Algebra 1

Published by Prentice Hall
ISBN 10: 0133500403
ISBN 13: 978-0-13350-040-0

Chapter 10 - Radical Expressions and Equations - 10-2 Simplifying Radicals - Practice and Problem-Solving Exercises: 48

Answer

a) height 150 ft = distance 15 miles height 225 ft = distance 18 miles height 300 ft = distance 21 miles b) Therefore as the height increases by 75 feet, the distance increases by about 3 miles.

Work Step by Step

$d = \sqrt \frac{3h}{2}$ a) Height : 150 ft We substitute the height into the equation. $d = \sqrt \frac{3(150)}{2}$ $\sqrt 225$ = 15 miles Height : 225 ft We substitute the height into the equation. $d = \sqrt \frac{3(225)}{2}$ $\sqrt 337.5$ =18 miles Height : 300 ft We substitute the height into the equation. $d = \sqrt \frac{3(300)}{2}$ $\sqrt 450$ =21 miles b) As the height goes from 150 to 225 the height increases by 75 feet, and the distances goes from 15 to 18 miles, the distance increases by about 3 miles. Therefore as the height increases by 75 feet, the distance increases by about 3 miles.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.