**Problem: Find \( x \).** The image depicts a triangle with three internal angles labeled as follows: 1. Angle 1: \((x + 40)^\circ\) 2. Angle 2: \((2x - 5)^\circ\) 3. Angle 3: \((3x - 17)^\circ\) **Explanation:** To find \( x \), use the fact that the sum of the angles in a triangle is always \( 180^\circ \). Set up the equation: \[ (x + 40) + (2x - 5) + (3x - 17) = 180 \] Combine like terms: \[ x + 40 + 2x - 5 + 3x - 17 = 180 \] \[ 6x + 18 = 180 \] Subtract 18 from both sides: \[ 6x = 162 \] Divide both sides by 6: \[ x = 27 \] **Solution:** \[ x = 27 \]

Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
ChapterP: Preliminary Concepts
SectionP.CT: Test
Problem 1CT
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**Problem: Find \( x \).**

The image depicts a triangle with three internal angles labeled as follows:

1. Angle 1: \((x + 40)^\circ\)
2. Angle 2: \((2x - 5)^\circ\)
3. Angle 3: \((3x - 17)^\circ\)

**Explanation:**

To find \( x \), use the fact that the sum of the angles in a triangle is always \( 180^\circ \). Set up the equation:

\[
(x + 40) + (2x - 5) + (3x - 17) = 180
\]

Combine like terms:

\[
x + 40 + 2x - 5 + 3x - 17 = 180
\]

\[
6x + 18 = 180
\]

Subtract 18 from both sides:

\[
6x = 162
\]

Divide both sides by 6:

\[
x = 27
\]

**Solution:**
\[ x = 27 \]
Transcribed Image Text:**Problem: Find \( x \).** The image depicts a triangle with three internal angles labeled as follows: 1. Angle 1: \((x + 40)^\circ\) 2. Angle 2: \((2x - 5)^\circ\) 3. Angle 3: \((3x - 17)^\circ\) **Explanation:** To find \( x \), use the fact that the sum of the angles in a triangle is always \( 180^\circ \). Set up the equation: \[ (x + 40) + (2x - 5) + (3x - 17) = 180 \] Combine like terms: \[ x + 40 + 2x - 5 + 3x - 17 = 180 \] \[ 6x + 18 = 180 \] Subtract 18 from both sides: \[ 6x = 162 \] Divide both sides by 6: \[ x = 27 \] **Solution:** \[ x = 27 \]
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