Answer:
Step-by-step explanation:
<u><em>The picture of the question in the attached figure</em></u>
we know that
The measure of the external angle is the semi-difference of the arches it covers.
so
![m\angle GET=\frac{1}{2}[arc\ TN-arc\ TG]](https://tex.z-dn.net/?f=m%5Cangle%20GET%3D%5Cfrac%7B1%7D%7B2%7D%5Barc%5C%20TN-arc%5C%20TG%5D)
Remember that the diameter divide the circle into two equal parts
In this problem
TN is a diameter
we have
----> because is half the circle (TN is a diameter)
---> is given
substitute
![m\angle GET=\frac{1}{2}[180^o-46^o]=67^o](https://tex.z-dn.net/?f=m%5Cangle%20GET%3D%5Cfrac%7B1%7D%7B2%7D%5B180%5Eo-46%5Eo%5D%3D67%5Eo)
Find the opposite of adding 15. So it would be subtracting 15. So subtract 15 on both sides. So it would have 2/3x on the left and then 2 on the other side. (2/3x=2) Now find the opposite of multiplying 2/3. It is dividing 23/3. So divide 23/3 on both sides. So multiply 2/3 times 1/2. It would equal 3. x = 3
Hope this helps. :-)
Answer: 42 degrees
Explanation: KFD=84 KFL is half of KFD(84)
84 divided by 2 = 42
Answer:
3.6 mph
Step-by-step explanation:
The total distance was 6 mi. and the total time was 5/3 hr.
6 ÷ 5/3 = 6(3/5) = 18/5 = 3.6 mph
