∠1 and ∠2 are supplementary // given∠3 and ∠4 are supplementary // given∠1 ≅ ∠3 // given m∠1 + m∠2 = 180° // definition of supplementary anglesm∠3 + m∠4 = 180° // definition of supplementary angles m∠1 + m∠2 = m∠3 + m∠4 // transitive property of equality m∠1 = m∠3 // definition of congruent angles m∠1 + m∠2 = m∠1 + m∠4 // substitution property of equality (replaced m∠3 with m∠1) m∠2 = m∠4 // subtraction property of equality (subtracted m∠1 from both sides) ∠2 ≅ ∠4 // definition of congruent angles
Answer:
x1=-4/5, x2=18/5 and x3=7/5
Step-by-step explanation:
![\left[\begin{array}{ccc|c}1&0&-3&-5\\3&1&2&4\\2&2&1&7\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%260%26-3%26-5%5C%5C3%261%262%264%5C%5C2%262%261%267%5Cend%7Barray%7D%5Cright%5D)
you can do linear combination between the rows:
2nd row=R2-3R1 and 3th row=R3-2R1
![\left[\begin{array}{ccc|c}1&0&-3&-5\\0&1&11&19\\0&2&7&17\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%260%26-3%26-5%5C%5C0%261%2611%2619%5C%5C0%262%267%2617%5Cend%7Barray%7D%5Cright%5D)
3th row=(3R2-R3)/15
![\left[\begin{array}{ccc|c}1&0&-3&-5\\0&1&11&19\\0&0&1&\frac{7}{5} \end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%260%26-3%26-5%5C%5C0%261%2611%2619%5C%5C0%260%261%26%5Cfrac%7B7%7D%7B5%7D%20%5Cend%7Barray%7D%5Cright%5D)
1st row=R1+3R3 and R2-11R3
![\left[\begin{array}{ccc|c}1&0&0&-4/5 \\0&1&0&18/5 \\0&0&1&7/5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%260%260%26-4%2F5%20%5C%5C0%261%260%2618%2F5%20%5C%5C0%260%261%267%2F5%5Cend%7Barray%7D%5Cright%5D)
x1=-4/5, x2=18/5 and x3=7/5
subsitute x = 6 into the lengths
3(6) -7
18 - 7
11cm
(6) + 5
11cm
therefore it is a square when x = 6 because the sides are all equal
no it is more than 4x it is 6x
