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If 2 ak = 4 and E bk = 42, find the following values. k= 1 k= 1 n b Σ 38 Σ Σ E (ak + bx). E (ak - bk). E (bx-Sak) 42 k = 1 k =1 k= 1 k = 1 k = 1 2 3ak k = 1 (Simplify your answer.) n b. Σ 42 k = 1 (Simplify your answer.) Σ k = 1 (Simplify your answer.) Σ k = 1 (Simplify your answer.) E (- 9ak) =U k=1 (Simplify your answer.)

If 2 ak = 4 and E bk = 42, find the following values. k= 1 k= 1 n b Σ 38 Σ Σ E (ak + bx). E (ak - bk). E (bx-Sak) 42 k = 1 k =1 k= 1 k = 1 k = 1 2 3ak k = 1 (Simplify your answer.) n b. Σ 42 k = 1 (Simplify your answer.) Σ k = 1 (Simplify your answer.) Σ k = 1 (Simplify your answer.) E (- 9ak) =U k=1 (Simplify your answer.)

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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If 2 ak = 4 and bx = 42, find the following values. k= 1 k=1 n b Σ 38 Σ 42 k = 1 E (ak + bx), E (ak - bR). E (bx - Sak) k = 1 Σ k =1 k = 1 k= 1 2 3ak k = 1 (Simplify your answer.) n b. Σ 42 k = 1 (Simplify your answer.) Σ k=1 (Simplify your answer.) Σ (ek - be) =0 k= 1 (Simplify your answer.) E (bx - 9ak) = D k = 1 (Simplify your answer.)
Transcribed Image Text:If 2 ak = 4 and bx = 42, find the following values. k= 1 k=1 n b Σ 38 Σ 42 k = 1 E (ak + bx), E (ak - bR). E (bx - Sak) k = 1 Σ k =1 k = 1 k= 1 2 3ak k = 1 (Simplify your answer.) n b. Σ 42 k = 1 (Simplify your answer.) Σ k=1 (Simplify your answer.) Σ (ek - be) =0 k= 1 (Simplify your answer.) E (bx - 9ak) = D k = 1 (Simplify your answer.)
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