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Select the Correct Graph of the Polar Equation r=74r = \frac { 7 } { 4 }

Question 39

Multiple Choice

Select the correct graph of the polar equation.Describe your viewing window.​ r=74r = \frac { 7 } { 4 } ​


A) ​ Select the correct graph of the polar equation.Describe your viewing window.​ r = \frac { 7 } { 4 } ​ A) ​ \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 5 \\ X _ { \max } = 5 \\ X _ { s c l } = 1 \\ Y _ { \min } = - 5 \\ Y _ { \max } = 5 \\ Y _ { s c l } = 1 \end{array} B) ​ \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 5 \\ X _ { \max } = 5 \\ X _ { s c l } = 1 \\ Y _ { \min } = - 5 \\ Y _ { \max } = 5 \\ Y _ { s c l } = 1 \end{array} C) ​ \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 5 \\ X _ { \max } = 5 \\ X _ { s c l } = 1 \\ Y _ { \min } = - 5 \\ Y _ { \max } = 5 \\ Y _ { s c l } = 1 \end{array} D) ​ \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 5 \\ X _ { \max } = 5 \\ X _ { s c l } = 1 \\ Y _ { \min } = - 5 \\ Y _ { \max } = 5 \\ Y _ { s c l } = 1 \end{array} E) ​ \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 5 \\ X _ { \max } = 5 \\ X _ { s c l } = 1 \\ Y _ { \min } = - 5 \\ Y _ { \max } = 5 \\ Y _ { s c l } = 1 \end{array} θmin⁡=0θmax⁡=2πθstep=π/24Xmin⁡=−5Xmax⁡=5Xscl=1Ymin⁡=−5Ymax⁡=5Yscl=1\begin{array} { l } \theta _ { \min } = 0 \\\theta _ { \max } = 2 \pi \\\theta _ { s t e p } = \pi / 24 \\X _ { \min } = - 5 \\X _ { \max } = 5 \\X _ { s c l } = 1 \\Y _ { \min } = - 5 \\Y _ { \max } = 5 \\Y _ { s c l } = 1\end{array}
B) ​ Select the correct graph of the polar equation.Describe your viewing window.​ r = \frac { 7 } { 4 } ​ A) ​ \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 5 \\ X _ { \max } = 5 \\ X _ { s c l } = 1 \\ Y _ { \min } = - 5 \\ Y _ { \max } = 5 \\ Y _ { s c l } = 1 \end{array} B) ​ \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 5 \\ X _ { \max } = 5 \\ X _ { s c l } = 1 \\ Y _ { \min } = - 5 \\ Y _ { \max } = 5 \\ Y _ { s c l } = 1 \end{array} C) ​ \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 5 \\ X _ { \max } = 5 \\ X _ { s c l } = 1 \\ Y _ { \min } = - 5 \\ Y _ { \max } = 5 \\ Y _ { s c l } = 1 \end{array} D) ​ \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 5 \\ X _ { \max } = 5 \\ X _ { s c l } = 1 \\ Y _ { \min } = - 5 \\ Y _ { \max } = 5 \\ Y _ { s c l } = 1 \end{array} E) ​ \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 5 \\ X _ { \max } = 5 \\ X _ { s c l } = 1 \\ Y _ { \min } = - 5 \\ Y _ { \max } = 5 \\ Y _ { s c l } = 1 \end{array} θmin⁡=0θmax⁡=2πθstep=π/24Xmin⁡=−5Xmax⁡=5Xscl=1Ymin⁡=−5Ymax⁡=5Yscl=1\begin{array} { l } \theta _ { \min } = 0 \\\theta _ { \max } = 2 \pi \\\theta _ { s t e p } = \pi / 24 \\X _ { \min } = - 5 \\X _ { \max } = 5 \\X _ { s c l } = 1 \\Y _ { \min } = - 5 \\Y _ { \max } = 5 \\Y _ { s c l } = 1\end{array}
C) ​ Select the correct graph of the polar equation.Describe your viewing window.