Question

Section 6.3 = Trigonometric Equations ? < 6.5.147 O Points: 0 of 1 The number of hours of daylight in City A is given by the following equation, where \(x\) is the number of days after January 1. Within a year, when does City A have 11.5 hours of daylight? Give your answer in days after January 1. \(y = 3 \sin \left[ \frac{2\pi}{365} (x - 79) \right] + 12\) days (Use a comma to separate answers as needed. Round to the nearest integer as needed.)

          Section 6.3
= Trigonometric
Equations
?
< 6.5.147
O Points:
0 of 1
The number of hours of daylight in City A is given by the following equation, where \(x\) is the
number of days after January 1. Within a year, when does City A have 11.5 hours
of daylight? Give your answer in days after January 1.
\(y = 3 \sin \left[ \frac{2\pi}{365} (x - 79) \right] + 12\)
days
(Use a comma to separate answers as needed. Round to the nearest integer as needed.)

Section 6.3
= Trigonometric
Equations
?
< 6.5.147
O Points:
0 of 1
The number of hours of daylight in City A is given by the following equation, where x is the
number of days after January 1. Within a year, when does City A have 11.5 hours
of daylight? Give your answer in days after January 1.
y = 3 sin[ (2π)/(365) (x - 79) ] + 12
days
(Use a comma to separate answers as needed. Round to the nearest integer as needed.)

Added by Alexander A.

Introductory and Intermediate Algebra for College Students 4th

Robert Blitzer 4th Edition

Instant Answer

Solved by Expert Arjun Singh

01/06/2022

Step 1

5 11.5 = 3 sin[2π/365(x-79)] + 12 Show more…

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Section 6.3 = Trigonometric Equations → 6.5.147 O Points: 0 of 1 The number of hours of daylight in City A is given by the following equation, where x is the number of days after January 1. Within a year, when does City A have 11.5 hours of daylight? Give your answer in days after January 1. 2π y=3 sin days 365 (x-79) +12 (Use a comma to separate answers as needed. Round to the nearest integer as needed.) Set y = 11.5 = 3 Subtract -0.5 = Divide b -0.166 To find a