Questions and Answers #1 Factors and multiples

Questions and Answers Sheet 1 Factors and multiples Question #1 Evaluate the following expression: 20!/(13! (20 − 13)!) Answer: 20!/(13! (20 − 13)!) = 20!/(13! 7!) (20𝑥𝑥19𝑥𝑥18𝑥𝑥17𝑥𝑥16𝑥𝑥15𝑥𝑥14𝑥𝑥13!)/(13! 7!) = (20𝑥𝑥19𝑥𝑥18𝑥𝑥17𝑥𝑥16𝑥𝑥15𝑥𝑥14)/(7𝑥𝑥6𝑥𝑥5𝑥𝑥4𝑥𝑥3𝑥𝑥2𝑥𝑥1) = (20𝑥𝑥19𝑥𝑥18𝑥𝑥17𝑥𝑥16𝑥𝑥15𝑥𝑥14)/((7𝑥𝑥2)(5𝑥𝑥4)(6𝑥𝑥3) = (20𝑥𝑥19𝑥𝑥18𝑥𝑥17𝑥𝑥16𝑥𝑥15𝑥𝑥14)/(14𝑥𝑥20𝑥𝑥18) = 19𝑥𝑥18𝑥𝑥16𝑥𝑥15 = 77520 Question #2 Multiply 1/9 and 24/25 Answer: 1/9𝑥𝑥24/25 = 24/225 = (24−: 3)/(225−: 3) = 8/75 Question #3 Find 𝑥𝑃(𝑥), 𝑥 2and 𝑥 2 𝑃(𝑥) when: 1. 𝑥 = 1; 𝑃(𝑥) = 0.25; 2. 𝑥 = 2; 𝑃(𝑥) = 0.28; 3. 𝑥 = 3; 𝑃(𝑥) = 0.41; 4. 𝑥 = 4; 𝑃(𝑥) = 0.06; Answer: 1. 𝑥𝑃(𝑥) = 1𝑥𝑥0.25 = 0.25 𝑥 2 = 12 = 1 𝑥 2 𝑃(𝑥) = 12 𝑥𝑥0.25 = 0.25 2. 𝑥𝑃(𝑥) = 2𝑥𝑥0.28 = 0.56 𝑥 2 = 22 = 4 𝑥 2 𝑃(𝑥) = 22 𝑥𝑥0.28 = 1.12 3. 𝑥𝑃(𝑥) = 3𝑥𝑥0.41 = 1.23 𝑥 2 = 32 = 9 𝑥 2 𝑃(𝑥) = 32 𝑥𝑥0.41 = 3.69 4. 𝑥𝑃(𝑥) = 4𝑥𝑥0.06 = 0.24 𝑥 2 = 42 = 16 𝑥 2 𝑃(𝑥) = 42 𝑥𝑥0.06 = 0.96 Question #4 What is the least common multiple of {120, 124, 165} ? Answer: To find the least common multiple, let's do a ' factorization on these numbers: 120 = 2 × 2 × 2 × 3 × 5 124 = 2 × 2 × 31 165 = 3 × 5 × 11 To find the lowest common multiple, we take the largest group of 's we can choose from. For instance, the first ' is 2. The 120 has 3 of them and that is the most that any of our other numbers have. So we need 3 2s. 2 × 2 × 2 ×... For 3, the next ', two of the numbers have a 3, so we need 1 also. 2 × 2 × 2 × 3 ×... We also have two numbers with a 5 each: 2 × 2 × 2 × 3 × 5 ×... And there is an 11 and a 31: 2 × 2 × 2 × 3 × 5 × 11 × 31 = 40,920 120 × 11 × 31 = 40,920 124 × 2 × 3 \𝑡𝑖𝑚𝑒 5 × 11 = 40,920 165 × 2 × 2 × 2 × 31 = 40,920 Question #5 How do you find the Least common multiple of 28𝑏 2 , 20𝑎𝑏 3 , 16𝑏 4 ? Answer: 𝐿𝐶𝑀 = 2 × 2 × 2 × 2 × 5 × 7 × 𝑎 × 𝑏 × 𝑏 × 𝑏 × 𝑏 = 560𝑎𝑏 4 Explanation: To find LCM, I like