Answer:

Step-by-step explanation:
see the attached figure to better understand the problem
step 1
Find the measure of angle KOM
In the triangle KOM
we have


Applying the law of cosines







step 2
Find the measure of the arc KM
we know that
----> by central angle
we have

so

step 3
Find the measure of angle KLM
we know that
The inscribed angle is half that of the arc comprising
![m\angle KLM=\frac{1}{2}[arc\ KM]](https://tex.z-dn.net/?f=m%5Cangle%20KLM%3D%5Cfrac%7B1%7D%7B2%7D%5Barc%5C%20KM%5D)
we have

substitute
![m\angle KLM=\frac{1}{2}[106.26^o]](https://tex.z-dn.net/?f=m%5Cangle%20KLM%3D%5Cfrac%7B1%7D%7B2%7D%5B106.26%5Eo%5D)

Answer:
Mochi
Step-by-step explanation:
Mochi
Each week theres a increase of 4 in Mochi's weight. So you will add 4 twice to get to the 5th week.
4th week : 17
5th week : 21
Kappa
In 4 weeks Kappa weighed 14, which is less that Mochi at the pace Kappa is going at.
on the 3rd week Kappa weighed 11
and 2nd week Kappa weighed 5
and 1st week Kappa weighed 3
Looking at the graph, we can give a rough estimate on the 5th week it's going to be 16 or 17 either way making Mochi heavier.
Answer:
Option A is the answer...
-7*-15*-5=-525
-7*-75=-525
-1*525=-525
The sequence is geometric, so

for some constant r. From this rule, it follows that

and we can determine the first term to be

Now, by substitution we have

and so on down to (D)

(notice how the exponent on r and the subscript on a add up to n)