Question

find the missing side. round to the nearest tenth.
38)
a) 35.6 b) 35.7
c) 23.3 d) 4.7
e) 24.4
39)
a) 16.3 b) 10.5 c) 7.4
d) 7.7 e) 6.8
find the slope of each line.
40)
a) $-\frac{2}{7}$ b) $-\frac{7}{2}$ c) $\frac{2}{7}$
d) $\frac{7}{2}$ e) 1

Explanation:

Problem 38

Step1: Identify triangle type and trigonometric ratio

This is a right triangle. We know one angle (70°), the adjacent side (13), and we need to find the hypotenuse \( x \). The cosine ratio is \( \cos(\theta)=\frac{\text{adjacent}}{\text{hypotenuse}} \), so \( \cos(70^\circ)=\frac{13}{x} \).

Step2: Solve for \( x \)

Rearrange the formula: \( x = \frac{13}{\cos(70^\circ)} \). Calculate \( \cos(70^\circ)\approx0.3420 \), then \( x=\frac{13}{0.3420}\approx37.95 \)? Wait, no, wait. Wait, maybe I mixed up adjacent and opposite. Wait, the right angle, the angle of 70°, so the side adjacent to 70° is 13, and the side we need is the hypotenuse? Wait, no, wait the triangle: the right angle, one angle 70°, the side opposite to 70°? Wait, no, let's re-examine. The triangle has a right angle, angle 70°, side 13 (let's say adjacent to 70°), and \( x \) is the hypotenuse? Wait, no, maybe it's tangent. Wait, no, maybe I made a mistake. Wait, the options are 35.6, 35.7, etc. Wait, maybe it's a right triangle with angle 70°, opposite side 13, and we need the adjacent? No, wait, no. Wait, let's use tangent. Wait, \( \tan(70^\circ)=\frac{x}{13} \), so \( x = 13\tan(70^\circ) \). \( \tan(70^\circ)\approx2.7475 \), so \( 13\times2.7475\approx35.7175 \), which rounds to 35.7. So that's option B.

Step1: Identify triangle type and trigonometric ratio

Right triangle, angle 34°, adjacent side 11, opposite side \( x \). So \( \tan(34^\circ)=\frac{x}{11} \).

Step2: Solve for \( x \)

\( x = 11\tan(34^\circ) \). \( \tan(34^\circ)\approx0.6745 \), so \( 11\times0.6745\approx7.4195 \), which rounds to 7.4. So option C.

Step1: Identify two points on the line

From the graph, let's find two points. Let's say the left point is (-3, -2) and the right point is (4, -3)? Wait, no, let's count the grid. Let's take two clear points. Let's say the first point is at (-2, -1) and the second at (5, -2)? Wait, no, looking at the graph, the line goes through two points. Let's assume the left point is (-3, -2) and the right point is (4, -3). Wait, no, better to use the slope formula \( m=\frac{y_2 - y_1}{x_2 - x_1} \). Let's pick two points: let's say (-2, -1) and (5, -2). Wait, no, maybe the points are (-3, -2) and (4, -3). Wait, no, let's look at the options. The options are \( -\frac{2}{7} \), \( -\frac{7}{2} \), etc. Let's take two points: suppose the left point is (-3, -2) and the right point is (4, -3). Then \( y_2 - y_1 = -3 - (-2) = -1 \), \( x_2 - x_1 = 4 - (-3) = 7 \), so slope is \( \frac{-1}{7}=-\frac{1}{7} \)? No, that's not matching. Wait, maybe the points are (-2, -1) and (5, -3). Then \( y_2 - y_1 = -3 - (-1) = -2 \), \( x_2 - x_1 = 5 - (-2) = 7 \), so slope is \( \frac{-2}{7}=-\frac{2}{7} \), which is option A.

Answer:

B) 35.7

Problem 39