Respuesta :
Well for part a all you have to do is add all of the intial calories toghther except for the quarter slab of ribs as for it isnt included in the intial meal
613+532+298+295= 1738 calories
for part b you do a simple y=mx+b equation (C=y and Q=x 1738=b)
C = 328Q + 1738
Plug that equation into a calculator to find a graph and sketch
The slope represents the amount of calories you add onto your initial amount of calories (1738), and the C-intercept represents your starting amount of calories from the original meal
For part E just use your graphing calculator to see where your linear equation is equal to 23 and use the X value at that point as the amount of Quarter slabs the teenager could ingest, assuming you do not have a calculator, the answer is around 2 quarter slabs
The domain and range at that point is (2,2394)
613+532+298+295= 1738 calories
for part b you do a simple y=mx+b equation (C=y and Q=x 1738=b)
C = 328Q + 1738
Plug that equation into a calculator to find a graph and sketch
The slope represents the amount of calories you add onto your initial amount of calories (1738), and the C-intercept represents your starting amount of calories from the original meal
For part E just use your graphing calculator to see where your linear equation is equal to 23 and use the X value at that point as the amount of Quarter slabs the teenager could ingest, assuming you do not have a calculator, the answer is around 2 quarter slabs
The domain and range at that point is (2,2394)