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

2024
Views:

84

Pages:

3
Author:

Bennett Wheeler

Question 1 To find the value of 11 ⋅ 48 11⋅ 48 . Answer 1 Rewrite 48 as a product of square factors: 11 ⋅ 48 = 11 ⋅ 16 ⋅ 3 11⋅ 48 =11⋅ 16⋅3 Use the property 𝑎 𝑏 = 𝑎 ⋅ 𝑏 ab = a ⋅ b : = 11 ⋅ 16 ⋅ 3 =11⋅ 16 ⋅ 3 Evaluate 16 = 4 16 =4: = 11 ⋅ 4 ⋅ 3 =11⋅4⋅ 3 Multiply: = 44 3 =44 3 So, the answer is 44 3 44 3 . Question 2 Solve the equation 5 𝑥 + 7 = 9 𝑥 − 17 5x+7=9x−17. Answer 1 Start with the equation: 5 𝑥 + 7 = 9 𝑥 − 17 5x+7=9x−17 Subtract 5x from both sides: 7 = 4 𝑥 − 17 7=4x−17 Add 17 to both sides: 24 = 4 𝑥 24=4x Divide by 4: 𝑥 = 6 x=6 So, the solution is 𝑥 = 6 x=6. Question 3 Prove that if 𝑛 n is a positive integer, then 𝑛 n is even if and only if 7 𝑛 + 4 7n+4 is even. Answer 1 To prove: 𝑛 n is even if and only if 7 𝑛 + 4 7n+4 is even. (If direction) Assume 𝑛 n is even. Then 𝑛 = 2 𝑘 n=2k for some integer 𝑘 k. Substitute 𝑛 = 2 𝑘 n=2k into 7 𝑛 + 4 7n+4: 7 𝑛 + 4 = 7 ( 2 𝑘 ) + 4 = 14 𝑘 + 4 7n+4=7(2k)+4=14k+4 Since 14 𝑘 14k and 4 4 are both even, 7 𝑛 + 4 7n+4 is the sum of two even numbers, which is also even. (Only if direction) Assume 7 𝑛 + 4 7n+4 is even. Suppose, for contradiction, that 𝑛 n is odd, so 𝑛 = 2 𝑘 + 1 n=2k+1 for some integer 𝑘 k. Substitute 𝑛 = 2 𝑘 + 1 n=2k+1 into 7 𝑛 + 4 7n+4: 7 𝑛 + 4 = 7 ( 2 𝑘 + 1 ) + 4 = 14 𝑘 + 7 + 4 = 14 𝑘 + 11 7n+4=7(2k+1)+4=14k+7+4=14k+11 Here, 14 𝑘 14k is even, but 11 11 is odd, so 7 𝑛 + 4 7n+4 would be odd, which contradicts the assumption that 7 𝑛 + 4 7n+4 is even. Thus, our supposition is false, and 𝑛 n must be even if 7 𝑛 + 4 7n+4 is even. Therefore, 𝑛 n is even if and only if 7 𝑛 + 4 7n+4 is even. Question 4 Given that 𝑥 𝑦 = 3 : 4 xy=3:4 and 𝑦 𝑧 = 5 : 8 yz=5:8, find the value of 2 𝑦 3 𝑥 − 𝑦 + 2 𝑧 3x−y+2z 2y . Answer 1 We are given: 𝑥 𝑦 = 3 4 and 𝑦 𝑧 = 5 8 xy= 4 3 andyz= 8 5 Express 𝑦 y in terms of 𝑥 x and 𝑧 z using these ratios: 𝑦 = 4 𝑥 3 and 𝑦 = 8 𝑧 5 y= 3 4x andy= 5 8z Substitute these values for 𝑦 y into 2 𝑦 3 𝑥 − 𝑦 + 2 𝑧 3x−y+2z 2y and solve accordingly. The problem solution requires further simplification based on these values for 𝑦 y. Question 5 Work each problem. Which of the following is equivalent to ln ⁡ ( 4 𝑥 ) − ln ⁡ ( 2 𝑥 ) ln(4x)−ln(2x) for 𝑥 > 0 x>0? A. 2 ln ⁡ 𝑥 2lnx B. ln ⁡ 2 𝑥 ln2x C. ln ⁡ 2 𝑥 ln x 2 D. ln ⁡ 2 ln2 Answer 1 To simplify ln ⁡ ( 4 𝑥 ) − ln ⁡ ( 2 𝑥 ) ln(4x)−ln(2x): Apply the quotient property of logarithms: ln ⁡ ( 4 𝑥 ) − ln ⁡ ( 2 𝑥 ) = ln ⁡ ( 4 𝑥 2 𝑥 ) ln(4x)−ln(2x)=ln( 2x 4x ) Simplify the fraction inside the logarithm: = ln ⁡ ( 4 2 ) = ln ⁡ 2 =ln( 2 4 )=ln2 So, the answer is D. ln ⁡ 2 ln2.