Answer:
c. m∠1 + m∠6 = m∠4 + m∠6
Step-by-step explanation:
Given: The lines l and m are parallel lines.
The parallel lines cut two transverse lines. Here we can use the properties of transverse and find the incorrect statements.
a. m∠1 + m∠2 = m∠3 + m∠4
Here m∠1 and m∠2 are supplementary angles add upto 180 degrees.
m∠3 and m∠4 are supplementary angles add upto 180 degrees.
Therefore, the statement is true.
b. m∠1 + m∠5 = m∠3 + m∠4
m∠1 + m∠5 = 180 same side of the adjacent angles.
m∠3 + m∠4 = 180, supplementary angles add upto 180 degrees.
Therefore, the statement is true.
Now let's check c.
m∠1 + m∠6 = m∠4 + m∠6
We can cancel out m∠6, we get
m∠1 = m∠4 which is not true
Now let's check d.
m∠3 + m∠4 = m∠7 + m∠4
We can cancel out m∠4, we get
m∠3 = m∠7, alternative interior angles are equal.
It is true.
Therefore, answer is c. m∠1 + m∠6 = m∠4 + m∠6
30 girls and 42 boys so idk this makes no sense and I'm older than you
The required proof is given in the table below:
![\begin{tabular}{|p{4cm}|p{6cm}|} Statement & Reason \\ [1ex] 1. $\overline{BD}$ bisects $\angle ABC$ & 1. Given \\ 2. \angle DBC\cong\angle ABD & 2. De(finition of angle bisector \\ 3. $\overline{AE}$||$\overline{BD}$ & 3. Given \\ 4. \angle AEB\cong\angle DBC & 4. Corresponding angles \\ 5. \angle AEB\cong\angle ABD & 5. Transitive property of equality \\ 6. \angle ABD\cong\angle BAE & 6. Alternate angles \end{tabular}](https://tex.z-dn.net/?f=%20%5Cbegin%7Btabular%7D%7B%7Cp%7B4cm%7D%7Cp%7B6cm%7D%7C%7D%20%0A%20Statement%20%26%20Reason%20%5C%5C%20%5B1ex%5D%20%0A1.%20%24%5Coverline%7BBD%7D%24%20bisects%20%24%5Cangle%20ABC%24%20%26%201.%20Given%20%5C%5C%0A2.%20%5Cangle%20DBC%5Ccong%5Cangle%20ABD%20%26%202.%20De%28finition%20of%20angle%20bisector%20%5C%5C%20%0A3.%20%24%5Coverline%7BAE%7D%24%7C%7C%24%5Coverline%7BBD%7D%24%20%26%203.%20Given%20%5C%5C%20%0A4.%20%5Cangle%20AEB%5Ccong%5Cangle%20DBC%20%26%204.%20Corresponding%20angles%20%5C%5C%0A5.%20%5Cangle%20AEB%5Ccong%5Cangle%20ABD%20%26%205.%20Transitive%20property%20of%20equality%20%5C%5C%20%0A6.%20%5Cangle%20ABD%5Ccong%5Cangle%20BAE%20%26%206.%20Alternate%20angles%0A%5Cend%7Btabular%7D)
![\begin{tabular}{|p{4cm}|p{6cm}|} 7. \angle AEB\cong\angle BAE & 7. Transitive property of equality \\ 8. \overline{EB}\cong\overline{AB} & 8. From 7. $\Delta ABE$ is isosceles \\ 9. EB = AB & 9. De(finition of congruence \\ 10. $\frac{AD}{DC}=\frac{EB}{BC}$ & 10. Triangle proportionality theorem \\ 11. $\frac{AD}{DC}=\frac{AB}{BC}$ & 11. Substitution Property of equality \\[1ex] \end{tabular} ](https://tex.z-dn.net/?f=%5Cbegin%7Btabular%7D%7B%7Cp%7B4cm%7D%7Cp%7B6cm%7D%7C%7D%0A7.%20%5Cangle%20AEB%5Ccong%5Cangle%20BAE%20%26%207.%20Transitive%20property%20of%20equality%20%5C%5C%0A8.%20%5Coverline%7BEB%7D%5Ccong%5Coverline%7BAB%7D%20%26%208.%20From%207.%20%24%5CDelta%20ABE%24%20is%20isosceles%20%5C%5C%0A9.%20EB%20%3D%20AB%20%26%209.%20De%28finition%20of%20congruence%20%5C%5C%2010.%20%24%5Cfrac%7BAD%7D%7BDC%7D%3D%5Cfrac%7BEB%7D%7BBC%7D%24%20%26%2010.%20Triangle%20proportionality%20theorem%20%5C%5C%0A11.%20%24%5Cfrac%7BAD%7D%7BDC%7D%3D%5Cfrac%7BAB%7D%7BBC%7D%24%20%26%2011.%20Substitution%20Property%20of%20equality%20%5C%5C%5B1ex%5D%20%0A%5Cend%7Btabular%7D%0A)
Answer:
A, (13/3 x 5/13)
Step-by-step explanation:
So first, you should change the mixed numbers into improper fractions. To convert, just multiply the whole number by the denominator and add the numerator. Do that to both fractions and you should get 13/3 ÷ 13/5. Since you need to switch to multiplication, switch the sign and flip that fraction/ That asnwer is 13/3 x 5/13. Hope that helps! :)
Luke: 1 practice 1 lap, 2 practices 2 laps, 3 practices 3 laps, 4 practices 4 laps
Alex: 1 practice 3 laps, 2 practices 6 laps, 3 practices 9 laps, 4 practices 12 laps
Luke ran 3 laps for 3 practices, and Alex ran 9 laps for 3 practices.
