3 sandwich choices 2 side item choices 4 beverage choices
Let's say the sandwich choices were PB&J, Ham, and Turkey Let's say the sides were fries and mandarin oranges Let's say the beverages were water, orange juice, milk, soda
PB&J Ham Turkey
Fries Mandarin Oranges
Water OJ Milk Soda
The combinations can be as follows:
PB&J, Fries, Water PB&J, Fries, OJ PB&J, Fries, Milk PB&J, Fries, Soda PB&J, Mandarin Oranges, Water PB&J, Mandarin Oranges, OJ PB&J, Mandarin Oranges, Milk PB&J, Mandarin Oranges, Soda Ham, Fries, Water Ham, Fries, OJ Ham, Fries, Milk Ham, Fries, Soda Ham, Mandarin Oranges, Water Ham, Mandarin Oranges, OJ Ham, Mandarin Oranges, Milk Ham, Mandarin Oranges, Soda Turkey, Fries, Water Turkey, Fries, OJ Turkey, Fries, Milk Turkey, Fries, Soda Turkey, Mandarin Oranges, Water Turkey, Mandarin Oranges, OJ Turkey, Mandarin Oranges, Milk Turkey, Mandarin Oranges, Soda
So your answer is correct: There's 24 lunch combinations that can be made.
<span>First thing you'll need to know is that the value for this equation is actually an approximation 'and' it is imaginary, so, one method is via brute force method.
You let f(y) equals to that equation, then, find the values for f(y) using values from y=-5 to 5, you just substitute the values in you'll get -393,-296,-225,... till when y=3 is f(y)=-9; y=4 is f(y)=48, so there is a change in </span><span>signs when 'y' went from y=3 to y=4, the answer is between 3 and 4, you can work out a little bit deeper using 3.1, 3.2... You get the point. The value is close to 3.1818...
The other method is using Newton's method, it is similar to this but with a twist because it involves differentiation, so </span><span> where 'n' is the number you approximate, like n=0,1,2... etc. f(y) would the equation, and f'(y) is the derivative of f(y), now what you'll need to do is substitute the 'n' values into 'y' to find the approximation.</span>