A 30% cranberry juice drink is mixed with a 100% cranberry juice drink. The function f(x)=(6)(1.0)+x(0.3)6+x models the concentration of cranberry juice in the drink after x gallons of the 30% drink are added to 6 gallons of pure juice. What will be the concentration of cranberry juice in the drink if 2 gallons of 30% drink are added? Give the answer as a percent.

  82.5%

Step-by-step explanation:

It helps to start with the correct formula:

  f(x) = ((6)(1.0) +x(0.3))/(6 +x) . . . . parentheses are required

Then f(2) is ...

  f(2) = (6 +.3(2))/(6+2) = 6.6/8

  f(2) = 82.5%

option 2 is correct.

step-by-step explanation:

we are given two inequality equation

x+y> 5 and -3x+2y< -2

first we draw the graph of both function using t-table method

equation: x+y> 5

for x=0, y=5

for x=-5, y=0

table:

x   :   0     5

y   :   5     0

test point (0,0) , 0> 5

false

equation: -3x+2y< -2

for x=0, y=-1

for x=2, y=4

table:

x   :   0     2

y   : -1       4

test point (0,0) , 0< -2

true

now we draw the graph. see the attachment for graph.

common region would be solution set.

option 2 is correct.