6 calandre years hope y get it correct and mark me brainliest plsssss
First blank: sin. C = h/a
The ratio for sin. is opposite (the height) over hypotenuse (a).
Second blank: Area = (1/2)b(a sin. C)
In this substitution, we are substituting the value of h into the traditional area of a triangle formula.
Third Blank: Area = (1/2)ab(sin C)
We are using the commutative property to move the position of a.
Hope this helps and good luck in your classes!
Answer: b. 50
Step-by-step explanation:
Hence the employee made an error for 50 serving size.
Let

In order to prove this by induction, we first need to prove the base case, i.e. prove that P(1) is true:

So, the base case is ok. Now, we need to assume
and prove
.
states that

Since we're assuming
, we can substitute the sum of the first n terms with their expression:

Which terminates the proof, since we showed that

as required