Questions and Answers #3 Algebra Foundations

Questions and Answers Sheet 3 Algebra Foundations Question #1 Translate the given phrase into algebraic expression. 3 A number subtracted from − 8 Answer: In order to write the algebraic expression for the given phrase, first let's assume that a number is x 3 Thus, we can reprase the given phrase as: x subtracted from − 8 Here the keyword is subtracted, which tells us the operation is subtraction. 3 Now we look for the words from in the phrase to identify the number to be subtracted from − . In the given expression we can see thet the numbers that is being subtracted from − 3 8 8 is x . So the given phrase in algebraic notation can be written as, 3 3 A number subtracted from − = 𝑥 subtracted from − 8 8 3 = − minus𝑥 8 3 =− −𝑥 8 3 Thus, the algebraic expression for the given phrase is − − 𝑥. 8 Question #2 To translate given phrase into an algebraic expression using x to represent the number. 1 The given phrase is divide a number by −6 . 11 Answer: Let us suppose the number be x. The phrase ''Divide a number by −6 1 11 .'' means divide the number x by −6 1 11 The following algebraic expression will be formed: 1 𝑥 ÷ (−6 ) 11 67 = 𝑥 ÷ (− ) 11 11 = 𝑥 × (− ) 67 {changing ÷ 𝑡𝑜 ÷ by inverting the fraction 11 =− 𝑥 67 The algebraic expression is − 11 67 𝑥. Question #3 To translate given phrase into an algebraic expression using x to represent the number. 3 The given phrase is the sum of 8 and a number. 4 Answer: Let us suppose the number be x. 3 3 The phrase ''The sum of 8 .'' means add the number x to 8 4 4 The following algebraic expression will be formed: 3 𝑥+8 4 3 =𝑥+8 4 3 The algebraic expression is 𝑥 + 8 . 4 Question #4 To factorize: 5𝑥 3 𝑦 − 15𝑥 2 𝑦 + 10𝑥𝑦 Answer: Let us consider the given algebraic expression and can be factorize as shown. 5𝑥 3 𝑦 − 15𝑥 2 𝑦 + 10𝑥𝑦 = 5𝑥𝑦(𝑥 2 − 3𝑥 + 2) = 5𝑥𝑦(𝑥 2 − 2𝑥 − 𝑥 + 2) = 5𝑥𝑦{𝑥(𝑥 − 2) − (𝑥 − 2)} = 5𝑥𝑦(𝑥 − 2)(𝑥 − 1) Question #5 To simplify: The given expression. (𝑥 14 )23 Answer: Consider the expression. (𝑥 14 )23 = 𝑥 14×23 = 𝑥 322 The expression can be solved by multiply the exponents. Thus, the given expression in simplified form is 𝑥 322 Question #6 The value of the logarithmic expression log 2 16 Answer: Step 1 Concept Used: Power property of logarithm log 𝑏 (𝑎𝑟 ) = 𝑟 log 𝑏 (𝑎) log 𝑏 𝑏 𝑥 = 𝑥 To find the value of the