Homework Help is Here – Start Your Trial Now!
Math
Geometry
1041 10. Find m/BCE. 11x=3=7x+5 7x nx YX-3=5 27+21=51 (7x + 5)° E (11x-3)° C

1041 10. Find m/BCE. 11x=3=7x+5 7x nx YX-3=5 27+21=51 (7x + 5)° E (11x-3)° C

Elementary Geometry For College Students, 7e
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
See similar textbooks
icon
Related questions
Question
100%

Yes number 10 says find angle BCE I HAVE SOLVED WHAT WAS GIVEN BY YOUR EXPERT BUT HOW DO I FIND THE ANGLE 

**Text Transcription for Educational Website:** --- **Problem 10: Find \( m\angle BCE \).** Given: - \( \angle AEB = (7x + 5)^\circ \) - \( \angle DEC = (11x - 3)^\circ \) Solution: 1. Set the angles equal: \( 11x - 3 = 7x + 5 \). 2. Rearrange the equation: \( 11x - 3 - 7x = 5 \). 3. Simplify: \( 4x - 3 = 5 \). 4. Add 3 to both sides: \( 4x = 8 \). 5. Divide by 4: \( x = 2 \). --- **Problem 12: Find \( m\angle XZW \).** Given: - Quadrilateral \( WXYZ \) with diagonals \( \angle XZR = (x + 20)^\circ \) and \( \angle YRX = (5x - 8)^\circ \). Solution: 1. Set up the equation: \( x + 20 + 5x - 8 = 90 \). 2. Simplify: \( 6x + 12 = 90 \). 3. Subtract 12 from both sides: \( 6x = 78 \). 4. Divide by 6: \( x = 13 \). 5. Substitute \( x \) into \( \angle XZW = (x + 20) \). 6. Calculate: \( x + 20 = 13 + 20 = 33^\circ \). --- **Diagram Description:** - For Problem 10, there is a parallelogram \( ABCD \) with diagonal \( AC \). The angles formed by the diagonals are labeled \( \angle AEB \) and \( \angle DEC \). - For Problem 12, there is quadrilateral \( WXYZ \) with diagonals forming \( \angle XZW \). The given expressions represent the angles formed by these diagonals. ---
Transcribed Image Text:**Text Transcription for Educational Website:** --- **Problem 10: Find \( m\angle BCE \).** Given: - \( \angle AEB = (7x + 5)^\circ \) - \( \angle DEC = (11x - 3)^\circ \) Solution: 1. Set the angles equal: \( 11x - 3 = 7x + 5 \). 2. Rearrange the equation: \( 11x - 3 - 7x = 5 \). 3. Simplify: \( 4x - 3 = 5 \). 4. Add 3 to both sides: \( 4x = 8 \). 5. Divide by 4: \( x = 2 \). --- **Problem 12: Find \( m\angle XZW \).** Given: - Quadrilateral \( WXYZ \) with diagonals \( \angle XZR = (x + 20)^\circ \) and \( \angle YRX = (5x - 8)^\circ \). Solution: 1. Set up the equation: \( x + 20 + 5x - 8 = 90 \). 2. Simplify: \( 6x + 12 = 90 \). 3. Subtract 12 from both sides: \( 6x = 78 \). 4. Divide by 6: \( x = 13 \). 5. Substitute \( x \) into \( \angle XZW = (x + 20) \). 6. Calculate: \( x + 20 = 13 + 20 = 33^\circ \). --- **Diagram Description:** - For Problem 10, there is a parallelogram \( ABCD \) with diagonal \( AC \). The angles formed by the diagonals are labeled \( \angle AEB \) and \( \angle DEC \). - For Problem 12, there is quadrilateral \( WXYZ \) with diagonals forming \( \angle XZW \). The given expressions represent the angles formed by these diagonals. ---
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 3 steps with 3 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer