Answer:
![m(\widehat {RS})+m(\widehat{ST})+m(\widehat{TQ}) + m(major\ arc RQ) = 360\ degree](https://tex.z-dn.net/?f=m%28%5Cwidehat%20%7BRS%7D%29%2Bm%28%5Cwidehat%7BST%7D%29%2Bm%28%5Cwidehat%7BTQ%7D%29%20%2B%20m%28major%5C%20arc%20RQ%29%20%3D%20360%5C%20degree)
Step-by-step explanation:
As we can see in the figure that
The R, S, T ,and Q are the points on the circle O.
Also
The measurement of the circular arc is equivalent to the measurement of the angle at the center of the arc
So by this
![m(\widehat{RS})=m(\angle ROS)](https://tex.z-dn.net/?f=m%28%5Cwidehat%7BRS%7D%29%3Dm%28%5Cangle%20ROS%29)
![m(\widehat{ST})=m(\angle SOT)](https://tex.z-dn.net/?f=m%28%5Cwidehat%7BST%7D%29%3Dm%28%5Cangle%20SOT%29)
![m(\widehat{TQ})=m(\angle TOQ)](https://tex.z-dn.net/?f=m%28%5Cwidehat%7BTQ%7D%29%3Dm%28%5Cangle%20TOQ%29)
m(major arc RQ) = m(∠QOR)
So,
![m(\widehat {RS})+m(\widehat{ST})+m(\widehat{TQ}) + m(major\ arc RQ) = m(\angle ROS)+m(\angle SOT)+m(\angle TOQ)+m(\angle QOR)](https://tex.z-dn.net/?f=m%28%5Cwidehat%20%7BRS%7D%29%2Bm%28%5Cwidehat%7BST%7D%29%2Bm%28%5Cwidehat%7BTQ%7D%29%20%2B%20m%28major%5C%20arc%20RQ%29%20%3D%20m%28%5Cangle%20ROS%29%2Bm%28%5Cangle%20SOT%29%2Bm%28%5Cangle%20TOQ%29%2Bm%28%5Cangle%20QOR%29)
And as we know that
All angles sum = 360°
Therefore
![m(\widehat {RS})+m(\widehat{ST})+m(\widehat{TQ}) + m(major\ arc RQ) = 360\ degree](https://tex.z-dn.net/?f=m%28%5Cwidehat%20%7BRS%7D%29%2Bm%28%5Cwidehat%7BST%7D%29%2Bm%28%5Cwidehat%7BTQ%7D%29%20%2B%20m%28major%5C%20arc%20RQ%29%20%3D%20360%5C%20degree)
Kidjov the answer I came up with for you is 792 I hope this helps
First, apply the distributive property to the left side of the inequality. Multiply each of the two numbers inside the parentheses by 6 and then add those products. Next, subtract -24 from both sides. Then, to get \begin{align*}x\end{align*} by itself on one side of the inequality, you need to divide both sides by 6.