given logarithmic expression first we will rewrite 16 in exponent form as an exponent of base 2 and then use the power property of the logarithm. So we get log 2 16 = log 2 24 = 4 log 2 (2) Using power property Now to simplify further, we use the same base formula log 𝑏 𝑏 𝑥 = 𝑥, to get log 2 16 = log 2 24 = 4 log 2 (2) Using power property = 4⋅1 =4 Hence, the value of the given logarithmic expressionis 4. Question #7 The total cost of the fencing. Cost of fencing per feet/foot =$2 Answer: From exercise 20, 182 feet of fencing is needed to enclose the garden. So, we need to find cost of fencing for 182 feet. Total cost of fencing = cost of fencing per foot ⋅ total foot of fencing} = 2 ⋅ total foot of fencing = 2 ⋅ 182 = 364 Therefore, total cost of fencing is $364 Question #8 Complete the table by the use of indicated operations. Monomials Add Subtract Multiply 6 x, 3 x 6 x + 3x = 9 x 6 x − 3x = 3x 6 x  3x = 18 x 2 −12 x 2 − 2 x, can t be simplified −12 x 2  2 x = −24 x 3 −12 x 2 , 2 x Divide 6x =2 3x −12 x 2 = −6 x 2x 5a, 15a 4 y 2 , 4 y3 −3 y 5 , 9 y 4 −14 x 2 , 2 x 2 Answer: Given: the given monomial is −3𝑦 5 and 9𝑦 4 Addition: −3𝑦 5 + 9𝑦 4 Subtraction: −3𝑦 5 − 9𝑦 4 Multiplication: −3𝑦 5 ⋅ 9𝑦 4 = −27𝑦 5+4 = −27𝑦 9 Division: −3𝑦 5 9𝑦 4 =− 𝑦 3 Hence the completed table is given below. Monomials −3 y 5 , 9 y 4 Add Subtract Multiply −3 y 5 + 9 y 4cant be simplified −3 y 5 − 9 y 4 cantbesimplified −3 y 5  9 y 4 = −27 y 5+4 = −27 y 9 Question #9 The multiplication of the expression (8x-3)(2x-4) Answer: Concept used: The multiplication of the expression (8x-3)(2x-4) is calculated as, (8x-3)(2x-4)=8x(2x-4)-3(2x-4) Divide −3 y 5 y =− 4 9y 3 = (8𝑥) ⋅ (2𝑥) + (8𝑥) ⋅ (−4) + (−3) ⋅ (2𝑥) + (−3) ⋅ (−4) = 16𝑥 2 − 32𝑥 − 6𝑥 + 12 = 16𝑥 2 − 38𝑥 + 12 Thus, the multiplication of the expression (8x-3)(2x-4) is 16𝑥 2 − 38𝑥 + 12 Question #10 To find: The nuclear waste left after 10 century. The nuclear waste from an atomic energy plant decays at arate of 3\% each century. If 150 pounds of nuclear waste is disposed initially. Answer: Formula Used: 𝑦 = 𝑎(1 − 𝑟) 𝑥 Where 𝑟 ⇒ Final amount 𝑥 ⇒ Number of days 𝑟 ⇒ rate of decay per day 𝑎 ⇒ Initial amount Initial amount of uranium avaible, a=150 pounds Rate of decay per day, 𝑟 = 3% ≈ 0.03 Number of days left, x=10 centuries Substitute the value into formula and solve for y 𝑦 = 150(1 − 0.03)10 𝑦 = 30(0.97)10 𝑦 = 30(0.7374) 𝑦 = 110.6 Pounds Hence the amount of uranium left after 10 centuries is 110.6 Pounds. Question #11 The amount owed at the end of 5 years. An amount of $3000 is loaned at a rate of 10\% compounded quarterly. Answer: Step 1 Formula used: 𝑟 𝑛𝑡 The compound interest formula, 𝐴 = 𝑃 (1 + ) 𝑛 Here A is the amount after t years, P is the principal amount, r is the annual interest rate and n is number of periods the interest compounding. Step 2 To find the amount owed at the end of 5 years, we will substitute 𝑃 = 3000, 𝑟 = 10% = 10 100 = 0.1, 𝑡 = 5 in the above compound interest formula and solve for A. Since the interest rate is compounded quarterly, so we substitute n=4 . So, we get 𝑟 𝑛𝑡 𝐴 = 𝑃 (1 + ) 𝑛 0.1 4×5 = 3000 (1 + ) 4 = 3000(1 + 0.025)20 = 3000(1.025)20 ≈ $4915.85 Thus, the amount owed is approximately $4915.85 at the end of 5 years. Question #12 a b   ; a, b  Q} . One can prove that R with the usual 0 a 0 b  matrix addition and multiplication is a ring. Consider J = {  ; b  Q} be a 0 0 Let R= { subset of the ring R. (a) Prove that the subset J is an ideal of the ring R; (b) Prove that the quotient ring R/J is isomorphic to Q. Answer: Claim ⇒ 𝑘𝑒𝑟(∅) = 𝐽 If a b  )  ker (), a a   = ( then ∅(𝛼) = 0 a b  0 = (  ) = a a a  Then, 𝑘𝑒𝑟(∅) ⊂ 𝐽 − 1 on the other hard. If 𝛽 = 0 b  0 0 ∈ 𝐽, then   ∅(𝛽) = 0 ⇒ 𝐽 ⊂ 𝑘𝑒𝑟(∅) − 2 From 1 × 2 𝑘𝑒𝑟∅ = 𝐽 . Since, for any 𝛼 ∈ 𝑄, we can have such thet a 0  0 a   R   a 0  (  ) = a. 0 a Then, ∅ is surgective. By the 1st isomorphic theorem for the rings, 𝑅/𝐽 ≅ 𝑄 Hence, the Quotient ring R/J is isomorphic to Q. (Proved). 1st Isomorphic Theorem for Rings: Let \emptysetask be a rinh homomorphism from R to S. Then the mapping from 𝑅/𝑘𝑒𝑟∅ to ∅(𝑅), given by 𝑟 + 𝑘𝑒𝑟∅ ⇒ ∅(𝑟) is an isomorphism 𝐼 × 𝑒, 𝑅/𝑘𝑒𝑟∅ ≅ ∅𝑅 ) . Question #13 For 𝑎, 𝑏 ∈ 𝑍, let 𝐵(𝑎, 𝑏) ∈ 𝑀(2, 𝑍) be defined by  a 3b  B ( a, b) =  . b a  Let 𝑆 = {𝐵(𝑎, 𝑏); 𝑎, 𝑏 ∈ 𝑍} ⊆ 𝑀(2, 𝑍). Show that 𝑆 ≅ 𝑍[√3] = {𝑎 + 𝑏√3; 𝑎, 𝑏 ∈ 𝑍}. Answer: 𝑆 = {𝐵(𝑎, 𝑏): 𝑎, 𝑏 ∈ 𝑍} ⊆ 𝑀(2, 𝑍) 𝐵 = (𝑎, 𝑏) a 3b  =  Z [ 3] = {a + b 3 : a, b  z}. b a  Show that 𝑆 ≅ 𝑍(√3) Defined A function f: 𝑆 ⇒ 𝑧(√3)𝑏𝑦𝑓(𝐵(𝑎, 𝑏)) = 𝑎 + 𝑏(√3). ie  a 3b  f ( B(a, b)) = f (  ) = a + b 3 b a  Clearly f is well defined. f is one-one Kerf = {𝐵(𝑎, 𝑏)| 𝑓(𝐵(𝑎, 𝑏)) = 0} = {𝐵(𝑎, 𝑏)|𝑎 + 𝑏√3 = 0} = {𝐵(𝑎, 𝑏)|𝑎 = 0𝐵𝑏 = 0}  0 3 x0  0 0  {B(0,0)} = { } = {  0 0} = {0} 0 0    Kerf = {0} . since Kerf = {0} . Then f is one-one. f is onto:- det p +𝑞√3 ∈ 𝑧[√3], 𝑝, 𝑞 ∈ 𝑧  d 3q  B ( p, q ) =    S such that 𝑓(𝐵(𝑝, 𝑞)) = 𝑝 + 𝑞√3. q p  This show that f is onto. 