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A. y = -3 sin (2x -) B. y=2 sin (x +프)-4 C. y =cos 4x D. y = -2 cos(2x - n) E. y=tan(금x +) -2 F. y = tan(x +) G. y = 2 csc(x – T) H. y = csc x +3 1. y =sec 2x – 3 J. y = - sec(x- n) y = cot (2x +) - 1 L. y = -3(cot x – 2) + 1 К. |

A. y = -3 sin (2x -) B. y=2 sin (x +프)-4 C. y =cos 4x D. y = -2 cos(2x - n) E. y=tan(금x +) -2 F. y = tan(x +) G. y = 2 csc(x – T) H. y = csc x +3 1. y =sec 2x – 3 J. y = - sec(x- n) y = cot (2x +) - 1 L. y = -3(cot x – 2) + 1 К. |

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Study the properties of the given circular functions then answer the following questions. (Choices are in the picture given below) Which of the functions stated below (picture) exhibit/s the given property? There may be multiple answers in a question and answers may be repeated. 1. Stretches horizontally 2. Compresses vertically 3. No horizontal shift/No phase shift 4. Range: [ -2,4 ] 5. Range is real numbers Thank you!
A. y = -3 sin (2x -) B. y = 2 sin (x +) – 4 C. y = cos 4x D. y = -2 cos(2x - T) E. y = -tan (x +) - 2 F. y = tan(x +) G. y = 2 csc(x - 1) H. y = cscx+3 + 1 I. y = sec 2x – 3 J. y = - sec(;x– n) K. y = cot (2x +) – 1 L. y = -3(cot x – 2)
Transcribed Image Text:A. y = -3 sin (2x -) B. y = 2 sin (x +) – 4 C. y = cos 4x D. y = -2 cos(2x - T) E. y = -tan (x +) - 2 F. y = tan(x +) G. y = 2 csc(x - 1) H. y = cscx+3 + 1 I. y = sec 2x – 3 J. y = - sec(;x– n) K. y = cot (2x +) – 1 L. y = -3(cot x – 2)
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