randRange( 2, 10 ) randRangeExclude( 2, 15, [ N1 ] ) localeToFixed( N1 * randRange( 80, 199 ) / 100, 2) "\\dfrac{" + N1 + "}{\\$" + C + "} = \\dfrac{" + N2 + "}{x}"

`N1` plural_form(deskItem(0), N1) cost `\$C`.

Which equation would help determine the cost of `N2` plural_form(deskItem(0), N2)?

`SOLUTION`

- `\dfrac{N2}{\$C} = \dfrac{x}{N1}`
- `\dfrac{N2}{N1} = \dfrac{\$C}{x}`
- `\dfrac{N1}{N2} = \dfrac{x}{\$C}`
- `\dfrac{x}{N2} = \dfrac{N1}{\$C}`
- `\dfrac{N2}{x} = \dfrac{\$C}{N1}`

We can write the fact that `N1` plural_form(deskItem(0), N1) cost `\$C` as a proportion:

`\qquad \dfrac{N1}{\$C}`

Let `x` represent the unknown cost of `N2` plural_form(deskItem(0), N2). Since `N2` plural_form(deskItem(0), N2) cost `x`, we have the following proportion:

`\qquad \dfrac{N2}{x}`

The cost changes along with the number of plural_form(deskItem(0)) purchased, and so the two proportions are equivalent.

`\qquad SOLUTION`

"\\dfrac{" + N2 + "}{x} = \\dfrac{" + N1 + "}{\\$" + C + "}"

Let `x` represent the unknown cost of `N2` plural_form(deskItem(0), N2). Since `N2` plural_form(deskItem(0), N2) cost `x`, we have the following proportion:

`\qquad \dfrac{N2}{x}`

We can write the fact that `N1` plural_form(deskItem(0), N1) cost `\$C` as a proportion:

`\qquad \dfrac{N1}{\$C}`

The cost changes along with the number of plural_form(deskItem(0)) purchased, and so the two proportions are equivalent.

`\qquad SOLUTION`

"\\dfrac{" + N1 + "}{" + N2 + "} = \\dfrac{\\$" + C + "}{x}"

We know the cost of `N1` plural_form(deskItem(0), N1). We want to know the cost of `N2` plural_form(deskItem(0), N2). We can write the numbers of plural_form(deskItem(0)) as a proportion:

`\qquad \dfrac{N1}{N2}`

We know `N1` plural_form(deskItem(0), N1) costs `\$C`. We can let `x` represent the unknown cost of `N2` plural_form(deskItem(0), N2). The proportion of these costs can be expressed as:

`\qquad \dfrac{\$C}{x}`

The cost changes along with the number of plural_form(deskItem(0)) purchased, and so the two proportions are equivalent.

`\qquad SOLUTION`

"\\dfrac{x}{" + N2 + "} = \\dfrac{\\$" + C + "}{" + N1 + "}"

If we let `x` represent the cost of `N2` plural_form(deskItem(0), N2), we have the following proportion:

`\qquad \dfrac{x}{N2}`

We have to pay `\$C` for `N1` plural_form(deskItem(0), N1), and that can be written as a proportion:

`\qquad \dfrac{\$C}{N1}`

Since the price per deskItem(0) stays the same, these two proportions are equivalent.

`\qquad SOLUTION`