Holt Mcdougal Larson Pre-algebra: Student Edition 2012

1st Edition

ISBN: 9780547587776

Author: HOLT MCDOUGAL

Publisher: HOLT MCDOUGAL

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Concept explainers

Sequence and Series

A sequence is defined as a collection of things. Series is defined to sum the things one by one in the sequence. It was invented by German mathematician Carl Friedrich Gauss.

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Question

Chapter 1.6, Problem 34E

(a).

To determine

To Calculate: The change in temperature during various production stages of ice cream

(a).

Expert Solution

Answer to Problem 34E

The absolute values of changes in temperature are:

Pasteurization (Stage I) - Aging (Stage II): 85oC

Aging (Stage II) − Hardening (Stage III): 35oC

Hardening (Stage III) − Storage (Stage-IV): 25oC

Explanation of Solution

Given Information:

Pasteurization (Stage I) temperature is 80oC .
Aging (Stage II) temperature is −5oC .
Hardening (Stage III) temperature is −40oC .
Storage (Stage IV) temperature is −15oC .

Method:

The change in a quantity is found by subtracting the initial value of the quantity from the value after change.
To subtract an integer, add its opposite.

Stage I-II: The change in the temperature is −5−(80) and the opposite of 80 is −80 .

Therefore, write the expression −5−(80) as −5+(−80) .

Further write the expression as −(5+80) and 5+80 is 85 .

So, the change in temperature is −85oC and therefore, the absolute value of the change in temperature is 85oC .

Stage II-III: The change in the temperature is −40−(−5) and the opposite of −5 is 5 .

Therefore, write the expression −40−(−5) as −40+5 .

Meanwhile −40+5 is equal to −35 .

So, the change in temperature is −35oC and therefore, the absolute value of the change in temperature is 35oC .

Stage II-III: The change in the temperature is −15−(−40) and the opposite of −40 is 40 .

Therefore, write the expression −15−(−40) as −15+40 .

Meanwhile −15+40 is equal to 25 .

So, the change in temperature is 25oC and therefore, the absolute value of the change in temperature is also 25oC .

(b).

To determine

To Compare: The absolute values of changes in temperature during various stages of ice cream production and identify the greatest absolute change.

(b).

Expert Solution

Answer to Problem 34E

The absolute change in temperature is the maximum between Stage I(Pasteurization) and Stage II(Aging). The value of the change is 85oC .

Explanation of Solution

Given Information: The absolute values of the temperature calculated from the previous sub-part.

Stage I-II: 85oC

Stage II-III: 35oC

Stage III-IV: 25oC

Among 85, 35, and 25, the integer 85 is the highest. Therefore, the highest absolute change in temperature of 85oC is between the corresponding consecutive processes of Pasteurization (Stage I) and Aging (Stage II).

(c).

To determine

To Calculate: The difference in the temperature of mouth and the temperature of ice cream just out of freezer.

(c).

Expert Solution

Answer to Problem 34E

The temperature of the mouth is greater than the temperature of the ice cream, just out of freezer, by about 52oC .

Explanation of Solution

Given Information: The temperature of the mouth is 99oF . The storage temperature of ice cream is −15oC .

The formula to convert temperature form Fahrenheit scale to Celsius scale is

C=5(F−32)9

Method:

Put the value of the temperature of mouth given in Fahrenheit scale in the above formula to convert it to Celsius scale.
To find about how much warmer is the mouth compared to the stored ice cream, subtract the temperature of the stored ice cream from the temperature of mouth.

Put the given value of the mouth temperature 99oF in the formula to get

C=5(99−32)9=37oC(approximately)

The difference in temperature is then given by 37−(−15) .

Meanwhile, the opposite of −15 is 15 .

So, write 37−(−15) as 37+15 and 37+15 is in fact equal to 52 .

Therefore, the temperature of the mouth is greater than the temperature of stored ice cream by about approximately 52oC .

Chapter 1.6, Problem 33E

Chapter 1.6, Problem 35E

Chapter 1 Solutions

Holt Mcdougal Larson Pre-algebra: Student Edition 2012

Ch. 1.1 - Prob. 1C Ch. 1.1 - Prob. 2C Ch. 1.1 - Prob. 3C Ch. 1.1 - Prob. 4C Ch. 1.1 - Prob. 1E Ch. 1.1 - Prob. 2E Ch. 1.1 - Prob. 3E Ch. 1.1 - Prob. 4E Ch. 1.1 - Prob. 5E Ch. 1.1 - Prob. 6E

Ch. 1.1 - Prob. 7E Ch. 1.1 - Prob. 8E Ch. 1.1 - Prob. 9E Ch. 1.1 - Prob. 10E Ch. 1.1 - Prob. 11E Ch. 1.1 - Prob. 12E Ch. 1.1 - Prob. 13E Ch. 1.1 - Prob. 14E Ch. 1.1 - Prob. 15E Ch. 1.1 - Prob. 16E Ch. 1.1 - Prob. 17E Ch. 1.1 - Prob. 18E Ch. 1.1 - Prob. 19E Ch. 1.1 - Prob. 20E Ch. 1.1 - Prob. 21E Ch. 1.1 - Prob. 22E Ch. 1.1 - Prob. 23E Ch. 1.1 - Prob. 24E Ch. 1.1 - Prob. 25E Ch. 1.1 - Prob. 26E Ch. 1.1 - Prob. 27E Ch. 1.1 - Prob. 28E Ch. 1.1 - Prob. 29E Ch. 1.1 - Prob. 30E Ch. 1.1 - Prob. 31E Ch. 1.1 - Prob. 32E Ch. 1.1 - Prob. 33E Ch. 1.1 - Prob. 34E Ch. 1.1 - Prob. 35E Ch. 1.1 - Prob. 36E Ch. 1.1 - Prob. 37E Ch. 1.1 - Prob. 38E Ch. 1.1 - Prob. 39E Ch. 1.1 - Prob. 40E Ch. 1.1 - Prob. 41E Ch. 1.1 - Prob. 42E Ch. 1.1 - Prob. 43E Ch. 1.1 - Prob. 44E Ch. 1.1 - Prob. 45E Ch. 1.1 - Prob. 46E Ch. 1.1 - Prob. 47E Ch. 1.1 - Prob. 48E Ch. 1.1 - Prob. 49E Ch. 1.1 - Prob. 50E Ch. 1.1 - Prob. 51E Ch. 1.1 - Prob. 52E Ch. 1.1 - Prob. 53E Ch. 1.1 - Prob. 54E Ch. 1.1 - Prob. 55E Ch. 1.1 - Prob. 56E Ch. 1.1 - 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