The answer is 20!
you need to change the subtraction sign to an addition sign and change -4 to a 4. then, solve!
hope this helps!!
Answer:
square root 12^5
Step-by-step explanation:
To work out the distance of the path, you need to find out the circumference.
Formula for circumference = pi multiplied by the diameter
So you simply do:
pi multiplied by 1.4 miles
And you should get:
4.39822971503... miles
Simply round the result to an appropriate value (e.g. 3 decimal places).
Answer: 4.398 miles (to 3dp)
Hope the answer helps :)
Answer:
y=-2x+7
enter the point into point slope form to find the slope and then use one of the points in point slope form and rearrange it to have y by itself
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)
