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 16P
To determine

To Find: The rate of running and swimming

Expert Solution & Answer
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Answer to Problem 16P

The rate of swimming is 4 km/h and the rate of running is 18km/h.

Explanation of Solution

Given Information: It is given that Kevin covers a total of 9 km by swimming for 45 minutes and running for 20 minutes

And next day he covers a total of 14km by swimming for 30 minutes and running for 40 minutes

Formula used:

  Distance = speed×time1 hour = 60 minutes

Calculation:

Let the rate of swimming = x km/h

And rate of running = y km/h

So, on the first day

The distance covered by swimming

  ds=x×(4560)=3x4km

Similarly, the distance covered by running

  dr=y×(2060)=y3km

Since, the total distance covered = 9km

So, 3x4+y3=9   (1)

On the next day,

The distance covered by swimming.

  dw2=x×(3060)=x2km

The distance covered by running:

  dr=y×(4060)=2y3km

Since the total distance covered = 14km

So, x2+2y3=14   (2)

Solving (1) and (2), we have

  9x+4y=108  (3)

  3x+4y=84  (4)

Subtracting (4) from (3), we have

  6x=24

  x=4

Now, putting this value in (4), we have

  12+4y=84

  4y=72

  y=18

Thus, the rate of swimming is 4 km/h and the rate of running is 12 km/h.

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Chapter 3.6, Problem 15P
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Chapter 3.6, Problem 17P

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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