Write down the first 6 elements of the following sequence (where n ∈ Z+), then give a recursive definition for an. Do not forget the base case. (You do not need to prove it is correct). a. an - 3n - 10
b. an= (1+(-1)^n)^n
c. an= 2n! (2)

a. The first six terms are:

-7, -4, -1, 2, 5, 8

b. The first six terms are:

0, 2, 0, 2, 0, 2.

c. The first six terms are:

4, 8, 24, 96, 480, 2880

Step-by-step explanation:

a. an - 3n - 10

For n = 1

a1 = 3(1) - 10

= -7

For n = 2

a2 = 3(2) - 10

= -4

For n = 3

a3 = 3(3) - 10

= -1

For n = 4

a4 = 3(4) - 10

= 2

For n = 5

a5 = 3(5) - 10

= 5

For n = 6

a6 = 3(6) - 10

= 8

The first six terms are:

-7, -4, -1, 2, 5, 8

b. an= (1+(-1)^n)^n

For n = 1

a1 = (1+(-1)^1)^1

= 0

For n = 2

a2 = (1+(-1)^2)^1

= 2

For n = 3

a3 = (1+(-1)^3)^1

= 0

For n = 4

a4 = (1+(-1)^4)^1

= 2

For n = 5

a5 = (1+(-1)^5)^1

= 0

For n = 6

a6 = (1+(-1)^6)^1

= 2

The first six terms are:

0, 2, 0, 2, 0, 2.

c. an= 2n! (2)

For n = 1

a1 = 2(1!)(2)

= 4

For n = 2

a2 = 2(2!)(2)

= 8

For n = 3

a3 = 2(3!)(2)

= 24

For n = 4

a4 = 2(4!)(2)

= 96

For n = 5

a5 = 2(5!)(2)

= 480

For n = 6

a6 = 2(6!)(2)

= 2880

The first six terms are:

4, 8, 24, 96, 480, 2880

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:

x -

median = 1.25

range = 12-14.5

interquartile = q12-q14.5

y -

median = 1

range = 15-17

interquartile = q15-q17