Please help again, sorry for asking…

QAmmunity.org

My account
Edit my Profile
Private messages
My favorites

Ask a Question
Questions
Unanswered
Tags
Categories
Ask a Question

asked Jan 1, 2022 226k views

1 vote

Please help again, sorry for asking…

Mathematics
college

Mangled Deutz asked

by Mangled Deutz

3.9k points

0 Comments

Please log in or register to add a comment.

Please log in or register to answer this question.

1 Answer

4 votes

$\text{BC}=6\text{x}+4\\\\\text{AC}=52\\\\\text{AB}=4\text{x}+8\\\\\text{AB}+\text{BC}=\text{AC}\\\\4\text{x}+8+6\text{x}+4=52\\\\10\text{x}+12=52 \ \ /-12\\\\10\text{x}=40 \ \ /:10\\\\\text{x}=4\\\\\text{BC}=6\cdot4+4=24+4=\boxed{28}$

Efim Rozovsky answered Jan 7, 2022

by Efim Rozovsky

4.3k points

0 Comments

Please log in or register to add a comment.

Ask a Question

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

4.5m questions

5.8m answers

0 Comments

Please log in or register to add a comment.

Please log in or register to answer this question.

1 Answer

0 Comments

Please log in or register to add a comment.

Other Questions