Based on the table, which best predicts the end behavior of the graph of f(x)? х -4 -3 -2 -1 o 1 2 3 f(x) 18 9 6 3 0 -3 -6 -9 As x-f(x) +, and as x>-00, f(x) — . As x— , f(x), and as x -f(x) — As x .f(x) - and as x -- FOX) As X -f(x) -00, and as x -- (x)

7.25

step-by-step explanation:

we know that 7 equals to 7.0. we also know that 1/4=0.25. so we can add the two together and we get 7.25

answer:   35 % is the probability of a late departure or an early arrival.

step-by-step explanation:

let l represents the late departure and e represents the early departure.

then, according to the question,

p(l) = 12 %,   p(e) = 27 % and = 4 %

and, we have to find out,

but, we know that,

which is the required equation that will use to find the probability of a late departure or an early departure.

therefore,   the probability of a late departure or an early arrival =   35 %

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.

a sample of water is taken and kept in a beaker in a freezer at a constant temperature of 0°c. if the system is at dynamic equilibrium, which of these statements is true?