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Select the Correct Graph of the Polar Equation r=π4r = \frac { \pi } { 4 }

Question 95

Multiple Choice

Select the correct graph of the polar equation.Describe your viewing window. r=π4r = \frac { \pi } { 4 }


A) Select the correct graph of the polar equation.Describe your viewing window. r = \frac { \pi } { 4 } A) \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 2 \\ X _ { \max } = 2 \\ X _ { s c l } = 0.5 \\ Y _ { \min } = - 2 \\ Y _ { \max } = 2 \\ Y _ { s c l } = 0.5 \end{array} B) \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 2 \\ X _ { \max } = 2 \\ X _ { s c l } = 0.5 \\ Y _ { \min } = - 2 \\ Y _ { \max } = 2 \\ Y _ { s c l } = 0.5 \end{array} C) \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 2 \\ X _ { \max } = 2 \\ X _ { s c l } = 0.5 \\ Y _ { \min } = - 2 \\ Y _ { \max } = 2 \\ Y _ { s c l } = 0.5 \end{array} D) \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 2 \\ X _ { \max } = 2 \\ X _ { s c l } = 0.5 \\ Y _ { \min } = - 2 \\ Y _ { \max } = 2 \\ Y _ { s c l } = 0.5 \end{array} E) \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 2 \\ X _ { \max } = 2 \\ X _ { s c l } = 0.5 \\ Y _ { \min } = - 2 \\ Y _ { \max } = 2 \\ Y _ { s c l } = 0.5 \end{array} θmin⁡=0θmax⁡=2πθstep=π/24Xmin⁡=−2Xmax⁡=2Xscl=0.5Ymin⁡=−2Ymax⁡=2Yscl=0.5\begin{array} { l } \theta _ { \min } = 0 \\\theta _ { \max } = 2 \pi \\\theta _ { s t e p } = \pi / 24 \\X _ { \min } = - 2 \\X _ { \max } = 2 \\X _ { s c l } = 0.5 \\Y _ { \min } = - 2 \\Y _ { \max } = 2 \\Y _ { s c l } = 0.5\end{array}
B) Select the correct graph of the polar equation.Describe your viewing window. r = \frac { \pi } { 4 } A) \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 2 \\ X _ { \max } = 2 \\ X _ { s c l } = 0.5 \\ Y _ { \min } = - 2 \\ Y _ { \max } = 2 \\ Y _ { s c l } = 0.5 \end{array} B) \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 2 \\ X _ { \max } = 2 \\ X _ { s c l } = 0.5 \\ Y _ { \min } = - 2 \\ Y _ { \max } = 2 \\ Y _ { s c l } = 0.5 \end{array} C) \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 2 \\ X _ { \max } = 2 \\ X _ { s c l } = 0.5 \\ Y _ { \min } = - 2 \\ Y _ { \max } = 2 \\ Y _ { s c l } = 0.5 \end{array} D) \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 2 \\ X _ { \max } = 2 \\ X _ { s c l } = 0.5 \\ Y _ { \min } = - 2 \\ Y _ { \max } = 2 \\ Y _ { s c l } = 0.5 \end{array} E) \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 2 \\ X _ { \max } = 2 \\ X _ { s c l } = 0.5 \\ Y _ { \min } = - 2 \\ Y _ { \max } = 2 \\ Y _ { s c l } = 0.5 \end{array} θmin⁡=0θmax⁡=2πθstep=π/24Xmin⁡=−2Xmax⁡=2Xscl=0.5Ymin⁡=−2Ymax⁡=2Yscl=0.5\begin{array} { l } \theta _ { \min } = 0 \\\theta _ { \max } = 2 \pi \\\theta _ { s t e p } = \pi / 24 \\X _ { \min } = - 2 \\X _ { \max } = 2 \\X _ { s c l } = 0.5 \\Y _ { \min } = - 2 \\Y _ { \max } = 2 \\Y _ { s c l } = 0.5\end{array}
C) Select the correct graph of the polar equation.Describe your viewing window. r = \frac { \pi } { 4 } A) \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 2 \\ X _ { \max } = 2 \\ X _ { s c l } = 0.5 \\ Y _ { \min } = - 2 \\ Y _ { \max } = 2 \\ Y _ { s c l } = 0.5 \end{array} B) \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 2 \\ X _ { \max } = 2 \\ X _ { s c l } = 0.5 \\ Y _ { \min } = - 2 \\ Y _ { \max } = 2 \\ Y _ { s c l } = 0.5 \end{array} C) \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 2 \\ X _ { \max } = 2 \\ X _ { s c l } = 0.5 \\ Y _ { \min } = - 2 \\ Y _ { \max } = 2 \\ Y _ { s c l } = 0.5 \end{array} D) \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 2 \\ X _ { \max } = 2 \\ X _ { s c l } = 0.5 \\ Y _ { \min } = - 2 \\ Y _ { \max } = 2 \\ Y _ { s c l } = 0.5 \end{array} E) \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 2 \\ X _ { \max } = 2 \\ X _ { s c l } = 0.5 \\ Y _ { \min } = - 2 \\ Y _ { \max } = 2 \\ Y _ { s c l } = 0.5 \end{array} θmin⁡=0θmax⁡=2πθstep=π/24Xmin⁡=−2Xmax⁡=2Xscl=0.5Ymin⁡=−2Ymax⁡=2Yscl=0.5\begin{array} { l } \theta _ { \min } = 0 \\\theta _ { \max } = 2 \pi \\\theta _ { s t e p } = \pi / 24 \\X _ { \min } = - 2 \\X _ { \max } = 2 \\X _ { s c l } = 0.5 \\Y _ { \min } = - 2 \\Y _ { \max } = 2 \\Y _ { s c l } = 0.5\end{array}
D) Select the correct graph of the polar equation.Describe your viewing window. r = \frac { \pi } { 4 } A) \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 2 \\ X _ { \max } = 2 \\ X _ { s c l } = 0.5 \\ Y _ { \min } = - 2 \\ Y _ { \max } = 2 \\ Y _ { s c l } = 0.5 \end{array} B) \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 2 \\ X _ { \max } = 2 \\ X _ { s c l } = 0.5 \\ Y _ { \min } = - 2 \\ Y _ { \max } = 2 \\ Y _ { s c l } = 0.5 \end{array} C) \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 2 \\ X _ { \max } = 2 \\ X _ { s c l } = 0.5 \\ Y _ { \min } = - 2 \\ Y _ { \max } = 2 \\ Y _ { s c l } = 0.5 \end{array} D) \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 2 \\ X _ { \max } = 2 \\ X _ { s c l } = 0.5 \\ Y _ { \min } = - 2 \\ Y _ { \max } = 2 \\ Y _ { s c l } = 0.5 \end{array} E) \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 2 \\ X _ { \max } = 2 \\ X _ { s c l } = 0.5 \\ Y _ { \min } = - 2 \\ Y _ { \max } = 2 \\ Y _ { s c l } = 0.5 \end{array} θmin⁡=0θmax⁡=2πθstep=π/24Xmin⁡=−2Xmax⁡=2Xscl=0.5Ymin⁡=−2Ymax⁡=2Yscl=0.5\begin{array} { l } \theta _ { \min } = 0 \\\theta _ { \max } = 2 \pi \\\theta _ { s t e p } = \pi / 24 \\X _ { \min } = - 2 \\X _ { \max } = 2 \\X _ { s c l } = 0.5 \\Y _ { \min } = - 2 \\Y _ { \max } = 2 \\Y _ { s c l } = 0.5\end{array}
E) Select the correct graph of the polar equation.Describe your viewing window. r = \frac { \pi } { 4 } A) \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 2 \\ X _ { \max } = 2 \\ X _ { s c l } = 0.5 \\ Y _ { \min } = - 2 \\ Y _ { \max } = 2 \\ Y _ { s c l } = 0.5 \end{array} B) \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 2 \\ X _ { \max } = 2 \\ X _ { s c l } = 0.5 \\ Y _ { \min } = - 2 \\ Y _ { \max } = 2 \\ Y _ { s c l } = 0.5 \end{array} C) \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 2 \\ X _ { \max } = 2 \\ X _ { s c l } = 0.5 \\ Y _ { \min } = - 2 \\ Y _ { \max } = 2 \\ Y _ { s c l } = 0.5 \end{array} D) \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 2 \\ X _ { \max } = 2 \\ X _ { s c l } = 0.5 \\ Y _ { \min } = - 2 \\ Y _ { \max } = 2 \\ Y _ { s c l } = 0.5 \end{array} E) \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 2 \\ X _ { \max } = 2 \\ X _ { s c l } = 0.5 \\ Y _ { \min } = - 2 \\ Y _ { \max } = 2 \\ Y _ { s c l } = 0.5 \end{array} θmin⁡=0θmax⁡=2πθstep=π/24Xmin⁡=−2Xmax⁡=2Xscl=0.5Ymin⁡=−2Ymax⁡=2Yscl=0.5\begin{array} { l } \theta _ { \min } = 0 \\\theta _ { \max } = 2 \pi \\\theta _ { s t e p } = \pi / 24 \\X _ { \min } = - 2 \\X _ { \max } = 2 \\X _ { s c l } = 0.5 \\Y _ { \min } = - 2 \\Y _ { \max } = 2 \\Y _ { s c l } = 0.5\end{array}

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