Explanation A plane figure is one which is drawn with the use of lines and curves in a plane surface, connecting points and establishing relations between them. Plane figures might be open, for example a line, a ray, a line segment, an angle or a curve which do not enclose an area on the plane. On the contrary, plane figures such as a circle, a triangle, a quadrilateral or a polygon enclose area on the plane and hence are called closed figures. For those figures which enclose an area, there exists a distinct boundary that demarcates whether a given point on the plane is inside or outside the area enclosed. The length of this outline is defined as the perimeter Calculation: Opposite sides of rectangle are equal in length. Perimeter formula for rectangle = 2 ( L + B )

Prealgebra & Introductory Algebra (5th Edition) (What's New in Developmental Math)

5th Edition

ISBN: 9780134707631

Author: Elayn Martin-Gay

Publisher: PEARSON

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Chapter 8.2, Problem 17E

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The perimeter of a football field is 53 yards wide and 120 yards long.

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Chapter 8.2, Problem 16E

Chapter 8.2, Problem 18E

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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 8 Solutions

Prealgebra & Introductory Algebra (5th Edition) (What's New in Developmental Math)

Ch. 8.1 - Prob. 1CC Ch. 8.1 - Prob. 1P Ch. 8.1 - Prob. 2P Ch. 8.1 - Prob. 3P Ch. 8.1 - Prob. 4P Ch. 8.1 - Prob. 5P Ch. 8.1 - Prob. 6P Ch. 8.1 - Prob. 7P Ch. 8.1 - Prob. 8P Ch. 8.1 - Prob. 1VRVC

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The perimeter of a football field is 53 y a r d s wide and 120 y a r d s long.

To find: The perimeter of a football field is 53 yards wide and 120 yards long.

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Chapter 8 Solutions

To find:

The perimeter of a football field is 53 yards wide and 120 yards long.