Algebra and Trigonometry (6th Edition)
6th Edition
ISBN: 9780134463216
Author: Robert F. Blitzer
Publisher: PEARSON
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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](https://content.bartleby.com/qna-images/question/66fa70cb-bd96-49f7-bbde-9ceb23da10ec/a9cb5331-24f5-488a-9eae-0477204376db/u3h0hu_processed.png)
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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but why did we have to multiply alpha by 2? was it just an intuitive approach? thank you!
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