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ISBN: 9780357540244

Author: EPP

Publisher: CENGAGE L

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Chapter 6.1, Problem 21ES

Let C i = { i , − i } for each nonnegative integer i.

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Chapter 6.1, Problem 20ES

Chapter 6.1, Problem 22ES

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Decomposition geometry: Mary is making a decorative yard space with dimensions as shaded in green (ΔOAB).Mary would like to cover the yard space with artificial turf (plastic grass-like rug). Mary reasoned that she could draw a rectangle around the figure so that the point O was at a vertex of the rectangle and that points A and B were on sides of the rectangle. Then she reasoned that the three smaller triangles resulting could be subtracted from the area of the rectangle. Mary determined that she would need 28 square meters of artificial turf to cover the green shaded yard space pictured exactly.

1. Matrix Operations Given: A = [ 33 ]A-[3-321] -3 B = [342]B-[3-41-2] (a) A² A2 Multiply A× A: -3 = (3 x 32x-3) (3 x 22 x 1) | = |[19–63 |-9-3 -6+21] = A² = 33 33 1-3×3+1x-3) (-3×2+1x1) [12]A2=[3-321][3-321]=[(3×3+2x-3)(-3×3+1x-3)(3×2+2×1)(-3×2+1×1)]=[9-6-9-36+2-6+1 ]=[3-128-5] (b) | A ||A| Determinant of A | A | (3 × 1) (2 x-3)=3+ 6 = 9|A|=(3×1)-(2x-3)=3+6=9 (c) Adjoint of A Swap diagonal elements and change sign of off-diagonals: A = [33], so adj (A) = |¯²]A=[3-321], so adj(A)=[13–23] -3 (d) B-¹B-1 First find | B ||B|: |B | (3x-2)- (1 × -4) = -6 + 4 = −2|B|=(3x-2)-(1x-4)=-6+4=-2 Then the adjoint of B: adj (B) = [² 3 adj(B)=[-24-13] Now, B-1 1 = |B| · adj (B) = 1 [²¯¯³¹³] = [2₂ B 0.5 |B-1=|B|1-adj(B)=-21[-24-13]=[1-20.5-1.5] 2. (a) Matrix Method: Solve (2x + 3y = 6 (2x-3y=14 {2x+3y=62x-3y=14 Matrix form: 22 33-22 = [223-3][xy]=[614] Find inverse of coefficient matrix: Determinant: | M | (2x-3) - (3 x 2) = -6 -6 = -12|M|=(2x-3)-(3×2)=-6-6=-12 Adjoint: adj(M) = [3]adj(M)-[-3-2-32] So…

Let the region R be the area enclosed by the function f(x)= = 3x² and g(x) = 4x. If the region R is the base of a solid such that each cross section perpendicular to the x-axis is an isosceles right triangle with a leg in the region R, find the volume of the solid. You may use a calculator and round to the nearest thousandth. y 11 10 9 00 8 7 9 5 4 3 2 1 -1 -1 x 1 2

Chapter 6 Solutions

WEBASSIGN F/EPPS DISCRETE MATHEMATICS

Ch. 6.1 - The notation is read”______” and means that___Ch. 6.1 - To use an element argument for proving that a set...Ch. 6.1 - Prob. 3TY Ch. 6.1 - An element x is in AB if , and only if,_______Ch. 6.1 - An element x in AB if, and only if,______Ch. 6.1 - An element x is in B-A if, and only if,______Ch. 6.1 - An elements x is in Acif, and only if.______Ch. 6.1 - The empty set is a set with ______Ch. 6.1 - The power set of a set A is _____Ch. 6.1 - Prob. 10TY

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Let C i = { i , − i } for each nonnegative integer i .

Solution Summary: The author calculates that i=0,i,-i are both equal to zero.

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