Questions and Answers #2 Algebra Foundations

Questions and Answers Sheet 2 Algebra Foundations Question #1 It takes you 50 minutes to get to campus. You spend t minutes walking to the bus stop and the rest of the time riding the bus. Your walking rate is 0.06 mile per minute and the bus travels at a rate of 0.5 mile per minute. The total distance walking and traveling by bus is given by the algebraic expression 0.06t + 0.5(50 - t) . a. Simplify the algebraic expression. b. Use each form of the algebraic expression to determine the total distance that you travel if you spend 20 minutes walking to the bus stop. Answer: Step 1 Given: It takes you 50 minutes to get to campus. You spend t minutes walking to the bus stop and the rest of the time riding the bus. Your walking rate is 0.06 mile per minute and the bus travels at a rate of 0.5 mile per minute. The total distance walking and traveling by bus is given by the algebraic expression 0.06t+0.5(50-t) Step 2 a) To simplify the given algebraic expression. ⇒ 0.06𝑡 + 0.5(50 − 𝑡) = 0.06𝑡 + 0.5(50) − 0.5𝑡 ⇒ 0.06𝑡 + 0.5(50 − 𝑡) = 0.06𝑡 + 25 − 0.5𝑡 b) To determine the total distance that you travel if you spent 20 minutes walking to the bus stop. Put t = 20 in the given algebraic expression. Let D=0.06t+0.5(50-t) ⇒ 0.06(20) + 0.5(50 − 20) = 12 + 0.5(30) = 12 + 15 = 27 ⇒ 𝐷 = 27 miles Question #2 A person on an assembly line produces P items per day after t days of training, 400 where P(t)= How many days of training will it take for a person to 1+23e produce 340 items? (Find algebraically.) Answer: Step 1 400 1 − 23𝑒 −𝑡 No. of item P(t)=340 𝑃(𝑡) = Step 2 400 1 + 23𝑒 −𝑡 400 340 = 1 + 23𝑒 −𝑡 340(1 + 23𝑒 −𝑡 ) = 400 340 + 7820𝑒 −𝑡 = 400 7820𝑒 −𝑡 = 60 𝑃(𝑡) = 60 7820 Take ln each sides 60 ln(𝑒 −𝑡 ) = ln ( ) 7820 3 −𝑡 ln(𝑒) = ln ( ) 391 −𝑡(1) = −4.87 𝑡 = 5 Therefore, 5 days of training 𝑒 −𝑡 = Question #3 Write the English phrase as an algebraic expression. Then, simplify the expression. Let x represent the number. The difference between the product of three and a number and two times the number. What is the algebraic expression? 3x-2x (Do not simplify.) What is the simplified expression? Answer: Step 1 given that the x is the number a) So from given phase Product of three and a number =3x And two times of the number =2x Now difference =3x-2x So final answer for (a) part 3x-2x (1) Step 2 b) from solution of part a) we find that The algebraic expression as follows 3x-2x Now simplified form of above algebraic expression is = x Question #4 Determine solutions for the equation 5 sin2 𝑥 + 2 sin 𝑥 = 0 in the interval 𝑥 = [−2𝜋, 2𝜋] algebraically. Please round your irrational answers to 3 digits after the decimal point (for example x=2.378 or x=0.001 ). Please use appropriate notation for your final answer please give me all solutions between [−2𝜋, 2𝜋] for this equation. Answer: The given trigonometric equation is 5 sin2 𝑥 + 2 sin 𝑥 = 0 in the interval 𝑥 ∈ [−2𝜋, 2𝜋] We have, 5 sin2 𝑥 + 2 sin 𝑥 = 0 sin 𝑥 (5 sin 𝑥 + 2) = 0 sin 𝑥 = 0 or 5 sin 𝑥 + 2 = 0 −2 𝑥 = −2 𝜋, −𝜋, 0,2 𝜋 or \𝑠𝑖𝑛𝑥 = 5 𝑥 = −2 𝜋, −𝜋, 0, 𝜋, 2 𝜋 or 𝑥 = −0.421, −2.73, 3.553, 5.872 𝑥 = −6.283, −3.142, 0, 3.142, 6.283 or 𝑥 = −0.421, −2.73, 3.553, 5.872 ⇒ 𝑥 ∈ {−6.283, −3.142, −2.73, −0.421, 0, 3.142, 3.553, 5.872, 6.283} Hence all solution between [−2𝜋, 2𝜋] are 𝑥 = −6.283, −3.142, −2.730, −0.421, 0, 3.142, 3.553, 5.872, 6.283 Question #5 Suppose V is finite-dimensional with dim 𝑉 ≥ 2 . Prove that there exist S,T in L(V,V) such that ST \ne TS . Answer: According to question given that C is finite dimensional with dim 𝑉 ≥ 2 Let dim V=2 ∴ 𝑆, 𝑇 ∈ 𝐿(𝑉) 𝑆(𝑥, 𝑦) = (𝑥, 𝑥) 𝑇(𝑥, 𝑦) = (𝑜, 𝑥 + 𝑦) Therefore (𝑆𝑇)(𝑥, 𝑦) = 𝑆{𝑇(𝑥, 𝑦)} = 𝑆(𝑜, 𝑥 + 𝑦) = (0,0) (𝑇𝑆)(𝑛, 𝑦) = 𝑇{𝑆(𝑛, 𝑦)} = 𝑇(𝑥, 𝑥, ) = (0,2𝑥) ∴ 𝑆𝑇 ≠ 𝑇𝑆 Question #6 A student bought a calculator and a textbook for a course in algebra. He told his friend that the total cost was $145(without tax) and that the calculator cost $20 more than four times the cost of the textbook. What was the cost of each item? Let x= the cost of a calculator and y= the cost of the textbook. The corresponding modeling system is x+y=145 x=4y+20 Solve the system by using the method of substitution and enter the ordered pair Answer: Given: x+y=145 (1) x=4y+20 (2) Subtitute the value of x from equation (2) in (1) (4𝑦 + 20) + 𝑦 = 145 ⇒ 5𝑦 = 145 − 20 ⇒ 5𝑦 = 125 125 ⇒𝑦= = 25 5 ∴ 𝑥 = 4 × 25 + 20 =120 Hence, (x, y)=$(120,25) Question #7 Factor each of the algebraic expressions completely. 3𝑥 4 + 81𝑥 Answer: Step 1 To do factor of the algebraic expressions: 3𝑥 4 + 81𝑥 Step 2 Factor of 3𝑥 4 + 81𝑥 = 𝑥(3𝑥 3 + 81) = 𝑥. 3. (𝑥 3 + 27) = 3𝑥(𝑥 3 + 27) = 3𝑥(𝑥 3 + 33) = 3𝑥(𝑥 + 3)(𝑥 2 − 𝑥. 3 + 32) = 3𝑥(𝑥 + 3)(𝑥 2 − 3𝑥 + 9) Step 3 Factor of 3𝑥 4 + 81𝑥 = 3𝑥(𝑥 + 3)(𝑥 2 − 3𝑥 + 9) Question #8 Write the expression as an algebraic expression in x for x > 0 . sin(tan−1 𝑥) Answer: sin(tan−1 𝑥) Now let 𝑡 = tan−1 𝑥 ⇒ tan(𝑡) = 𝑥 𝑥 ⇒ sin(𝑡) = √1 + 𝑥 2 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 (because tan 𝑡 = ⇒ 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 = 𝑥 and base = 1 then hypotenuse = √1 + 𝑥 2 a𝑙𝑠𝑜 sin 𝑡 = 𝑏𝑎𝑠𝑒 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 ) ⇒ 𝑡 = sin−1 So, tan −1 𝑥 √1 + 𝑥 2 𝑥 = sin−1 𝑥 √1+𝑥 2 ⇒ sin(tan−1 𝑥) = sin (sin−1 = 𝑥 𝑥 √1 + 𝑥 2 ) √1 + 𝑥 2 Question #9 Factor each of the algebraic expressions completely. 27𝑥 3 − 8𝑦 3 Answer: 27𝑥 3 − 8𝑦 3 The formula for this expression is: 𝑎3 − 𝑏 3 = (𝑎 − 𝑏)(𝑎2 + 𝑎𝑏 + 𝑏 2 ) where a=3x and b=2y (3𝑥)3 − (2𝑦)3 = (3𝑥 − 2𝑦)((3𝑥)2 + (3𝑥) × (2𝑦) + (2𝑦)2 ) = (3𝑥 − 2𝑦)(9𝑥 2 + 6𝑥𝑦 + 4𝑦 2 ) Question #10 Write the trigonometric expression as an algebraic expression in u. sec(cos −1 𝑢) Answer: Inside of the braces can be rewritten cos 𝜃 = 𝑢 and from Pythagoras theorem, we have another side as √1 − 𝑢2 1 =𝑢 sec 𝜃 1 sec 𝜃 = 𝑢 1 Answer: sec 𝜃 = 𝑢 Question #11 Solve the equation algebracially: 5-3x=8x-20 Answer: Solve the equation algebraically and graphically: 5-3x=8x-20 Solving algebraically: 5-3x=8x-20 5+25=8x+3x 11x=30 30 𝑥= 11 Question #12 Factorize the following algebraic expressions: (a) 6x+24 (b) 8x-4x Answer: Given, Factorize (a) 6x+24 (b) 8x-4x Part a 6x+24=6(x+4) Part b 8x-4x=4x Question #13 Multiply the algebraic expression using a Special Product Formula, and simplify. (2 + 𝑦 3 )2 Answer: Step 1 Given: (2 + 𝑦 3 )2 Step 2 (2 + 𝑦 3 )2 = 4 + 𝑦 6 + 4𝑦 3 = 4 + (𝑦 2 )3 + 4𝑦 3 = 4 + (𝑦 2 + 4𝑦)(𝑦 4 − 4𝑦 3 + 16𝑦 2 ) Question #14 Multiply the algebraic expression using a Special Product Formula, and simplify. (2 − 7𝑦)2 Answer: Step 1 The given expression is (2 − 7𝑦)2 Step 2 (2 − 7𝑦)2 = 22 − 2 ⋅ 2 ⋅ 7𝑦 + (7𝑦)2 [(𝑎 − 𝑏)2 = 𝑎2 − 2𝑎𝑏 + 𝑏 2 ] = 4 − 28𝑦 + 49𝑦 2 = 49𝑦 2 − 28𝑦 + 4 Question #15 Multiply the algebraic expression using a Special Product Formula, and simplify. (3𝑦 − 1)2 Answer: To simplify: (3𝑦 − 1)2 Solution: We know, (𝑎 − 𝑏)2 − 𝑎2 − 2𝑎𝑏 + 𝑏 2 (1) Given ex[ression is: (3𝑦 − 1)2 , ⇒ 𝑎 = 3𝑦, 𝑏 = 1 Simplifying further using (1) we get: (3𝑦 − 1)2 = (3𝑦)2 − 2(3𝑦)(1) + 12 = 9𝑦 2 − 6𝑦 + 1 ⇒ (3𝑦 − 1)2 = 9𝑦 2 − 6𝑦 + 1 Answer: (3𝑦 − 1)2 = 9𝑦 2 − 6𝑦 + 1 Question #16 Dermine the solutions of the equation: 4𝑏 2 − 100 = 0 algebraically Answer: We will simply solve the equation to find the value of b. Solution of the above problem is the step below. 4𝑏 2 − 100 = 0 4𝑏 2 = 100 100 𝑏2 = 4 𝑏 2 = 25 𝑏 = ±(250.5 ) = ±5 Hence, solution is b=5,-5 Question #17 Simplify. Assume that variable a is nonzero. 𝑐 2 𝑎−3 𝑐 2 𝑎−3 = square (Use positive exponents only. Simplify your answer.) Answer: Solution: ∵ by law of indices 1 𝑎−𝑚 = 𝑚 𝑎 1 ∴ 𝑐 2 𝑎−3 = 𝑐 2 3 𝑎 2 𝑐 ∴ 𝑐 2 𝑎−3 = 3 𝑎 Question #18 Solve the following equation. 3 − 4(𝑥 + 1) − 3 = 3 + 2(4 − 2𝑥) − 16 Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The solution is x = square . (Simplify your answer.) B. Every real number is a solution. C. There is no solution. Answer: 3-4(x+1)-3=3+2(4-2x)-16 3 − 4x − 4 − 3 = 3 + 8 − 4x −16 -4x-4=-4x-5 −4 x + 4 x = −5 + 4 0=-1 Clearly it is the case of no solution. Question #19  4 0 5 A= ,  −1 3 2 1 1 1  B= , 3 5 7  Let  2 −3 C=  0 1  Find: 3A-B , and 𝐶 × 𝐵 + 𝐴 Answer: 4 0 A=  −1 3 1 1 B= 3 5 5 2  1 7   2 −3 C=  0 1   4 0 5  1 1 1  (3 A − B) =  −   −1 3 2  3 5 7  12 0 15 1 1 1  B= −   −3 9 6  3 5 7   11 −1 14  (3 A − B) =    −6 4 1   2 −3  2 C =   =  0 1   −3  2 0  1 1 CB =    −3 1  3 5 t 0 1  1   2 2 2 = 7   0 2 4  2 2 2  4 0 5  6 2 7  CB+ A=  + =   0 2 4   −1 3 2   −1 5 6  Question #20 To translate given phrase into an algebraic expression using x to represent the number. 9 The given phrase is multiply 1 by a number. 10 Answer: Let us suppose the number be x 9 The phrase ''Multiply 1 by a number'' means multiply the number x to 1 10 The following algebraic expression will be formed: 9 𝑥 × (1 ) 10 19 =𝑥× 10 19 = 𝑥 10 19 The algebraic expression is 𝑥. 10 9 10