first to do a ' factorization of the numbers in question: 28𝑏 2 = 2 × 14 × 𝑏 × 𝑏 = 2 × 2 × 7 × 𝑏 × 𝑏 20𝑎𝑏 3 = 2 × 10 × 𝑎 × 𝑏 × 𝑏 × 𝑏 = 2 × 2 × 5 × 𝑎 × 𝑏 × 𝑏 × 𝑏 16𝑏 4 = 2 × 8 × 𝑏 × 𝑏 × 𝑏 × 𝑏 = 2 × 2 × 4 × 𝑏 × 𝑏 × 𝑏 × 𝑏 = 2 × 2 × 2 × 2 × 𝑏 × 𝑏 × 𝑏 × 𝑏 . The LCM will include all the 's listed (and all the variables too) in the biggest quantity from any given number we're working with. For instance, we have 2s. There are 4 of them in the 16𝑏 4 , so our LCM will have 4 as well: 𝐿𝐶𝑀 = 2 × 2 × 2 × 2 × ? We also have a 5 in the 20𝑎𝑏 3 . 𝐿𝐶𝑀 = 2 × 2 × 2 × 2 × 5 ×? There's a 7 in the 28𝑏 2 . 𝐿𝐶𝑀 = 2 × 2 × 2 × 2 × 5 × 7 ×? That's it for numbers. We have an a in 20𝑎𝑏 3 and we have 4 b in 16𝑏 4 . 𝐿𝐶𝑀 = 2 × 2 × 2 × 2 × 5 × 7 × 𝑎 × 𝑏 × 𝑏 × 𝑏 × 𝑏 All of this gives us: 𝐿𝐶𝑀 = 2 × 2 × 2 × 2 × 5 × 7 × 𝑎 × 𝑏 × 𝑏 × 𝑏 × 𝑏 = 560𝑎𝑏 4 . Question #6 Multiply (−6𝑥 − 5)(𝑥 3 − 𝑥 + 2) Answer: Step 1 Given: (−6𝑥 − 5)(𝑥 3 − 𝑥 + 2) Step 2 = −6𝑥(𝑥 3 − 𝑥 + 2) − 5(𝑥 3 − 𝑥 + 2) = −5𝑥 4 + 6𝑥 2 − 12𝑥 − 5𝑥 3 + 5𝑥 − 10 = −6𝑥 4 − 5𝑥 3 + 6𝑥 2 − 7𝑥 + 10 Question #7 How to factor 𝑥 2 − 2𝑥 − 24 ? Answer: Then 4 − 6 = 2 and 4 ∗ −6 = −24 we can factor this equation as: (𝑥 + 4)(𝑥 − 6) Question #8 What is 89.1 ∗ 0.47 ? Answer: Stack the numbers in a straight line. 89.1 × .47 Use multiplication to multiply as if there are no decimals. 38493 Then count how many decimal places there are total in the equation. For example, 3.5 has one decimal place. Place a decimal point in your answer counting from right to left. 38.493 Question #9 How to factor 3𝑥 2 = 16𝑥 + 12 ? Answer: 3𝑥 2 = 16𝑥 + 12 Subtract (16𝑥 + 12) from each side. 3𝑥 2 − (16𝑥 + 12) = 0 Remove the brackets and make it simpler. A negative is the result of a positive and a negative: 3𝑥 2 − 16𝑥 − 12 = 0 Factorise. 3𝑥 2 − 18𝑥 + 2𝑥 − 12 = 0 3𝑥(𝑥 − 6) + 2(𝑥 − 6) = 0 (3𝑥 + 2)(𝑥 − 6) = 0 3𝑥 + 2 = 0 or 𝑥 − 6 = 0 𝑥 = −2/3 or 𝑥 = 6