Algebra and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN: 9780134463216
Author: Robert F. Blitzer
Publisher: PEARSON
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Question
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Please explain how the solution became this

A
-4 -6
-1
3
Aa=
-2
2
-12
-4
6
-0-0-0----
JA-A-A-
=
1
2
-4 -6 -12]
-2 -1 -4
2 3 6
= 32a =
A
0
O1 4:41 PM
32
2 = 2a. Thus,
A5a = A4 (Aa) = A4 (2a) = 2A4a = 2A³ (Aa)
= 24³ (2a) = 44³a = 4A² (Aa) = 44² (2a)
= 8A²a = 8A(Aa) = 8A(2a) = 16Aa= 16(2a)
[-64]
1
expand button
Transcribed Image Text:A -4 -6 -1 3 Aa= -2 2 -12 -4 6 -0-0-0---- JA-A-A- = 1 2 -4 -6 -12] -2 -1 -4 2 3 6 = 32a = A 0 O1 4:41 PM 32 2 = 2a. Thus, A5a = A4 (Aa) = A4 (2a) = 2A4a = 2A³ (Aa) = 24³ (2a) = 44³a = 4A² (Aa) = 44² (2a) = 8A²a = 8A(Aa) = 8A(2a) = 16Aa= 16(2a) [-64] 1
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but why did we have to multiply alpha by 2? was it just an intuitive approach? thank you!

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