Explanation Given: The number is 3.7 + 1.22 . Calculation: Consider the three flats of a base-ten block, and seven longs for 3.7 . Figure-(1) Here, one flat represents one unit and each long represents 1 / 10 of a unit. Calculate the sum of number 3.7 . 3 + 7 10 Consider the one flat of a base-ten block, two longs and two cubes for 1.22 , Figure-(2) Here, one flat represents one unit, each long represents 1 / 10 of a unit and each cube represents 1 / 100 of a unit. Calculate the sum of number 1.22 . 1 + 2 10 + 2 100 Calculate the number of flats as shown in Figure (3)

A Problem Solving Approach to Mathematics for Elementary School Teachers (12th Edition)

12th Edition

ISBN: 9780321987297

Author: Rick Billstein, Shlomo Libeskind, Johnny Lott

Publisher: PEARSON

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Chapter 7.2B, Problem 1A

To determine

To find:

The sum of the 3.7+1.22 using base-ten block.

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Chapter 7.2A, Problem 40A

Chapter 7.2B, Problem 2A

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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…

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