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