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Chapter Sequences; Induction; The Binomial Theorem, Question P
College Algebra 10 by Michael Sullivan
College Algebra 10 Edition
Author:
Michael Sullivan
Edition:
10
ISBN:
9780321979476
Publisher:
PEARSON
Subject:
Algebra 2
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Chapter Sequences; Induction; The Binomial Theorem, Question Problem 13AYU

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RECALL: (1) For any natural number , (2) Use the definition in (1) above to obtain: \require{cancel} \dfrac{3!7!}{4!} \\=\dfrac{(3\cdot2\cdot1)(7(6)(5)(4)(3)(2)(1))}{(4)(3)(2)(1)} \\=\dfrac{(3\cdot2\cdot1)(7(6)(5)\cancel{(4)(3)(2)(1))}}{\cancel{(4)(3)(2)(1)}} \\=\dfrac{6 \cdot (42(5))}{1} \\= \dfrac{6\cdot210}{1} \\=1,260
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An equation for the given graph. ( x − 1 ) 2 9 + ( y + 1 ) 2 16 = 1 . Given: Concept used: The equation of the form ( x − h ) 2 b 2 + ( y − k ) 2 a 2 = 1 represents a vertical ellipse with center at ( h , k ) , the vertices at ( h , k ± a ) , and the co-vertices at ( h ± b , k ) . Calculation: The given conic is a vertical ellipse with center at ( 1 , − 1 ) , the vertices at ( 1 , 3 ) , ( 1 , − 5 ) , and the co-vertices at ( − 2 , − 1 ) , ( 4 , − 1 ) . So, ( h , k ) = ( 1 , − 1 ) The vertices are, ( h , k + a ) = ( 1 , 3 ) and ( h , k − a ) = ( 1 , − 5 ) ( 1 , − 1 + a ) = ( 1 , 3 ) and ( 1 , − 1 − a ) = ( 1 , − 5 ) − 1 + a = 3 and − 1 − a = − 5 a = 4 And, the co-vertices are ( h − b , k ) = ( − 2 , − 1 ) and ( h + b , k ) = ( 4 , − 1 ) ( 1 − b , − 1 ) = ( − 2 , − 1 ) and ( 1 + b , − 1 ) = ( 4 , − 1 ) 1 − b = − 2 and 1 + b = 4 b = 3 Hence the equation is, ( x − 1 ) 2 3 2 + ( y − ( − 1 ) ) 2 4 2 = 1 ( x − 1 ) 2 9 + ( y + 1 ) 2 16 = 1
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Question Text
Chapter Sequences; Induction; The Binomial Theorem, Question Problem 13AYU
TopicAll Topics
SubjectAlgebra 2
ClassClass 11
Answer TypeText solution:1