y _ { s c l } = 3\end{array}
C) Select the correct graph of the polar equation.Describe your viewing window. r = \frac { 9 \pi } { 2 } A) \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 18 \\ X _ { \max } = 18 \\ X _ { s c l } = 3 \\ Y _ { \min } = - 18 \\ Y _ { \max } = 18 \\ Y _ { s c l } = 3 \end{array} B) \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 18 \\ X _ { \max } = 18 \\ X _ { s c l } = 3 \\ Y _ { \min } = - 18 \\ Y _ { \max } = 18 \\ Y _ { s c l } = 3 \end{array} C) \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 18 \\ X _ { \max } = 18 \\ X _ { s c l } = 3 \\ Y _ { \min } = - 18 \\ Y _ { \max } = 18 \\ Y _ { s c l } = 3 \end{array} D) \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 18 \\ X _ { \max } = 18 \\ X _ { s c l } = 3 \\ Y _ { \min } = - 18 \\ Y _ { \max } = 18 \\ Y _ { s c l } = 3 \end{array} E) \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 18 \\ X _ { \max } = 18 \\ X _ { s c l } = 3 \\ Y _ { \min } = - 18 \\ Y _ { \max } = 18 \\ Y _ { s c l } = 3 \end{array} θmin⁡=0θmax⁡=2πθstep=π/24Xmin⁡=−18Xmax⁡=18Xscl=3Ymin⁡=−18Ymax⁡=18Yscl=3\begin{array} { l } \theta _ { \min } = 0 \\\theta _ { \max } = 2 \pi \\\theta _ { s t e p } = \pi / 24 \\X _ { \min } = - 18 \\X _ { \max } = 18 \\X _ { s c l } = 3 \\Y _ { \min } = - 18 \\Y _ { \max } = 18 \\Y _ { s c l } = 3\end{array}
D) Select the correct graph of the polar equation.Describe your viewing window. r = \frac { 9 \pi } { 2 } A) \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 18 \\ X _ { \max } = 18 \\ X _ { s c l } = 3 \\ Y _ { \min } = - 18 \\ Y _ { \max } = 18 \\ Y _ { s c l } = 3 \end{array} B) \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 18 \\ X _ { \max } = 18 \\ X _ { s c l } = 3 \\ Y _ { \min } = - 18 \\ Y _ { \max } = 18 \\ Y _ { s c l } = 3 \end{array} C) \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 18 \\ X _ { \max } = 18 \\ X _ { s c l } = 3 \\ Y _ { \min } = - 18 \\ Y _ { \max } = 18 \\ Y _ { s c l } = 3 \end{array} D) \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 18 \\ X _ { \max } = 18 \\ X _ { s c l } = 3 \\ Y _ { \min } = - 18 \\ Y _ { \max } = 18 \\ Y _ { s c l } = 3 \end{array} E) \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 18 \\ X _ { \max } = 18 \\ X _ { s c l } = 3 \\ Y _ { \min } = - 18 \\ Y _ { \max } = 18 \\ Y _ { s c l } = 3 \end{array} θmin⁡=0θmax⁡=2πθstep=π/24Xmin⁡=−18Xmax⁡=18Xscl=3Ymin⁡=−18Ymax⁡=18Yscl=3\begin{array} { l } \theta _ { \min } = 0 \\\theta _ { \max } = 2 \pi \\\theta _ { s t e p } = \pi / 24 \\X _ { \min } = - 18 \\X _ { \max } = 18 \\X _ { s c l } = 3 \\Y _ { \min } = - 18 \\Y _ { \max } = 18 \\Y _ { s c l } = 3\end{array}
E) Select the correct graph of the polar equation.Describe your viewing window. r = \frac { 9 \pi } { 2 } A) \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 18 \\ X _ { \max } = 18 \\ X _ { s c l } = 3 \\ Y _ { \min } = - 18 \\ Y _ { \max } = 18 \\ Y _ { s c l } = 3 \end{array} B) \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 18 \\ X _ { \max } = 18 \\ X _ { s c l } = 3 \\ Y _ { \min } = - 18 \\ Y _ { \max } = 18 \\ Y _ { s c l } = 3 \end{array} C) \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 18 \\ X _ { \max } = 18 \\ X _ { s c l } = 3 \\ Y _ { \min } = - 18 \\ Y _ { \max } = 18 \\ Y _ { s c l } = 3 \end{array} D) \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 18 \\ X _ { \max } = 18 \\ X _ { s c l } = 3 \\ Y _ { \min } = - 18 \\ Y _ { \max } = 18 \\ Y _ { s c l } = 3 \end{array} E) \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 18 \\ X _ { \max } = 18 \\ X _ { s c l } = 3 \\ Y _ { \min } = - 18 \\ Y _ { \max } = 18 \\ Y _ { s c l } = 3 \end{array} θmin⁡=0θmax⁡=2πθstep=π/24Xmin⁡=−18Xmax⁡=18Xscl=3Ymin⁡=−18Ymax⁡=18Yscl=3\begin{array} { l } \theta _ { \min } = 0 \\\theta _ { \max } = 2 \pi \\\theta _ { s t e p } = \pi / 24 \\X _ { \min } = - 18 \\X _ { \max } = 18 \\X _ { s c l } = 3 \\Y _ { \min } = - 18 \\Y _ { \max } = 18 \\Y _ { s c l } = 3\end{array}

Correct Answer:

verifed

Verified

Unlock this answer now
Get Access to more Verified Answers free of charge

Related Questions

Q87: Select the correct graph of the

Q88: A point (a,b) shown in below

Q89: Consider the equation

Q90: Select the correct graph of the

Q91: Select the correct graph of the

Q93: Select the correct graph of the

Q94: Select the correct graph of the

Q95: Select the correct graph of the

Q96: Select the correct graph of the

Q97: Select the graph of the equation.

Unlock this Answer For Free Now!

View this answer and more for free by performing one of the following actions

qr-code

Scan the QR code to install the App and get 2 free unlocks

upload documents

Unlock quizzes for free by uploading documents