A restaurant sells pizzas for which you can choose from nine toppings.
(a) How many different types of pizza can the restaurant serve?
different types
(b) What is the minimum number of toppings the restaurant must provide if it wishes to offer at least 2,000 different types of pizza?
toppings

A) 362880

B) 4

Step-by-step explanation:

Given that a customer can choose from nine toppings

To find different types of pizza the restaurant serve, we will permutate it. That is 9 factorial

9! = 362880 different types of pizza

B) if it wishes to offer at least 2,000 different types of pizza, then,

9 permutation x is equal to 2000. That is 9Px = 2000

Let's find x

9!/(9-x)! = 2000

(9 - x)! = 9!/2000

(9-x)! = 181.44

Let assume x = 4

(9 - 4)! = 181.44

5! = 181.44

120 = 181.44

The two numbers are very close. Therefore, 4 is the minimum number of toppings the restaurant must provide if it wishes to offer at least 2,000 different types of pizza?

toppings

the transverse axis of a hyperbola is the line that contains the two vertices and the two focuses.

in this example (hyperbola of equation x^2/2 - y^2/4 = 1), the transverse axis is the  x-axis  . 

step-by-step explanation:

the correct functions are

h(x) = 3x + 2  

g(x) = 2^x

step-by-step explanation:

i proceed to calculate the different values ​​of k (x)

case 1)  k(x) = g(x) ∘ h(x)

g(x) ∘ h(x)=g(h(x))

k(x) = [2]^[3x + > is the option  4. k(x) = 2^(3x + 2 )

case 2)  k(x) = g(x) + h(x)

k(x) = [2^x]+[3x + 2]=2^x+3x+> is the option  1.) k(x) = 2^x + 3x + 2

case 3)k(x) = h(x) ÷ g(x)

k(x) =  [3x + 2]/[2^x]=3x/(2^x)+2/(2^x)=3x*(2^-x)+2^(1-x)

is the option  3.)   k(x) = (3x)2^(-x) + 2^(-x + 1 )

case 4)k(x) = g(x) - h(x)

k(x) =  [2^x]-[3x + 2]=2^x-3x-> is the option  2.) k(x) = 2^x - 3x - 2 

case 5)  k(x) = g(x) × h(x)

k(x) = [2^x]*[3x + > is the option  5.) k(x) = 2^x(3x + 2)

case 6)k(x) = h(x) ∘ g(x)

h(x) ∘ g(x)=h(g(x))

k(x)=3*[2^x]+> is the option  6. k(x) = 3(2^x) +2