Captain William has a ship, the H. M.S Crimson Lynx. The ship is five furlongs from the dread pirate Emily and her merciless band of thieves.
If his ship hasn't already been hit, Captain William has probability \dfrac{1}{3}
3
1

start fraction, 1, divided by, 3, end fraction of hitting the pirate ship. If his ship has been hit, Captain William will always miss.
If her ship hasn't already been hit, dread pirate Emily has probability \dfrac{1}{7}
7
1

start fraction, 1, divided by, 7, end fraction of hitting the Captain's ship. If her ship has been hit, dread pirate Emily will always miss.
If the Captain and the pirate each shoot once, and the Captain shoots first, what is the probability that the Captain hits the pirate ship, but the pirate misses?

(1, –1), (16, –4), (36, –6)

step-by-step explanation:

given that the function is

this can also be written as

we have to graph this function and check which list of points lie on the graph

see attachment for graph:

we find that the points lying are (1,-1) (16,-4) and (36,-6)

since square root is there we find that x can only be posiitve but y can be since square root both positive and negative.

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.