Question

The table shows the number of candies packed by Machine J. The equation shows the number of candies packed by Machine K. In both representations, x is a measure of the number of minutes and y is a measure of the number of candies packed.
Machine J Candy Packing
x
(minutes) y
(candies)
3 90
6 180
9 270
12 360


Machine K: y = 26x

How many more candies could machine J pack than machine K in 11 minutes?
30
44
90
330

Answer 1

Answer:

B

Step-by-step explanation:

Answer 2

To find out the answer, treat both machines as lines. The answer can be found when you have equations for each machine. The first step of that would be to make an equation for "line" J, so find the slope using two sets of its given coordinates, I'll use (3,90) and (6,180).

$\frac{y_2 - y_1}{x_2 - x_1}$   Substitute in the x and y values
$\frac{180 - 90}{6 - 3}$   Subtract
$\frac{90}{3}$   Divide
30

Now that you know 30 is the slope, you can plug that and one set of coordinates, I'll use (3,90) into point-slope form.

y - y_1 = m(x - x_1)   Substitute in your x and y coordinates and slope
  y - 90 = 30(x - 3)     Use the Distributive Property
  y - 90 = 30x - 90      Add 90 to both sides
         y = 30x   

Now you have the equations for both machines. You want to find out how many candies each machine packages in 11 minutes, and x represent minutes, so plug 11 into the x value for both equations.

Machine J

y = 30x         Substitute
y = 30 (11)   Multiply
y = 330
 
Machine J packs 330 candies in 11 minutes.

Machine K

y = 26x         Substitute
y = 26 (11)    Multiply
y = 286

Machine K packs 286 candies in 11 minutes.

Finally, to find out how many more candies Machine J packs than Machine K, subtract the two values.

330 - 286 = 44

Machine J can pack 44 more candies than Machine K in 11 minutes.