Question

1. which equation represents a line which is perpendicular to the line 7x - 8y = 16?

a. y = (8/7)x - 1
b. y = (7/8)x + 8
c. y = - (8/7)x - 7
d. y = - (7/8)x + 4

1. identify two angles that are marked congruent to each other on the diagram below. (diagram is not to scale.)

diagram of a polygon with points f, g, h, i, d, and angles marked at h and e

1. point b is on line segment \\(\overline{ac}\\). given \\(ab = 4x - 6\\), \\(ac = 2x + 10\\), and \\(bc = 4\\), determine the numerical length of \\(\overline{ac}\\).

diagram of line segment ac with b between a and c, labeled ab = 4x - 6, bc = 4, ac = 2x + 10

1. in \\(\triangle xyz\\), \\(m\angle x = 15^\circ\\) and \\(m\angle y = 81^\circ\\). which statement about the sides of \\(\triangle xyz\\) must be true?

a. \\(xy > yz > zx\\)
b. \\(yz > xy > zx\\)
c. \\(zx > xy > yz\\)
d. \\(zx > yz > xy\\)
e. \\(xy > zx > yz\\)
f. \\(yz > zx > xy\\)
diagram of \\(\triangle xyz\\) with angles marked 15° at x and 81° at y

Explanation:

Question 12

Step1: Find slope of given line

Rewrite \(7x - 8y = 16\) in slope - intercept form \(y=mx + b\) (where \(m\) is the slope).
Subtract \(7x\) from both sides: \(-8y=-7x + 16\).
Divide by \(-8\): \(y=\frac{7}{8}x-2\). So the slope of the given line is \(m_1 = \frac{7}{8}\).

Step2: Find slope of perpendicular line

If two lines are perpendicular, the product of their slopes is \(- 1\). Let the slope of the perpendicular line be \(m_2\). Then \(m_1\times m_2=-1\).
\(\frac{7}{8}\times m_2=-1\), so \(m_2=-\frac{8}{7}\).

Step3: Check the slopes of the options

• Option A: \(y = \frac{8}{7}x-1\), slope \(m=\frac{8}{7}

eq-\frac{8}{7}\).

• Option B: \(y=\frac{7}{8}x + 8\), slope \(m=\frac{7}{8}

eq-\frac{8}{7}\).

• Option C: \(y =-\frac{8}{7}x-7\), slope \(m =-\frac{8}{7}\).
• Option D: \(y=-\frac{1}{8}x + 4\), slope \(m=-\frac{1}{8}

eq-\frac{8}{7}\).

In the diagram, angle at \(H\) (marked with the same arc - like congruence mark as angle at \(E\)) and angle at \(E\) are marked congruent. Also, from the markings (the tick marks on the segments and the right - angle and congruence marks), \(\angle JHI\) and \(\angle KED\) are marked congruent (as per the hand - written notes and the diagram's congruence markings).

Step1: Use segment addition postulate

Since \(B\) is on \(\overline{AC}\), \(AC=AB + BC\).
We know that \(AB = 4x-6\), \(AC=2x + 10\), and \(BC = 4\). So, \(2x + 10=(4x-6)+4\).

Step2: Solve for \(x\)

Simplify the right - hand side: \(2x + 10=4x-6 + 4=4x-2\).
Subtract \(2x\) from both sides: \(10 = 2x-2\).
Add 2 to both sides: \(12 = 2x\).
Divide by 2: \(x = 6\).

Step3: Find the length of \(AC\)

Substitute \(x = 6\) into the expression for \(AC\): \(AC=2x + 10\).
\(AC=2\times6+10=12 + 10=22\).

Answer:

C. \(y =-\frac{8}{7}x-7\)

Question 13