Mathematics, 15.07.2020 02:01 ihatedevin12

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. Assuming all choices are available, how many different possible meals does the restaurant offer?

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Answer from: tashaylinm02

377 choices

Step-by-step explanation:

The following values were given in the question:

The restaurant offered

6 choices of appetizer

8 choices of main meal

5 choices of dessert.

We are also told in the question that the 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

B = Main meal = 8

C = Dessert = 5

a) The 3 choices together

ABC=6 × 8 × 5=240 choices

b) AB= Appetizer and Main meal

= 6 × 8 = 48 choices

c) AC= Appetizer and Dessert

= 6 × 5 = 30 choices

d) BC = Main meal × Dessert

= 8 × 5 = 40 choices

e) A,B,C = the customer having each of the choices only

Appetizer + Main meal + Dessert

= 6 + 8 + 5

= 19 choices

The number of possible meals is calculated as:

240 choices + 48 choices + 30 choices + 40 choices + 19 choices

= 377 choices

Answer from: Quest

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.