Algebra and Trigonometry: Structure and Method, Book 2
Algebra and Trigonometry: Structure and Method, Book 2
2000th Edition
ISBN: 9780395977255
Author: MCDOUGAL LITTEL
Publisher: McDougal Littell
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Concept explainers

Unitary Method
The word “unitary” comes from the word “unit”, which means a single and complete entity. In this method, we find the value of a unit product from the given number of products, and then we solve for the other number of products.
Speed, Time, and Distance
Imagine you and 3 of your friends are planning to go to the playground at 6 in the evening. Your house is one mile away from the playground and one of your friends named Jim must start at 5 pm to reach the playground by walk. The other two friends are 3 …
Profit and Loss
The amount earned or lost on the sale of one or more items is referred to as the profit or loss on that item.
Units and Measurements
Measurements and comparisons are the foundation of science and engineering. We, therefore, need rules that tell us how things are measured and compared. For these measurements and comparisons, we perform certain experiments, and we will need the experime…
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Question
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Chapter 3.6, Problem 20P
To determine

To calculate:The speed of the current in the given situation.

Expert Solution & Answer
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Answer to Problem 20P

The speed of the current is 1 mile/hour.

Explanation of Solution

Given Information: It is given that canoeists paddled upstream for 1 hour and then paddled downstream to the starting point and the log covers a distance of 2 miles at the same time.

Formula used: Distance = speed × time

Calculation:

Let the speed of canoeists in still water= x miles/hr.

Speed of the current = y miles/hr.

Upstream speed=( x − y )miles/hr.

Downstream speed = ( x + y ) miles/hr.

Distance covered by canoeist in 1 hour = ( x - y ) miles

And taken by log to reach the starting point is

Time = distancespeed

Time = 2y hours

Now time is taken to paddle downstream = ( 2y−1 ) hours

Downstream distance covered = (x+y)×(2−yy)

Since from the given situation,it can be said that the difference between the downstream distance covered and upstream distance covered in 1 hour is 2 miles.

Therefore,

  (x+y)×(2−yy)−(x−y) = 2

  2x+2y−xy−y2−xy+y2y=2

  2x+2y−xy−y2−xy+y2 = 2y

  2x−2xy = 0

  2x(1−y)=0

  x=0

  y=1

So, the speed of current = 1 mile/hr

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Chapter 3.6, Problem 19P
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Chapter 3.7, Problem 1OE

