University:
Purdue University
Course:
MA 15300 | College Algebra
Academic year:

2024
Views:

500

Pages:

4
Author:

Delphinixxc3

Question 1 Write the domain of the function in interval notation. Write numbers as integers or simplified fractions. 𝑔 ( 𝑥 ) = 2 𝑥 − 3 10 𝑥 2 − 33 𝑥 − 7 g(x)= 10x 2 −33x−7 2x−3 Answer 1 Step 1: To find the domain, identify where the function is undefined by setting the denominator equal to zero. 10 𝑥 2 − 33 𝑥 − 7 = 0 10x 2 −33x−7=0 Solve for 𝑥 x: ( 5 𝑥 + 1 ) ( 2 𝑥 − 7 ) = 0 (5x+1)(2x−7)=0 So, 𝑥 = − 1 5 x=− 5 1 and 𝑥 = 7 2 x= 2 7 . The function is undefined for 𝑥 = − 1 5 x=− 5 1 and 𝑥 = 7 2 x= 2 7 . Step 2: The domain includes all real numbers except these values. Domain : ( − ∞ , − 1 5 ) ∪ ( − 1 5 , 7 2 ) ∪ ( 7 2 , ∞ ) Domain:(−∞,− 5 1 )∪(− 5 1 , 2 7 )∪( 2 7 ,∞) Question 2 Find the partial fraction decomposition: 𝑥 2 + 𝑥 − 6 ( 𝑥 2 + 1 ) ( 𝑥 − 1 ) (x 2 +1)(x−1) x 2 +x−6 Answer 1 Step 1: Express the function as a sum of partial fractions: 𝑥 2 + 𝑥 − 6 ( 𝑥 2 + 1 ) ( 𝑥 − 1 ) = 𝐴 𝑥 + 𝐵 𝑥 2 + 1 + 𝐶 𝑥 − 1 (x 2 +1)(x−1) x 2 +x−6 = x 2 +1 Ax+B + x−1 C Step 2: Multiply both sides by ( 𝑥 2 + 1 ) ( 𝑥 − 1 ) (x 2 +1)(x−1) and equate coefficients: 𝑥 2 + 𝑥 − 6 = ( 𝐴 𝑥 + 𝐵 ) ( 𝑥 − 1 ) + 𝐶 ( 𝑥 2 + 1 ) x 2 +x−6=(Ax+B)(x−1)+C(x 2 +1) Expand and collect like terms. Solve for 𝐴 A, 𝐵 B, and 𝐶 C: 𝐴 = 3 , 𝐵 = 4 , 𝐶 = 2 A=3, B=4, C=2 Step 3: The partial fraction decomposition is: 𝑥 2 + 𝑥 − 6 ( 𝑥 2 + 1 ) ( 𝑥 − 1 ) = 3 𝑥 + 4 𝑥 2 + 1 + 2 𝑥 − 1 (x 2 +1)(x−1) x 2 +x−6 = x 2 +1 3x+4 + x−1 2 Question 3 Please explain the addition property of equality with fractions. Answer 1 Step 1: The addition property of equality states that if you add the same value to both sides of an equation, the equation remains balanced. Step 2: For example, if you have: 𝑥 − 1 4 = 3 4 x− 4 1 = 4 3 Add 1 4 4 1 to both sides: 𝑥 − 1 4 + 1 4 = 3 4 + 1 4 x− 4 1 + 4 1 = 4 3 + 4 1 𝑥 = 1 x=1 Thus, the addition property of equality is satisfied. Question 4 How to add or subtract fractions with unlike denominators? Answer 1 Step 1: To add or subtract fractions with different denominators, first find the least common denominator (LCD). Step 2: For example, to add 1 2 2 1 and 1 3 3 1 : The LCD of 2 and 3 is 6. 1 2 = 3 6 , 1 3 = 2 6 2 1 = 6 3 , 3 1 = 6 2 Now add: 3 6 + 2 6 = 5 6 6 3 + 6 2 = 6 5 Question 5 Convert the following decimals to percents, fractions to percents, and percents to fractions. Also, solve for portions of given values. Answer 1 Step 1: Convert decimals to percents: 0.88 = 88 % 0.88=88% Step 2: Convert fractions to percents: 13 12 = 1.0833 = 83.33 % 12 13 =1.0833=83.33% 1 12 = 0.0833 = 8.33 % 12 1 =0.0833=8.33% Step 3: Convert percents to fractions and simplify: 17.4 % = 17.4 100 = 87 500 17.4%= 100 17.4 = 500 87 Step 4: Solve for portions: 23.7 % of 150 = 0.237 × 150 = 35.55 23.7% of 150=0.237×150=35.55 17.4 % of 900 = 0.174 × 900 = 156.6 17.4% of 900=0.174×900=156.6