Solutions to Chapter 4-6 | Answer Key - Edubirdie

University:
California State University, Northridge
Course:
MATH 210 | Basic Number Concepts
Academic year:

2024
Views:

207

Pages:

11
Author:

Leland Mcdonald

SOLUTIONS TO CHAP 4-6 REVIEW PS (a) n, n+2, n+4 (b) (25x+y+i00π) cents (c) (2k+2s) cm 2)(a) Method 1 Let x = James' amount($) [x+(x+60)+2(x+60)=600] So x+60= Peter's amt [X+X+60+2X+120=600] and 2(x+60) = John's amt [4x+180=600] [4x=420] [x=105] Solution James has $105 Peter has 105+60 or $165 John has 2 (105+60) or

330 Methodz Let t = Peter's amt() So 2t John's amt and t-60 Jame amt [t+2t+(t-60)=600] [4t-60=600] [4t=660] [t=165] Solution Peter has $165 John has 2(165) or $330 James has 165-60 or $105 (b) Let x = Jane's amt (in $) So 3x Mary's amt [3x-125=x-35] [3x=x+90] [2x=90] [x=45] Solution Jare had $45 to start Mary had 3($45) or $135 to start (a) a b AREA of square with side at b [(a+b)^(2)=(a+b)(a+b)] [=a^(2)+ab+ba^-+b^(2)] [a^(2)+ab+ab+ab+b^(2)] [a^(2)+2ab+b^(2)] bazab by Commutative property. of muit bb262 (b^(2)) [(a+b)^(2)=(\underline{a+b)(a+b)}=(a+b)}\underline{a}+(a+b)\underline{b}=a^(2)+ba+ab+b^(2)] [=a^(2)+2ab+b^(2)] since F (a) 31^2 = (30+1)^2 = 30^2 + 2(30)(1) + 1^2 = 900 + 60 + 1 = [961] (b) 23 x 17 = (20+3)(20-3) = 20^2 - 3^2 = 400 - 9 = [391] (c) 58^2 = (60-2)^2 = 60^2 - 2(60)(2) + 2^2 = 3600 - 240 + 4 = 3360 + 4 = [3364] (d) 66^2 - 34^2 = (66+34)(66-34) = 100.32 (a) 55,264 ones digit is 4 → DIV BY 2, 4/64 DIV BY 4. NOT DIV BY 5, NOT DIV BY 10 t number represented by last two digits 8/264 DIV BY 8 number represented by last three digits sum of digits (= 5+5+2+6+4 = 22) 3/22 NOT DIV BY 3 NOT DIV BY 3 NOT DIV BY 6 Sum of digits in even powers of 10 spots = 11 922 NOT DIV BY 9 sum of digits In odd powers of 10. 5+2+4 Does 11/(11-11)? Yes 11/0 10 spots = DIVBY IL 115+6 (b) 2,357,220 ones digit is O DIV BY 2, DIV BY 5, DIV BY 10 4/20 DIV BY 4 8/220 NOT DIV BY 8 sum of digits = 2+3+5+7+2+2+0=21 3/21 → DIV BY 3 9/21 NOT DIV BY 9 DIV BY BOTH 2 and 3 DIV BY 6 11(12-9) (that is, 11x3)→ NOT DIV BY II 1 \(3+7+2\) 2+5+2+0 (c) 5445 ones digit is 5 DIV BY 5, NOT DIV BY 2, NOT DIV BY 10 4/45 NOT DIV BY 4 8445 NOT DIV BY 8 sum of digits = 5+4+4+5=18 3/18 DIV BY 3 9/18 DIV BY9 11/(9-9) (that is, 11/0)→ DIV BY II ↑↑ (d) 11, 111 4+5 5+4 NOT DIV BY 2,3,4,5,6,8,9,10,11 (give reasons as above)

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