Amath club has 4 female members and 6 male members, 3 offices including a chairperson, a secretary and a treasurer must be filled. assume that each person can fill at most one office. in how many different ways can the three offices be filled so that both female and male are included?

576

Step-by-step explanation:

Number of female members = 4

Number of male members = 6

Number of positions = 3

Both female and male has to be included which means that either 2 females and 1 male are there or 1 male or 2 females.

So, Case - 1 - 2 females and 1 male

Possible cases = , where 3! is done to permutate the members to hold various positions.

So, Case - 2 - 1 female and 2 males

Possible cases = , where 3! is done to permutate the members to hold various positions.

Total ways = 216 + 360 = 576

for this pair of functions g(x)=x-2 and f(x)=|x| is (g o f)(a)= |a| -2

step-by-step explanation:

g(x)=x-2

f(x)=|x|

(g o f)(a)=?

(g o f)(x)=g(f(x))

(g o f)(x)=g(|x|)

(g o f)(x)=|x|-2

(g o f)(a)=|a|-2

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

the factor is 2

step-by-step explanation:

let f(x)=8(3^(x−2)+2)

to stretch a graph by 3, it would be 3f(x)

g(x) =3f(x)

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

x     −2     0     2     4     6

f(x) 1/64 1/16   1/4     1     4

1/64 * z * z = 1/16

i multiplies by z twice because it its 2 steps from -2 to 0

1/64 z^2 = 1/16

multiply by 64 to clear the fraction on the left side

1/4 * 64 z^2 =1/16 *64

z^2 = 4

take the square root of each side

sqrt(z^2) = ±sqrt(4)

z = ±2  

we are multiplying by 2 each time.