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Chapter 11: Problem 85

Solve each equation for the specified variable. $$ F=\frac{9}{5} C+32 \text { for } C $$

Short Answer

Expert verified
C = \( \frac{5}{9} (F - 32) \)

Step by step solution

01

Subtract 32 from both sides

To isolate the term with the variable on one side, start by subtracting 32 from both sides of the equation: \[ F - 32 = \frac{9}{5} C \]
02

Multiply both sides by 5/9

To solve for \( C \), multiply both sides by the reciprocal of \( \frac{9}{5} \), which is \( \frac{5}{9} \): \[ C = \frac{5}{9} (F - 32) \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Isolating Variables
Isolating a variable means getting the variable alone on one side of the equation. This is a crucial step in solving equations. In the given example, our goal is to solve for \( C \). This equation is initially written as \( F = \frac{9}{5} C + 32 \). To isolate \( C \), we must remove the other terms.

Here are the steps:
  • Subtract 32 from both sides: This will help remove the constant term attached to the variable on one side.
    So, we have \( F - 32 = \frac{9}{5} C \).
  • Multiply both sides by the reciprocal: The reciprocal of \( \frac{9}{5} \) is \( \frac{5}{9} \). This step helps to cancel out the coefficient multiplying the variable.
    So, we end up with \( C = \frac{5}{9} (F - 32) \).
The variable \( C \) is now isolated, and we have our solution!
Linear Equations
A linear equation is an equation where the highest power of the variable is one. These equations often appear as straight lines when graphed. In the provided exercise, we deal with the linear equation \( F = \frac{9}{5} C + 32 \).

To solve linear equations:
  • Identify the variable and the other terms involved.
  • Use basic arithmetic operations (addition, subtraction, multiplication, and division) to isolate the variable. For example, to solve \( F = \frac{9}{5} C + 32 \), we:
    • Subtract 32 from both sides to handle the constant term.
    • Multiply by the reciprocal to isolate the variable.
Linear equations are simpler compared to other types, like quadratic or exponential equations, making them a foundational concept in algebra.
Temperature Conversion
Temperature conversion equations are used to switch from one temperature scale to another. In our exercise, the equation \( F = \frac{9}{5} C + 32 \) is a formula to convert Celsius (\( C \)) to Fahrenheit (\( F \)).

When converting temperatures:
  • From Celsius to Fahrenheit: Use the formula \( F = \frac{9}{5} C + 32 \). This equation means you multiply the Celsius temperature by \( \frac{9}{5} \) and then add 32.
  • From Fahrenheit to Celsius: To do the reverse, isolate \( C \) as we did in the exercise.
    Starting from \( F = \frac{9}{5} C + 32 \), subtract 32 and then multiply by \( \frac{5}{9} \): \[ C = \frac{5}{9} (F - 32) \].
Temperature conversion formulas are practical in many real-world scenarios, such as when traveling or understanding different weather systems.

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Most popular questions from this chapter

Match each quadratic function with the description of the parabola that is its graph. (a) \(f(x)=(x-4)^{2}-2\) (b) \(f(x)=(x-2)^{2}-4\) (c) \(f(x)=-(x-4)^{2}-2\) (d) \(f(x)=-(x-2)^{2}-4\) A. Vertex \((2,-4),\) opens down B. Vertex \((2,-4),\) opens up C. Vertex \((4,-2),\) opens down D. Vertex \((4,-2),\) opens up

Determine the number that will complete the square to solve each equation, after the constant term has been written on the right side and the coefficient of the second-degree term is 1. Do not actually solve. \(3 w^{2}-w-24=0\)

Use the quadratic formula to solve each equation. (All solutions for these equations are real numbers.) $$ x^{2}+3 x-28=0 $$

Evaluate \(\sqrt{b^{2}-4 a c}\) for the given values of \(a, b,\) and \(c\). \(a=1, b=-6, c=9\)

For \(f(x)=a(x-h)^{2}+k,\) in what quadrant is the vertex if (a) \(h>0, k>0\) (b) \(h>0, k<0\) (c) \(h<0, k>0\) (d) \(h<0, k<0 ?\)
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