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Chapter 8: Problem 69

Suppose that a highway rises a distance of 135 feet in a horizontal distance of 2640 feet. Express the grade of the highway to the nearest tenth of a percent.

Short Answer

Expert verified
The grade of the highway is 5.1%.

Step by step solution

01

Understand the Grade Formula

The grade of a highway is the ratio of the vertical rise to the horizontal run, expressed as a percentage. The formula to find the grade is: \[ \text{Grade ( ext{%})} = \left( \frac{\text{Vertical Rise}}{\text{Horizontal Run}} \right) \times 100 \].
02

Substitute the Given Values

In this problem, the vertical rise is given as 135 feet and the horizontal run is 2640 feet. Substitute these values into the formula: \[ \text{Grade ( ext{%})} = \left( \frac{135}{2640} \right) \times 100 \].
03

Perform the Division

Calculate the division inside the parentheses to find the ratio: \( \frac{135}{2640} = 0.051136 \).
04

Convert to Percentage

Multiply the result from the division by 100 to express the ratio as a percentage: \( 0.051136 \times 100 = 5.1136 \text{%} \).
05

Round to the Nearest Tenth

Round the percentage to the nearest tenth of a percent. 5.1136% rounded to the nearest tenth is 5.1%. Therefore, the grade of the highway is 5.1%.

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Key Concepts

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

Grade Calculation
Grade calculation involves determining how steep a surface or slope is, often in terms of a percentage. It is especially useful in contexts like roads, railroads, or ramps. When calculating grade, you are finding a ratio that represents how much one thing rises compared to how much it extends along the ground.
The most common formula used is:
  • \( \text{Grade (\%)} = \left( \frac{\text{Vertical Rise}}{\text{Horizontal Run}} \right) \times 100 \)
To apply this, you need two key pieces of information:
  • The vertical rise, which is the height gained.
  • The horizontal run, which is the distance over the ground.
By dividing these values and multiplying by 100, you convert the fraction into a percentage. This percentage reflects the steepness of the slope. A higher percentage means a steeper incline.
Percentage Conversion
Converting a fraction or decimal to a percentage is a common task in math problems. Percentages are a way of expressing numbers as parts of a hundred, making it easier to understand proportions. To convert a decimal into a percentage, simply multiply the decimal by 100.
For example, if you have a calculated ratio of \(0.051136\), you multiply by 100 to get \(5.1136\) which is 5.1136%. This represents the proportion in terms of one hundred parts.
  • Multiply the decimal by 100.
  • Attach the \(\%\) symbol to express it as a percentage.
In the specific exercise, after calculating the division \( \frac{135}{2640} = 0.051136 \), we then converted this result to a percentage to better communicate the steepness of that segment of the highway.
Algebraic Formulas
Algebraic formulas are the foundation of solving mathematical problems effectively and systematically. They provide a structured way to deal with numbers and symbols, enabling you to find solutions to a wide range of problems.
When you see a problem, recognizing which algebraic formula applies can save time and ensure accuracy. For instance, the grade formula \( \text{Grade (\%)} = \left( \frac{\text{Vertical Rise}}{\text{Horizontal Run}} \right) \times 100 \) is specific to calculating slope steepness in percentage terms.
  • Identify the correct formula for the situation.
  • Substitute the given numbers into the formula.
  • Follow the operations, respecting the order of arithmetic processes (division first, then multiplication in this instance).
Steps like these turn potentially complex calculations into a series of more manageable tasks, making problems easier to solve and understand.

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Most popular questions from this chapter

Find the equation of the line that contains the two given points. Express equations in the form \(A x+B y=C\), where \(A, B\), and \(C\) are integers. (Objective \(1 \mathrm{~b}\) ) \((-1,-2)\) and \((-6,-7)\)

Determine the slope and \(y\) intercept of the line represented by the given equation, and graph the line. (Objective 2) $$-5 x-13 y=26$$

Find the equation of the line that contains the two given points. Express equations in the form \(A x+B y=C\), where \(A, B\), and \(C\) are integers. (Objective \(1 \mathrm{~b}\) ) \((2,3)\) and \((7,10)\)

Find the equation of the line with the given slope and \(y\) intercept. Leave your answers in slope-intercept form. (Objective 1a) \(m=-\frac{7}{12}\) and \(b=-\frac{2}{3}\)

Find the equation of the line that contains the two given points. Express equations in the form \(A x+B y=C\), where \(A, B\), and \(C\) are integers. (Objective \(1 \mathrm{~b}\) ) \((-2,0)\) and \((0,-9)\)
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