I need the answer as soon as possible How did he get the7value inside the line16(2+1)(4z-1)(2z-1)11285(+1)96z1685(2+1)5(2z-1)Using above in (7), we get11296z1632-85(2+1)85(2+1)5(27-1)17(4z-1)()()96225(-2+1)112z16z32z%D85(-2415(2z-1)17(4z-1)962+85(z+1)112z8z8z%3D85(+1)5(z-1/2)17(z-1/8)Since we havez- = sin (). Z-A= cos (4)and Z- = a*.%3D2+1%3D

How did he get the value inside the line 162 112 96z 16 - (2+1)(4z-1)(2z-1) 85(2+1) 85(2+1) 5(2z-1) Using above in (7), we get 112 85(2+1) 96z 32 17(4z-1) 16 + - 85(+1) 5(2z-1) 112z 962 16z 32z $5(2+1) 5(2z-1) 17(4z-1) () 112z 962 87 5(z-1/2) 87 17(z-1/8) %3D 85(2+1) 85(2+1) Since we have 2- -sin (불), 2-1=cox (블) and Z- = a. = cos (*) %3D %3D

How did he get the value inside the line 162 112 96z 16 - (2+1)(4z-1)(2z-1) 85(2+1) 85(2+1) 5(2z-1) Using above in (7), we get 112 85(2+1) 96z 32 17(4z-1) 16 + - 85(+1) 5(2z-1) 112z 962 16z 32z $5(2+1) 5(2z-1) 17(4z-1) () 112z 962 87 5(z-1/2) 87 17(z-1/8) %3D 85(2+1) 85(2+1) Since we have 2- -sin (불), 2-1=cox (블) and Z- = a. = cos (*) %3D %3D

Algebra and Trigonometry (6th Edition)

6th Edition

ISBN:9780134463216

Author:Robert F. Blitzer

Publisher:Robert F. Blitzer

ChapterP: Prerequisites: Fundamental Concepts Of Algebra

Section: Chapter Questions

Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...

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Question

I need the answer as soon as possible

How did he get the
7
value inside the line
16
(2+1)(4z-1)(2z-1)
112
85(+1)
96z
16
85(2+1)
5(2z-1)
Using above in (7), we get
112
96z
16
32
-
85(2+1)
85(2+1)
5(27-1)
17(4z-1)
()
()
9622
5(-2+1)
112z
16z
32z
%D
85(-241
5(2z-1)
17(4z-1)
962
+
85(z+1)
112z
8z
8z
%3D
85(+1)
5(z-1/2)
17(z-1/8)
Since we have
z- = sin (). Z-A= cos (4)
and Z- = a*.
%3D
2+1
%3D

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