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If 2 ak = 6 and E bk = 36, find the following values. %3D k= 1 k= 1 n bk E 5ak. E 36 k= 1 E (ax + bx) · E (ak - br). E (bk - 7ax) k=1 k= 1 k= 1 k=1 2 5ax k=1 (Simplify your answer.) n b Σ 36 k=1 (Simplify your answer.) E (ak + bx) k=1 (Simplify your answer.) k= 1 (Simplify your answer.)

If 2 ak = 6 and E bk = 36, find the following values. %3D k= 1 k= 1 n bk E 5ak. E 36 k= 1 E (ax + bx) · E (ak - br). E (bk - 7ax) k=1 k= 1 k= 1 k=1 2 5ax k=1 (Simplify your answer.) n b Σ 36 k=1 (Simplify your answer.) E (ak + bx) k=1 (Simplify your answer.) 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 = 6 and 2 bk = 36, find the following values. k= 1 k = 1 n bk Σ Σ E (Dx -7ax) 2 5ak. 36 (ak + bx). E (ak- br). k=1 k = 1 k= 1 k= 1 k= 1 %3D k = 1 (Simplify your answer.) n br Σ 36 k=1 (Simplify your answer.) k= 1 (Simplify your answer.) k= 1 (Simplify your answer.)
Transcribed Image Text:If 2 ak = 6 and 2 bk = 36, find the following values. k= 1 k = 1 n bk Σ Σ E (Dx -7ax) 2 5ak. 36 (ak + bx). E (ak- br). k=1 k = 1 k= 1 k= 1 k= 1 %3D k = 1 (Simplify your answer.) n br Σ 36 k=1 (Simplify your answer.) k= 1 (Simplify your answer.) k= 1 (Simplify your answer.)
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