Answer:
377 choices
Step-by-step explanation:
From the above question, we are told that
A restaurant offers 6 choices of appetizer, 8 choices of main meal and 5 choices of dessert. A customer can choose to eat just one course, or two different courses, or all three courses.
Let us represent each choice by :
A = Appetizer = 6
M = Main meal = 8
D = Dessert = 5
a) The combination of the 3 choices together
AMD=6 × 8 × 5=240
b) AM= Appetizer and Main meal
= 6 × 8 = 48
c) AD= Appetizer and Dessert
= 6 × 5 = 30
d) MD = Main meal × Dessert
= 8 × 5 = 40
e) A,M,D (each alone)=
Appetizer + Main meal + Dessert
= 6 + 8 + 5
= 19
Assuming all choices are available, how many different possible meals does the restaurant offer?
This is calculated as:
AMD + AM + AD + MD + A,M,D
240 + 48 + 30 + 40 + 19
= 377 choices
Divide by 4 to get 9 and a R of 2, then just put the remainder as the numerator, and you have 9 1/2 (simplified)
Answer:
Number 2 is A
Number 3 is (5, -4)
Number 4 is (5, -4)
Number 5 is (4, 2)
Step-by-step explanation:
Answer:
78°
Step-by-step explanation:
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D^2 + 17d + 72 = 0
( d + 8 ) ( d + 9 ) = 0
d = - 8 , d = - 9
