A menu offers a choice of 2 salads, 6 main dishes, and 5 desserts. How many different meals consisting of one salad, one main dish, and one dessert are possible?

Answer is 60

Step-by-step explanation:

The possibility of a salad meal is =2

2C1 = 2

The possibility of a main dish = 6

6C1 = 6

The possibility of a dessert is = 5

5C1 = 5

So the total possibility is the multiplication of the three i.e

2*6*5 = 60

The principle applied is the combination principle.

Note that C stands for combination

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

⭐dhruv⭐

2 x 503

step-by-step explanation:

its 1006