​ r = \frac { 7 } { 4 } ​ A) ​ \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 5 \\ X _ { \max } = 5 \\ X _ { s c l } = 1 \\ Y _ { \min } = - 5 \\ Y _ { \max } = 5 \\ Y _ { s c l } = 1 \end{array} B) ​ \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 5 \\ X _ { \max } = 5 \\ X _ { s c l } = 1 \\ Y _ { \min } = - 5 \\ Y _ { \max } = 5 \\ Y _ { s c l } = 1 \end{array} C) ​ \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 5 \\ X _ { \max } = 5 \\ X _ { s c l } = 1 \\ Y _ { \min } = - 5 \\ Y _ { \max } = 5 \\ Y _ { s c l } = 1 \end{array} D) ​ \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 5 \\ X _ { \max } = 5 \\ X _ { s c l } = 1 \\ Y _ { \min } = - 5 \\ Y _ { \max } = 5 \\ Y _ { s c l } = 1 \end{array} E) ​ \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 5 \\ X _ { \max } = 5 \\ X _ { s c l } = 1 \\ Y _ { \min } = - 5 \\ Y _ { \max } = 5 \\ Y _ { s c l } = 1 \end{array} θmin⁡=0θmax⁡=2πθstep=π/24Xmin⁡=−5Xmax⁡=5Xscl=1Ymin⁡=−5Ymax⁡=5Yscl=1\begin{array} { l } \theta _ { \min } = 0 \\\theta _ { \max } = 2 \pi \\\theta _ { s t e p } = \pi / 24 \\X _ { \min } = - 5 \\X _ { \max } = 5 \\X _ { s c l } = 1 \\Y _ { \min } = - 5 \\Y _ { \max } = 5 \\Y _ { s c l } = 1\end{array}
D) ​ Select the correct graph of the polar equation.Describe your viewing window.​ r = \frac { 7 } { 4 } ​ A) ​ \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 5 \\ X _ { \max } = 5 \\ X _ { s c l } = 1 \\ Y _ { \min } = - 5 \\ Y _ { \max } = 5 \\ Y _ { s c l } = 1 \end{array} B) ​ \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 5 \\ X _ { \max } = 5 \\ X _ { s c l } = 1 \\ Y _ { \min } = - 5 \\ Y _ { \max } = 5 \\ Y _ { s c l } = 1 \end{array} C) ​ \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 5 \\ X _ { \max } = 5 \\ X _ { s c l } = 1 \\ Y _ { \min } = - 5 \\ Y _ { \max } = 5 \\ Y _ { s c l } = 1 \end{array} D) ​ \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 5 \\ X _ { \max } = 5 \\ X _ { s c l } = 1 \\ Y _ { \min } = - 5 \\ Y _ { \max } = 5 \\ Y _ { s c l } = 1 \end{array} E) ​ \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 5 \\ X _ { \max } = 5 \\ X _ { s c l } = 1 \\ Y _ { \min } = - 5 \\ Y _ { \max } = 5 \\ Y _ { s c l } = 1 \end{array} θmin⁡=0θmax⁡=2πθstep=π/24Xmin⁡=−5Xmax⁡=5Xscl=1Ymin⁡=−5Ymax⁡=5Yscl=1\begin{array} { l } \theta _ { \min } = 0 \\\theta _ { \max } = 2 \pi \\\theta _ { s t e p } = \pi / 24 \\X _ { \min } = - 5 \\X _ { \max } = 5 \\X _ { s c l } = 1 \\Y _ { \min } = - 5 \\Y _ { \max } = 5 \\Y _ { s c l } = 1\end{array}
E) ​ Select the correct graph of the polar equation.Describe your viewing window.​ r = \frac { 7 } { 4 } ​ A) ​ \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 5 \\ X _ { \max } = 5 \\ X _ { s c l } = 1 \\ Y _ { \min } = - 5 \\ Y _ { \max } = 5 \\ Y _ { s c l } = 1 \end{array} B) ​ \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 5 \\ X _ { \max } = 5 \\ X _ { s c l } = 1 \\ Y _ { \min } = - 5 \\ Y _ { \max } = 5 \\ Y _ { s c l } = 1 \end{array} C) ​ \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 5 \\ X _ { \max } = 5 \\ X _ { s c l } = 1 \\ Y _ { \min } = - 5 \\ Y _ { \max } = 5 \\ Y _ { s c l } = 1 \end{array} D) ​ \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 5 \\ X _ { \max } = 5 \\ X _ { s c l } = 1 \\ Y _ { \min } = - 5 \\ Y _ { \max } = 5 \\ Y _ { s c l } = 1 \end{array} E) ​ \begin{array} { l } \theta _ { \min } = 0 \\ \theta _ { \max } = 2 \pi \\ \theta _ { s t e p } = \pi / 24 \\ X _ { \min } = - 5 \\ X _ { \max } = 5 \\ X _ { s c l } = 1 \\ Y _ { \min } = - 5 \\ Y _ { \max } = 5 \\ Y _ { s c l } = 1 \end{array} θmin⁡=0θmax⁡=2πθstep=π/24Xmin⁡=−5Xmax⁡=5Xscl=1Ymin⁡=−5Ymax⁡=5Yscl=1\begin{array} { l } \theta _ { \min } = 0 \\\theta _ { \max } = 2 \pi \\\theta _ { s t e p } = \pi / 24 \\X _ { \min } = - 5 \\X _ { \max } = 5 \\X _ { s c l } = 1 \\Y _ { \min } = - 5 \\Y _ { \max } = 5 \\Y _ { s c l } = 1\end{array}

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