A restaurant offers 5 choices of appetizer, 10 choices of main meal and 4 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?

Hence there 329 different possible meals that the restaurant offer.

Step-by-step explanation:

Given:

A restaurant offers 5 choices of appetizer,10 of meal and 4 of dessert.

To Find:

All different possible ways that restaurant offers the meal.

Solution:

Consider ,

A=appetizer Course=5

M=main  course=10

D= Desert course=4.

For ,

All individual course=total courses=A+M+D

=5+10+4

=19

So there 19 ways to eat just one course from restaurant.

Now,

For 2 different choices

It is given by combination of 2 courses with each other

=AD+AM+DM

=5*4+5*10+10*4

=20+50+40

=110

So there 110 ways to eat 2 different courses.

Now

For 3 different choices

It is given by

=A*D*M

=5*10*4

=200

So there 200 ways to eat 3 different courses.

Now,

All choices available = (ways to eat one course only +ways to eat 2 different courses+ways to eat 3 different courses)

=19+110+200

=329

56.6

step-by-step explanation:

the ratio of the perimeters is equal to the scale.

the ratio of the areas is equal to the square of the scale.

34 / 20 = k

a / 19.6 = k²

substituting:

a / 19.6 = (34 / 20)²

a = 19.6 (34 / 20)²

a ≈ 56.6