𝑞√3𝑓: 𝑆 ⇒ 𝑧(√3) by 𝑓(𝐵(𝑎, 𝑏)) = 𝑎 + 𝑏√3 is esomorphism. ∴ 𝑆 ≅ 𝑧(√3) Question #14 Simplify. 8{[4(𝑝 − 4) + 19] − [2(3𝑝 − 2) + 5]} Answer: Ques. 8{[4(𝑝 − 4) + 19] − [2(3𝑝 − 2) + 5]} 8{(4𝑝 − 16 + 19) − (6𝑝 − 4 + 5)} 8{4𝑝 + 3 − 6𝑝 − 1} 8{−2𝑝 + 2} −16𝑝 + 16 Question #15 Foundations/Comprehension 1. Find the inverse (𝐴−1 ) of the following 3 × 3 matrix, using either the row-operation method or the adjugate method (6 points): 211 A = (311) 1 32 Answer: Invers of the following matrix by elementary slow bane formations 2 1 1 A =  3 1 1  1 3 2  2 1  Set A = IA  3 1  1 3  2 1 1  1    is... 3 1 1 = 0    1 3 2 0 1  1 0 0   2 1 1  1  = 0 1 0  3 1 1  2 0 0 1  1 3 2 0 0 1 0  A 0 1  (𝐴 = 𝐼𝐴) ⇒ by row bane formations, we will change it in identity matrix) 𝑅1 ⇒ 𝑅1 /2  1 1 1  1 2 2   2 0 0      3 1 1  =  0 1 0  A 1 3 2   0 0 1          R2  R2 − 3R3 ; 1  1  2  R3  R3 − R1 0 (1 − 9)  1 0 (3 − ) 2  R2 R2  −8 1 1  1  0 1 2 2   2    1 0 1 5  =  0    8 −8    0 5 3   − 1 0  2 2   2 1  1 2  5 R3  R3 − − R2 0 1  2  0 0  R3  (−16) 1   1 1 1  1    0 0 1 0 0       2 2 2 2 2        (1 − 6)  =  0 1 −3 A 0 −8 −5 =  0 1 −3 A    1   1 5 3  1 (2 − )   − 0 1  0 0 1  − 2   2 2 2  2     0  3 A 8  1  1  1 2  2   5  = 0 8  −1   1 − 16   2 1  1 2  R3 0 1  0 0   1  1 2 2   5  = 0 8  1   8   R2  R2 − 5 8 1  1  0 0   2 2    0  =  −5 3 1  A 1   8 −5 1        1  1 2  R3 0 1 0 0    0  −1 3  A 8 8 −5 −1  0  0  −1 3  A 8 8 5 1  16 16  0 1 2 1   1 0 2   3    R2 0 1 0  =  −5 0 0 1   8       1 R1  R1 − 2 1 0 0   −1 R3 0 1 0  =  −5 0 0 1   8 R1  R1 − −3  0 2  3 1A −5 −1   1 0 3 1  A −5 −1 2 1 1   0 ∴ Inverse of A = 3 1 1 1   1 3 2  −1 1 0   given by −5 3 1  as it becomes   8 −5 −1 (𝐼 = 𝐴−1 𝐴). Question #16 Foundations/Comprehension 1.The following limits are of the form 0/0. Using algebraic methods, simplify the expressions, then compute the limits (5 points each). a) lim 3𝑥 2 −23𝑥−70 𝑥⇒10 11𝑥 2 −1100 5(8+ℎ)2 −320 b) lim ℎ ℎ⇒0 Answer: a) lim 3𝑥 2 −23𝑥−70 𝑥⇒10 11𝑥 2 −1100 As 𝑥 ⇒ 10 then numerator and denominator both tends to zero. 0 𝐼 ⋅ 𝑒 limit is of the form form. 3𝑥 2 −23𝑥−70 Now lim 0 𝑥⇒10 11𝑥 2 −1100 2 3𝑥 − 30𝑥 + 7𝑥 − 70 𝑥⇒10 11𝑥 2 − 1100 3𝑥(𝑥−10)+7(𝑥−10) = lim (factoring numerator and denominator) 2 = lim 𝑥⇒10 11(𝑥 −100) (3𝑥 + 7)(𝑥 − 10) = lim (∵ 𝑎2 − 𝑏 2 = (𝑎 − 𝑏)(𝑎 + 𝑏)) 𝑥⇒10 11(𝑥 − 10)(𝑥 + 10) 3𝑥+7 = lim (concelling the common factor) = = 𝑥⇒10 11(𝑥+10) 3(10)+7 11(10+10) (finally compuling the limit) 3(10) + 7 11(+10) Final fons: lim 3𝑥 2 −23𝑥−70 𝑥⇒10 11𝑥 2 −1100 = 37 220 Ans. b) lim 5(8+ℎ)2 −320 ℎ ℎ⇒0 As ℎ ⇒ 0 , both numerator and denominator tends to zero, so limit is of Now lim lim 5(8+ℎ)2 −320 0 0 form. ℎ ℎ⇒0 5(82 +ℎ2 +16ℎ)−320 ℎ ℎ⇒0 320 + 5h 2 + 16h − 320 lim (simplifying the numerator) h 0 h 5ℎ2 + 80ℎ ℎ⇒0 ℎ ℎ(5ℎ+80) lim (taken common factors out) lim ℎ ℎ⇒0 lim (5ℎ + 80) (concelling common factor) ℎ⇒0 =5(0)+80=80 (finally compuling the limit) ∴ lim 5(8+ℎ)2 −320 ℎ⇒0 ℎ = 80 Ans Question #17 For the functions 𝑓(𝑥) = 10𝑥 − 8 and 𝑔(𝑥) = 5𝑥 + 7, find the following. (𝑎)(𝑓 + 𝑔)(𝑥) (𝑏)(𝑓 − 𝑔)(𝑥) Answer: Step 1 Let Given function 𝑓(𝑥) = 10𝑥 − 8 𝑎𝑛𝑑 𝑔(𝑥) = 5𝑥 + 7 (a)(f+9)(x) we know that, (f+9)(x)=f(x)+9(x) ⇒ (𝑓 + 9)(𝑥) = 𝑓(𝑥) + 9(𝑥) = (10𝑥 − 8) + (5𝑥 + 7) (𝑓 + 9)(𝑥) = 15𝑥 − 1 Step 2 (b)(f-9)(x) we know that, (f-9)(x)=f(x)-9(x) (𝑓 − 9)(𝑥) = 𝑓(𝑥) − 9(𝑥) = (10𝑥 − 8) − (5𝑥 + 7) = 10𝑥 − 8 − 5𝑥 − 7 (𝑓 − 9)(𝑥) = 5𝑥 − 15 Question #18 The folowing table gives the frequency distribution of the sales athletic shoes at a store in an outlet mall cach day during the past 30 days. Calculate the mode a algebraically. Namber of sales Namber of days 14 − 18 19 − 23 24 − 28 29 − 22 5 9 6 10 Answer: Solutoins: Class Frequency ( f ) (1) (2) 14 − 18 5 19 − 23 9 24 − 28 6 29 − 33 10 ... ... n = 30 To find Mode Class Here, maximum frequency is 10. ∴ The mode class is 28.5 − 33.5. ∴ 𝐿 = lower boundary point of made class = 28.5 ∴ 𝑓1 = frequensy of the mode class = 10 ∴ 𝑓0 = frequensy of the preceding class = 6 ∴ 𝑓2 = frequensy of the succedding class = 0 ∴ 𝑐 = class length of the mode class = 5 𝑓1 − 𝑓0 𝑍 =𝐿+( )⋅𝑐 2 ⋅ 𝑓1 − 𝑓0 − 𝑓2 10 − 6 = 28.5 + ( )⋅5 2 ⋅ 10 − 6 − 0 4 = 28.5 + ( ) ⋅ 5 14 = 28.5 + 1.4286 = 29.9286 Question #19 Simplify, 5{[4(𝑞 − 4) + 18] − [2(5𝑞 − 2) + 3]} Answer: = 5{[4(𝑞 − 4) + 18] − [2(5𝑞 − 2) + 3]} = 5{[4𝑞 − 16 + 18] − [10𝑞 − 4 + 3]} = 5{4𝑞 + 2 − 10𝑞 − 1} = 5{1 − 6𝑞} = 5(1 − 6𝑞) Question #20 Find the sloope if it exist, 4x-y=8 Answer: 4x-y=8 We can wait this a y=4x-8(1) This is thee forn of y=mx+c(2) Compsing ∑ 𝑞(1) with ∑ 𝑞(2) m=4 where m is the slope