A is the answer to this question because if you plug in 8.6-5 it is approximately 3
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)
The lengths of sides a and c add to 9 cm, the length of side b.
In a triangle, the the sum of the lengths of any two sides must be greater than the the length of the third side. These dimensions do not make a triangle.
Is this really what you meant? Why did you use capital A for a side? Or is A an angle, and it measures 6 deg, not 6 cm?
A^4 * a^6
exponent rule : x^y * x^z = x^(y + z)
so, a^4 * a^6 = a^(4 + 6) = a^10
