First off, you have to subtract the 3rd number, then carry the number if so, then do the same to the second and first number, did i help in any way?
180-(180-49-48)
180-(83)
97
180-((180-141)+(180-76))
180-(39+104)
180-143
137
180-((180-101)+60)
180-(79+60)
180-139
41
360-147-79
213-79
139
180-(360-109*2)
180-(360-218)
180-142
38
(180-118)/2
62/2
31
Hope this helps :)
Answer:
vyuWYIRTG3ñwyiehg
Step-by-step explanation:
Answer:
option c 1000
Step-by-step explanation:
...................
Answer:
.
Step-by-step explanation:
Since repetition isn't allowed, there would be 
choices for the first donut, 
choices for the second donut, and 
choices for the third donut. If the order in which donuts are placed in the bag matters, there would be 
unique ways to choose a bag of these donuts.
In practice, donuts in the bag are mixed, and the ordering of donuts doesn't matter. The same way of counting would then count every possible mix of three donuts type 
times.
For example, if a bag includes donut of type 
, 
, and 
, the count would include the following 
arrangements:
Thus, when the order of donuts in the bag doesn't matter, it would be necessary to divide the count by to find the actual number of donut combinations:
.
Using combinatorics notations, the answer to this question is the same as the number of ways to choose an unordered set of 
objects from a set of distinct objects:
.
.
.
.
.
.
.