High School Math 2015 Common Core Algebra 1 Student Edition Grade 8/9
High School Math 2015 Common Core Algebra 1 Student Edition Grade 8/9
15th Edition
ISBN: 9780133281149
Author: Prentice Hall
Publisher: Prentice Hall
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Chapter 5.6, Problem 4P
To determine

Todetermine:The equation an architect can enter to represent a second beamwhose graph will pass through the corner at (0,10) and be parallel to the existing beam. To answer in slope intercept form.

Expert Solution & Answer
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Answer to Problem 4P

The architecture can enter the equation y=mx+10 to represent a second beamwhose graph passes through the corner at point (0,10) .

Explanation of Solution

Given information:

The graph of second beampasses through the corner at (0,10) and be parallel to the existing beam.

Formulaused:

The equation of line with slope m and that makes intercept of length c on y-axis is given by

  y=mx+c . The slopes of two or more parallel to lines are equal.

Calculations:

Let OA is the existing beam and CB the beam parallel to the existing beam but passing through corner (0,10) as shown in Figure-1.

Suppose equation of existing beam (line) that makes angle θ with horizontal is y=mx , where m=tanθ is known as slope of bean or line OA.

Then equation of beam (line) CB that is parallel to beam OA is y=mx+c , where c=OC .

Therefore, equation of second beam that architect can enter is y=mx+10 .

In the graphs shown in Figure-1, m=tanθ=2020=1.

Hence, for the structure shown in Figure-1, the architecture can enter the equation y=x+10 to represent a second beamwhose graph passes through the corner at point (0,10) .

  High School Math 2015 Common Core Algebra 1 Student Edition Grade 8/9, Chapter 5.6, Problem 4P

Conclusion:

For the structure shown in Figure-1, the architecture can enter the equation y=x+10 to represent a second beamwhose graph passes through the corner at point (0,10) .

Chapter 5 Solutions

High School Math 2015 Common Core Algebra 1 Student Edition Grade 8/9

Ch. 5.1 - Prob. 7LCCh. 5.1 - Prob. 8PPECh. 5.1 - Prob. 9PPECh. 5.1 - Prob. 10PPECh. 5.1 - Prob. 11PPECh. 5.1 - Prob. 12PPECh. 5.1 - Prob. 13PPECh. 5.1 - Prob. 14PPECh. 5.1 - Prob. 15PPECh. 5.1 - Prob. 16PPECh. 5.1 - Prob. 17PPECh. 5.1 - Prob. 18PPECh. 5.1 - Prob. 19PPECh. 5.1 - Prob. 20PPECh. 5.1 - Prob. 21PPECh. 5.1 - Prob. 22PPECh. 5.1 - Prob. 23PPECh. 5.1 - Prob. 24PPECh. 5.1 - Prob. 25PPECh. 5.1 - Prob. 26PPECh. 5.1 - Prob. 27PPECh. 5.1 - Prob. 28PPECh. 5.1 - Prob. 29PPECh. 5.1 - Prob. 30PPECh. 5.1 - Prob. 31PPECh. 5.1 - Prob. 32PPECh. 5.1 - Prob. 33PPECh. 5.1 - Prob. 34PPECh. 5.1 - Prob. 35PPECh. 5.1 - Prob. 36PPECh. 5.1 - Prob. 37PPECh. 5.1 - Prob. 38PPECh. 5.1 - Prob. 39PPECh. 5.1 - Prob. 40PPECh. 5.1 - Prob. 41PPECh. 5.1 - Prob. 42PPECh. 5.1 - Prob. 43PPECh. 5.1 - Prob. 44PPECh. 5.1 - Prob. 45PPECh. 5.1 - Prob. 46PPECh. 5.1 - Prob. 47PPECh. 5.1 - Prob. 48PPECh. 5.1 - Prob. 49PPECh. 5.1 - Prob. 50PPECh. 5.1 - Prob. 51PPECh. 5.1 - Prob. 52PPECh. 5.1 - Prob. 53PPECh. 5.1 - Prob. 54PPECh. 5.1 - 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