Question

1. (4 points) graph and label the image of a(7, -6) after a reflection in y = x followed by a reflection in x = -3.

1. (1 point) cooking a cook flips (reflects) a pancake over line j and then line k. what single transformation maps the pancake from p to p?

a. a reflection at 3.4 inches
b. a horizontal translation left 6.8 inches
c. a horizontal translation left 13.6 inches

1. (3 points) use the interior angle formula ( x = \frac{180(n - 2)}{n} ) to find an interior angle of a regular 20 - gon.

interior angle: ______°
will the regular 20 - gon tessellate the plane? explain your reasoning!

1. (2 points) what are the order and magnitude of symmetry for the figure shown?

order: 12
magnitude: ______°

Explanation:

1. Problem 10:

• Reflection over \(y = x\):
• The rule for reflection over \(y=x\) is \((x,y)\to(y,x)\). For the point \(A(7, - 6)\), after reflection over \(y = x\), the image \(A_1\) is \((-6,7)\).
• Reflection over \(x=-3\):
• The rule for reflection over the vertical line \(x = a\) is \((x,y)\to(2a - x,y)\). Here \(a=-3\), and for the point \((-6,7)\), we have \(x=-6\), \(y = 7\). Then \(2a-x=2\times(-3)-(-6)=-6 + 6=0\). So the image \(A_2\) is \((0,7)\).

1. Problem 11:

• When a figure is reflected over two parallel lines \(j\) and \(k\) (separated by a distance \(d\)), the composition of two reflections over parallel lines is a translation. The distance of the translation is \(2d\). Here \(d = 6.8\) inches.
• The direction of translation is perpendicular to the parallel lines (horizontal in this case, since the lines \(j\) and \(k\) are vertical). The distance of the translation \(T=2\times6.8=13.6\) inches to the left.
• So the single transformation is a horizontal translation left \(13.6\) inches. The answer is C.

1. Problem 12:

• Finding the interior angle:
• Given the formula \(x=\frac{180(n - 2)}{n}\), where \(n = 20\).
• Substitute \(n = 20\) into the formula: \(x=\frac{180\times(20 - 2)}{20}=\frac{180\times18}{20}=162^{\circ}\).
• Tessellation check:
• For a regular polygon to tessellate the plane, the interior angle \(x\) must divide \(360^{\circ}\) evenly (\(360\div x\) must be an integer). Since \(360\div162=\frac{360}{162}=\frac{20}{9}

otin\mathbb{Z}\), a regular 20 - gon will not tessellate the plane.

1. Problem 13:

• Order of symmetry:
• The order of symmetry of a regular \(n\) - petal flower - like figure (assuming it is a regular rotational symmetry figure) is the number of times the figure coincides with itself during a full \(360^{\circ}\) rotation. If we assume the figure has 12 congruent parts (from the order value given as 12), the order of symmetry \(n = 12\).
• Magnitude of symmetry:
• The magnitude of symmetry (the smallest angle of rotation) is given by \(\theta=\frac{360^{\circ}}{n}\). Since \(n = 12\), \(\theta=\frac{360^{\circ}}{12}=30^{\circ}\).

Summary of answers:

• Problem 10: The image of \(A(7,-6)\) after the two - step reflection is \((0,7)\).
• Problem 11: C. a horizontal translation left \(13.6\) inches.
• Problem 12: Interior angle: \(162^{\circ}\). A regular 20 - gon will not tessellate the plane because \(360\div162\) is not an integer.
• Problem 13: Order: \(12\). Magnitude: \(30^{\circ}\).

Answer:

1. Problem 10:

• Reflection over \(y = x\):
• The rule for reflection over \(y=x\) is \((x,y)\to(y,x)\). For the point \(A(7, - 6)\), after reflection over \(y = x\), the image \(A_1\) is \((-6,7)\).
• Reflection over \(x=-3\):
• The rule for reflection over the vertical line \(x = a\) is \((x,y)\to(2a - x,y)\). Here \(a=-3\), and for the point \((-6,7)\), we have \(x=-6\), \(y = 7\). Then \(2a-x=2\times(-3)-(-6)=-6 + 6=0\). So the image \(A_2\) is \((0,7)\).

1. Problem 11:

• When a figure is reflected over two parallel lines \(j\) and \(k\) (separated by a distance \(d\)), the composition of two reflections over parallel lines is a translation. The distance of the translation is \(2d\). Here \(d = 6.8\) inches.
• The direction of translation is perpendicular to the parallel lines (horizontal in this case, since the lines \(j\) and \(k\) are vertical). The distance of the translation \(T=2\times6.8=13.6\) inches to the left.
• So the single transformation is a horizontal translation left \(13.6\) inches. The answer is C.

1. Problem 12:

• Finding the interior angle:
• Given the formula \(x=\frac{180(n - 2)}{n}\), where \(n = 20\).
• Substitute \(n = 20\) into the formula: \(x=\frac{180\times(20 - 2)}{20}=\frac{180\times18}{20}=162^{\circ}\).
• Tessellation check:
• For a regular polygon to tessellate the plane, the interior angle \(x\) must divide \(360^{\circ}\) evenly (\(360\div x\) must be an integer). Since \(360\div162=\frac{360}{162}=\frac{20}{9}

otin\mathbb{Z}\), a regular 20 - gon will not tessellate the plane.

1. Problem 13:

• Order of symmetry:
• The order of symmetry of a regular \(n\) - petal flower - like figure (assuming it is a regular rotational symmetry figure) is the number of times the figure coincides with itself during a full \(360^{\circ}\) rotation. If we assume the figure has 12 congruent parts (from the order value given as 12), the order of symmetry \(n = 12\).
• Magnitude of symmetry:
• The magnitude of symmetry (the smallest angle of rotation) is given by \(\theta=\frac{360^{\circ}}{n}\). Since \(n = 12\), \(\theta=\frac{360^{\circ}}{12}=30^{\circ}\).

Summary of answers:

• Problem 10: The image of \(A(7,-6)\) after the two - step reflection is \((0,7)\).
• Problem 11: C. a horizontal translation left \(13.6\) inches.
• Problem 12: Interior angle: \(162^{\circ}\). A regular 20 - gon will not tessellate the plane because \(360\div162\) is not an integer.
• Problem 13: Order: \(12\). Magnitude: \(30^{\circ}\).