B is the only one that would not be perpendicular to the first line. The first line would have a point-slope intercept form of y=-6x+3 and so any line perpendicular to it would require a slope of 1/6. All of them have a slope of 1/6 except B, which has a slope of 6. Let me know if you need a complete show of work as I can show you the math behind my answer.
Answer:
I hope this makes sense, sorry for the lack of proofs.
Step-by-step explanation:
It is given that S is the midpoint of line QT. The definition of a midpoint is that it bisects the line it is on. So, line QS and line TS are congruent, or the same.
Now, we also know that line QR and line TU are parallel. Because they are parallel, it means that they form congruent corners with lines QS and TS..I think the proof here is "angles opposite to congruent sides are congruent in a triangle." But I'm not sure if this is right. Anyways, this means angles RQS and UTS are congruent.
There's also some proof that when two lines cross, the opposite angles are congruent. This means that angles TSU and QSR are congruent.
Therefore, by ASA (angle-side-angle) ΔQRS ≅ ΔTUS.
Given:
To find:
The value of 
.
Solution:
We know that,
...(i)
Where n is an integer.
We have,
Using (i), it can be written as
It is given that .
Therefore, the value of the given function expression is 
.
Answer:
arc AB = 49°
arc ABC = 253°
arc BAC = 156°
arc ACB = 311°
Step-by-step explanation:
arc AB sees the center angle and it is equal to the center angle it sees
arc ABC is equal to 360 - 107 = 253° because it is also sees center angle
arc BAC is equal to 107 + 49 = 156°
arc ACB is equal to 360 - 49 = 311°
