Two trucks leave chicago at the same time traveling in opposite directions. one travels at an average of 64mph and the other at 78mph. in how many hours will they be 284 miles apart? let x =time of travel for the first truck and y =time of travel for the second truck. then the system that models the problelm is x=y 64x+78y=284

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.

third choice down

step-by-step explanation:

the first inequality is based on the number of adults and campers, and the second inequality is based on the cost of feeding both the adult and the camper.

we know that there is room for 525 people max.   it's ok to have less than 525, but it's not ok to have more than 525, so the inequality is less than or equal to.   the number combination of adults (x) and campers (y) cannot exceed this 525 max.   therefore:

x + y ≤ 525

now we talk about cost.   it costs $9 to feed one adult, so depending upon the number of adults you have (x) depends upon how much it will cost to feed that number of adults.   same with the campers, who cost $5 to feed.   the adults can be represented by 9x and the campers can be represented by 5y.   since the budget cannot exceed 1,200, the second inequality is:

9x + 5y ≤ 1200