The Celsius to Fahrenheit formula: F=95C+32


Solve this formula for C. Check all answers that are true.

(4 points)
Responses

First step is to multiply by 9 on each side.

First step is to add 32 to both sides.

The next step from F - 32 = 9/5 C, could be to multiply both sides by 5

The next step from F - 32 = 9/5 C, could be to divide both sides by 5

Solution: 5/9F - 32 = C

To solve this equation for C, you would have to distribute at some point.

1 answer

Let's start with the formula for converting Celsius to Fahrenheit:

\[ F = \frac{9}{5}C + 32 \]

We want to solve for \( C \). We'll go through the steps to rearrange the equation:

  1. Subtract 32 from both sides: \[ F - 32 = \frac{9}{5}C \]

  2. Now, multiply both sides by \( \frac{5}{9} \) to solve for \( C \): \[ C = \frac{5}{9}(F - 32) \]

This can also be written as: \[ C = \frac{5}{9}F - \frac{5 \times 32}{9} \]

Now, let’s look at the statements and identify the true ones based on our steps:

  1. First step is to multiply by 9 on each side.
    False. The first step is to subtract 32 from both sides.

  2. First step is to add 32 to both sides.
    False. We need to subtract 32 from both sides.

  3. The next step from F - 32 = \(\frac{9}{5}C\), could be to multiply both sides by 5.
    False. The next step should be to multiply both sides by \(\frac{5}{9}\) to isolate \( C \).

  4. The next step from F - 32 = \(\frac{9}{5}C\), could be to divide both sides by 5.
    False. We should not divide both sides by 5; instead, we multiply by \(\frac{5}{9}\).

  5. Solution: \(\frac{5}{9}F - 32 = C\)
    False. This expression is incorrect as it does not correctly represent \( C \). The correct representation is \( C = \frac{5}{9}(F - 32) \).

  6. To solve this equation for C, you would have to distribute at some point.
    True. After multiplying by \(\frac{5}{9}\), you would distribute when calculating \( C \) if you were rearranging that expression further.

So, the only true statement from the options given is:
To solve this equation for C, you would have to distribute at some point.