How many different menus showing 10 main courses can a restaurant make if it has 15 main courses from which to choose? 360,360 3,003 6

3003

Step-by-step explanation:

The number of differents menus containing 10 main courses that the restaurant can make if it has 15 main courses from which to chose is calculated through the combination: 15C10. The formula of the combination is: nCr = n! / ((r!) x(n - r)!) 

Where r=10 and n=15

Substituting the values to the equation: 15C10 = 15! / (10!)x(10 - 5)! = 3003

Then there are 3003 different menus that a restaurant can makeif it has 15 main courses from which to choose.

3,003 different menus

Step-by-step explanation:

We must calculate the number of ways in which a restaurant can make a menu of 10 main courses by selecting between 15 dishes.

In this case, the order in which the 10 dishes of the menu are found is not relevant. That is, for this case it is the same to have: {Dish 1, Dish 2 ...} that have {Dish 2, Dish 1 ...}

So this is a problem of combinations.

We must calculate:

Where n is the number of objects you have and select r from them.

In this problem:

So:

Therefore the answer is  3003 different menus

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.

1. 2

2. 11

3. 5

4. 6

5. 7

6. 5

7. -1

8. 1

step-by-step explanation: