The local diner offers a meal combination consisting of an appetizer, a soup, a main course, and a dessert. There are three appetizers, five soups, three main courses, and three desserts. Your diet restricts you to choosing between a dessert and an appetizer. (You cannot have both.) Given this restriction, how many three-course meals are possible

number of ways to chose a three course meal is 90

Step-by-step explanation:

Given the data in the question;

Number of appetizer = 3

Number of soups = 5

Number of main courses = 3

Number of desserts = 3

Now,

Number of ways that appetizer can be chosen = ³C₁ = 3!/(1!(3-1)!) = 3

Number of ways that soups can be chosen = ⁵C₁ = 5!/(1!(5-1)!) = 5

Number of ways that main courses can be chosen = ³C₁ = 3!/(1!(3-1)!) = 3

Number of ways that desserts can be chosen = ³C₁ = 3!/(1!(3-1)!) = 3

So,

First we chose an appetizer, soup and main course.

Number of ways will be;

⇒ 3 × 5 × 3 = 45

Next,  we chose a dessert, a soup & a main course.

Number of ways will be;

⇒ 3 × 5 × 3 = 45

Total number of ways to chose a three course meal

⇒ 45 + 45 = 90 ways

Therefore, number of ways to chose a three course meal is 90

answer: no

step-by-step explanation: because the big angle will spread away the 2 side that it is on and make the opposite side even longer.

i am so sorry , i have no idea but i want points so i won't be a beginner anymore