# Chapter 4.1, Problem 57E

### Trigonometry (11th Edition)

11th Edition
Margaret L. Lial + 3 others
ISBN: 9780134217437

Chapter
Section

### Trigonometry (11th Edition)

11th Edition
Margaret L. Lial + 3 others
ISBN: 9780134217437

(a)

To determine

To calculate: The temperature on March 15 (74th day).

Solution: 30.942oF

Explanation

Given: The function of temperature(T) with the number of days(x) is given by

T(x)=37+21sin[2π365(x91)]

where x=1 corresponds to January 1.

Calculation: Now, the corresponding value for March 15 (74th day) will be x=74.

Therefore,

T(74)=37+21sin[2π365(7491)]T(74)=37+21sin[2π365(17)]T(74)=376.058=30.942oF

The temperature on March 15 (74th day) is 30.942oF.

(b)

To determine

To calculate: The temperature on April 5 (95th day).

Solution: 38.44oF

Explanation

Given: The function of temperature(T) with the number of days(x) is given by

T(x)=37+21sin[2π365(x91)]

where x=1 corresponds to January 1.

Calculation: Now, the corresponding value for April 5 (95th day) will be x=95.

Therefore,

T(95)=37+21sin[2π365(9591)]T(95)=37+21sin[2π365(4)]T(95)=37+1.444=38.44oF

The temperature on April 5 (95th day) is 38.44oF.

(c)

To determine

To calculate: The temperature on day 200.

Solution: 57.027oF

Explanation

Given: The function of temperature(T) with the number of days(x) is given by

T(x)=37+21sin[2π365(x91)]

where x=1 corresponds to January 1.

Calculation: Therefore, the corresponding value for day 200 will be x=200.

Therefore,

T(200)=37+21sin[2π365(20091)]T(200)=37+21sin[2π365(109)]T(200)=37+20.027=57.027oF

The temperature on day 200is 57.027oF.

(d)

To determine

To calculate: The temperature on June 25 (176th day).

Solution: 57.878oF

Explanation

Given: The function of temperature(T) with the number of days(x) is given by

T(x)=37+21sin[2π365(x91)]

where x=1 corresponds to January 1.

Calculation: Now June 25 will be 176th day from January 1. Thus, the corresponding value for June25 will be x=176.

Therefore,

T(176)=37+21sin[2π365(17691)]T(176)=37+21sin[2π365(85)]T(176)=37+20.878=57.878oF

The temperature on June 25is 57.878oF.

(e)

To determine

To calculate: The temperature on October 1.

Solution: 36.8193oF

Explanation

Given: The function of temperature(T) with the number of days(x) is given by

T(x)=37+21sin[2π365(x91)]

where x=1 corresponds to January 1.

Calculation: Now October 1 will be 274thday from January 1. Thus, the corresponding value for October1 will be x=274.

Therefore,

T(274)=37+21sin[2π365(27491)]T(274)=37+21sin[2π365(183)]T(274)=370.1807=36.8193oF

The temperature on October1 is 36.8193oF.

(f)

To determine

To calculate: The temperature on December 31.

Solution: 16.0001oF

Explanation

Given: The function of temperature(T) with the number of days(x) is given by

T(x)=37+21sin[2π365(x91)]

where x=1 corresponds to January 1.

Calculation: Now December 31 will be 365th day from January 1. Thus, the corresponding value for December 1 will be x=365.

Therefore,

T(365)=37+21sin[2π365(36591)]T(365)=37+21sin[2π365(274)]T(365)=3720.999=16.001°F

The temperature on December31 is 16.001oF.

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