In the equation g = 312 ÷ α , the variable g can be described best as the 1. number of degrees that a skateboarder turns when making α rotations. 2. total number of groups, g, with α students each that can be made if there are 312 students to be grouped. 3. weight of a bag containing α grapefruits if each piece of fruit weighs 312 grams. 4. total number of goats that can graze on 312 acres if each acre can feed α goats. 5. number of grams of fuel, g, needed to raise the temperature of a solution, α, to a temperature of 312◦F.

Here we have a problem. The skateboarder must necessarily and randomly turn 312°, and its fractions. But in a circle, the rotation cannot follow this pattern.

False

2) The total number of groups, g, with "α" students each that can be made if there are 312 students to be grouped.

[tex]g=\frac{312}{a}[/tex]

1 group --------------- 312 students

2 groups ------------ 156 students

5 groups -------------62.4 students

Even though 312 is divisible for 1,2,3,4 it is not for 5,7,9, and the group is a countable, natural category.

False

3) Weight of a bag containing "α" grapefruits if each piece of fruit weighs 312 grams

[tex]g=\frac{312}{a}[/tex]

g=1 bag with 1 grapefruit-------------- 312 g

g=1 bag with 2 grapefruits ---------- 156 g

That doesn't make sense, since for this description. The best should be g=312a and not g=312/a.

False

4) The total number of goats that can graze on 312 acres if each acre can feed "α" goats.

Since there's a relation

1 acre can feed ----------------- 1 goat

312 acres can feed ----------------g

g= 312/1 = 312 acres can feed 1 goat (1 acre for 1 goat)

g=312/2= since 312 acres can feed 156 goats (1 acre for 2 goats)

g =312/3 = 312 acres can feed 104 (1 acre for 3 goats)

Clearly, this function g=312/a does not describe this since the ratio is not the same, as long as we bring more goats to graze on those 312 acres.

False

5) Number of grams of fuel, g, needed to raise the temperature of a solution, α, to a temperature of 312◦F

g= number of grams of a fuel

a= initial temperature of a solution

g=312/a

Let's pick a=100 F initial temperature

g=312/100

g=3.12 grams

Let's now pick 200F as our initial temperature.

g=312/200 g=1.56 grams of solution

The more heat needed to raise, the more fuel necessary. Then True