Select the correct answer from each drop-down menu. ∆ABC has vertices at A(11, 6), B(5, 6), and C(5, 17).
∆XYZ has vertices at X(-10, 5), Y(-12, -2), and Z(-4, 15).
∆MNO has vertices at M(-9, -4), N(-3, -4), and O(-3, -15).
∆JKL has vertices at J(17, -2), K(12, -2), and L(12, 7).
∆PQR has vertices at P(12, 3), Q(12, -2), and R(3, -2).

can be shown to be congruent by a sequence of reflections and translations.
can be shown to be congruent by a single rotation.

m∠cdf = 43° , m∠gdf = 114°, m∠gde = 23°, m∠edh = 23° , m∠cdh = 157°

step-by-step explanation:

since, it is given that de bisects the angle gdh ⇒ de divides the angle gdh into two equal angles.

⇒ m∠gde = m∠edh

⇒ 8·x   - 1 = 6·x + 5

⇒ x = 3

so, m∠gde = 8·x - 1 = 23°

and m∠edh = 23°

now, it is given ∠cde is an straight angle so, m∠cde = 180°

⇒ m∠cdf + m∠fdg + m∠gde = 180°

⇒ 43° + m∠fdg + 23° = 180°

⇒ m∠fdg = 114°

again, m∠cde = m∠cdh + m∠edh = 180°

m∠cdh + 23° = 180°

⇒ m∠cdh = 157°

answer: 7. if the functions had a negative leading coefficient and is of even degree, which statement about the graph is true?

step-by-step explanation: