There were 900 students enrolled in a high school in 2009 and 1,500 students enrolled in the same high school in 2012. the student enrollment of the high school, p, has increased at a constant rate each year, t, since 2009. a) write the given enrollment numbers as a pair of points in the form (t, p). b) find the slope of the line passing through the pair of points from part a and explain the information the slope gives about the situation. c) write an equation that relates the high school's student enrollment, p, to the number of years since 2009, t. d) predict the high school's enrollment in 2017.

(A)

(B)

(C)

(D)

The high school's enrollment in 2017 is 2500

Step-by-step explanation:

Let's assume time starts since 2009

so, In 2009 , t=0

P=900

In 2012,

t=2012-2009=3

P=1500

(A)

So, we have points as

(B)

we can use slope formula

we can plug values

we know that

slope is rate of change of P with respect to time

so, slope means increase in population is 200 per year

(C)

we can use point slope form of line

(D)

In 2017 ,

t=2017-2009=8

we can plug t=8

and we get

The high school's enrollment in 2017 is 2500

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.

the answer for your question is

step-by-step explanation: