University:
Jackson College
Course:
MAT-010 | Basic Mathematics
Academic year:

2024
Views:

223

Pages:

3
Author:

Bennett Wheeler

Question 1 Solve the equation for 𝑡 t: 𝐴 = 𝑝 + 𝑝 𝑟 𝑡 A=p+prt Answer 1 Start with: 𝐴 = 𝑝 + 𝑝 𝑟 𝑡 A=p+prt Subtract 𝑝 p from both sides: 𝐴 − 𝑝 = 𝑝 𝑟 𝑡 A−p=prt Divide both sides by 𝑝 𝑟 pr to isolate 𝑡 t: 𝑡 = 𝐴 − 𝑝 𝑝 𝑟 t= pr A−p So, the solution for 𝑡 t is: 𝑡 = 𝐴 − 𝑝 𝑝 𝑟 t= pr A−p Question 2 Use the Zero Product Property to solve the equation: ( 𝑥 + 5 ) ( 𝑥 + 8 ) = 0 (x+5)(x+8)=0 Answer 1 By the Zero Product Property, if ( 𝑥 + 5 ) ( 𝑥 + 8 ) = 0 (x+5)(x+8)=0, then either 𝑥 + 5 = 0 x+5=0 or 𝑥 + 8 = 0 x+8=0. Solve each equation separately: For 𝑥 + 5 = 0 x+5=0: 𝑥 = − 5 x=−5 For 𝑥 + 8 = 0 x+8=0: 𝑥 = − 8 x=−8 So, the solutions are 𝑥 = − 5 x=−5 and 𝑥 = − 8 x=−8. Question 3 A farmer has a 3% solution and an 8% solution of a pesticide. How much of each must he mix to get 2000 L of a 4% solution for his sprayer? Answer 1 Let 𝑥 x be the amount (in liters) of the 3% solution, so 2000 − 𝑥 2000−x will be the amount of the 8% solution. Set up the equation for the total concentration: 0.03 𝑥 + 0.08 ( 2000 − 𝑥 ) = 0.04 ( 2000 ) 0.03x+0.08(2000−x)=0.04(2000) Expand and solve for 𝑥 x: 0.03 𝑥 + 160 − 0.08 𝑥 = 80 0.03x+160−0.08x=80 − 0.05 𝑥 = − 80 −0.05x=−80 𝑥 = 1600 x=1600 So, the farmer needs 1600 L of the 3% solution and: 2000 − 1600 = 400 2000−1600=400 400 L of the 8% solution. Question 4 Solve for 𝑥 x and verify it: 9 𝑥 − 7 = 10 9x−7=10 Answer 1 Add 7 to both sides: 9 𝑥 = 17 9x=17 Divide by 9: 𝑥 = 17 9 x= 9 17 Verification: Substitute 𝑥 = 17 9 x= 9 17 back into the original equation: 9 ( 17 9 ) − 7 = 10 9( 9 17 )−7=10 17 − 7 = 10 17−7=10 10 = 10 10=10 The solution is verified as 𝑥 = 17 9 x= 9 17 . Question 5 Using a proof by contraposition, prove that if 𝑥 x and 𝑦 y are real numbers where 𝑥 + 𝑦 ≥ 2 x+y≥2, then 𝑥 ≥ 1 x≥1 or 𝑦 ≥ 1 y≥1. Answer 1 To prove by contraposition, assume the negation: 𝑥 < 1 x<1 and 𝑦 < 1 y<1. If 𝑥 < 1 x<1 and 𝑦 < 1 y<1, then: 𝑥 + 𝑦 < 1 + 1 = 2 x+y<1+1=2 This contradicts the given condition 𝑥 + 𝑦 ≥ 2 x+y≥2. Therefore, if 𝑥 + 𝑦 ≥ 2 x+y≥2, it must be true that 𝑥 ≥ 1 x≥1 or 𝑦 ≥ 1 y≥1. Thus, the proof by contraposition shows that if 𝑥 + 𝑦 ≥ 2 x+y≥2, then 𝑥 ≥ 1 x≥1 or 𝑦 ≥ 1 y≥1.