Chapter 3 Solutions

Algebra and Trigonometry: Structure and Method, Book 2

Ch. 3.1 - Prob. 1OECh. 3.1 - Prob. 2OECh. 3.1 - Prob. 3OECh. 3.1 - Prob. 4OECh. 3.1 - Prob. 5OECh. 3.1 - Prob. 6OECh. 3.1 - Prob. 7OECh. 3.1 - Prob. 8OECh. 3.1 - Prob. 9OECh. 3.1 - Prob. 10OE
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Prob. 2MRECh. 3.9 - Prob. 3MRECh. 3.9 - Prob. 4MRECh. 3.9 - Prob. 5MRECh. 3.9 - Prob. 6MRECh. 3.9 - Prob. 7MRECh. 3.9 - Prob. 8MRECh. 3.9 - Prob. 9MRECh. 3.9 - Prob. 10MRECh. 3.9 - Prob. 11MRECh. 3.9 - Prob. 12MRECh. 3.9 - Prob. 13MRECh. 3.10 - Prob. 1OECh. 3.10 - Prob. 2OECh. 3.10 - Prob. 3OECh. 3.10 - Prob. 4OECh. 3.10 - Prob. 5OECh. 3.10 - Prob. 6OECh. 3.10 - Prob. 7OECh. 3.10 - Prob. 8OECh. 3.10 - Prob. 9OECh. 3.10 - Prob. 10OECh. 3.10 - Prob. 11OECh. 3.10 - Prob. 12OECh. 3.10 - Prob. 13OECh. 3.10 - Prob. 14OECh. 3.10 - Prob. 15OECh. 3.10 - Prob. 16OECh. 3.10 - Prob. 17OECh. 3.10 - Prob. 18OECh. 3.10 - Prob. 19OECh. 3.10 - Prob. 20OECh. 3.10 - Prob. 21OECh. 3.10 - Prob. 22OECh. 3.10 - Prob. 23OECh. 3.10 - Prob. 24OECh. 3.10 - Prob. 1WECh. 3.10 - Prob. 2WECh. 3.10 - Prob. 3WECh. 3.10 - Prob. 4WECh. 3.10 - Prob. 5WECh. 3.10 - Prob. 6WECh. 3.10 - Prob. 7WECh. 3.10 - Prob. 8WECh. 3.10 - Prob. 9WECh. 3.10 - Prob. 10WECh. 3.10 - Prob. 11WECh. 3.10 - Prob. 12WECh. 3.10 - Prob. 13WECh. 3.10 - Prob. 14WECh. 3.10 - Prob. 15WECh. 3.10 - Prob. 16WECh. 3.10 - Prob. 17WECh. 3.10 - Prob. 18WECh. 3.10 - Prob. 19WECh. 3.10 - Prob. 20WECh. 3.10 - Prob. 21WECh. 3.10 - Prob. 22WECh. 3.10 - Prob. 23WECh. 3.10 - Prob. 24WECh. 3.10 - Prob. 25WECh. 3.10 - Prob. 26WECh. 3.10 - Prob. 27WECh. 3.10 - Prob. 28WECh. 3.10 - Prob. 29WECh. 3.10 - Prob. 30WECh. 3.10 - Prob. 31WECh. 3.10 - Prob. 32WECh. 3.10 - Prob. 33WECh. 3.10 - Prob. 34WECh. 3.10 - Prob. 35WECh. 3.10 - Prob. 36WECh. 3.10 - Prob. 37WECh. 3.10 - Prob. 1STCh. 3.10 - Prob. 2STCh. 3.10 - Prob. 3STCh. 3.10 - Prob. 4STCh. 3.10 - Prob. 5STCh. 3.10 - Prob. 6STCh. 3.10 - Prob. 7STCh. 3.10 - Prob. 8STCh. 3.10 - Prob. 9STCh. 3.10 - Prob. 1ECh. 3.10 - Prob. 2ECh. 3.10 - Prob. 3ECh. 3.10 - Prob. 4ECh. 3.10 - Prob. 5ECh. 3 - Prob. 1CRCh. 3 - Prob. 2CRCh. 3 - Prob. 3CRCh. 3 - Prob. 4CRCh. 3 - Prob. 5CRCh. 3 - Prob. 6CRCh. 3 - Prob. 7CRCh. 3 - Prob. 8CRCh. 3 - Prob. 9CRCh. 3 - Prob. 10CRCh. 3 - Prob. 11CRCh. 3 - Prob. 12CRCh. 3 - Prob. 13CRCh. 3 - Prob. 14CRCh. 3 - Prob. 15CRCh. 3 - Prob. 16CRCh. 3 - Prob. 17CRCh. 3 - Prob. 18CRCh. 3 - Prob. 1CTCh. 3 - Prob. 2CTCh. 3 - Prob. 3CTCh. 3 - Prob. 4CTCh. 3 - Prob. 5CTCh. 3 - Prob. 6CTCh. 3 - Prob. 7CTCh. 3 - Prob. 8CTCh. 3 - Prob. 9CTCh. 3 - Prob. 10CTCh. 3 - Prob. 11CTCh. 3 - Prob. 12CTCh. 3 - Prob. 13CTCh. 3 - Prob. 14CTCh. 3 - Prob. 15CTCh. 3 - Prob. 16CTCh. 3 - Prob. 1CLRCh. 3 - Prob. 2CLRCh. 3 - Prob. 3CLRCh. 3 - Prob. 4CLRCh. 3 - Prob. 5CLRCh. 3 - Prob. 6CLRCh. 3 - Prob. 7CLRCh. 3 - Prob. 8CLRCh. 3 - Prob. 9CLRCh. 3 - Prob. 10CLRCh. 3 - Prob. 11CLRCh. 3 - Prob. 12CLRCh. 3 - Prob. 13CLRCh. 3 - Prob. 14CLRCh. 3 - Prob. 15CLRCh. 3 - Prob. 16CLRCh. 3 - Prob. 17CLRCh. 3 - Prob. 18CLRCh. 3 - Prob. 19CLRCh. 3 - Prob. 20CLRCh. 3 - Prob. 21CLRCh. 3 - Prob. 22CLRCh. 3 - Prob. 23CLRCh. 3 - Prob. 24CLRCh. 3 - Prob. 25CLRCh. 3 - Prob. 26CLRCh. 3 - Prob. 27CLRCh. 3 - Prob. 28CLR

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