Factors and multiples #6 - Q&A

Question 1 Taj says the expression 3 – 2s cannot be factored using the GCF. Is he correct? Explain why or why not. Select the correct answer below and, if necessary, fill in the answer box to complete your choice. A. Taj is incorrect because all expressions with two terms can be factored. B. Taj is correct because 3 and –2s have no common factors. C. Taj is correct because the greatest common factor of 3 and –2s is 1 and factoring out 1 does not change the expression. D. Taj is incorrect because the greatest common factor of 3 and –2s is ?. The factored expression is ?. Answer 1 Step 1 Factorisation is Process of splitting an expression in the product of some other expressions. 𝐺 𝐶 𝐷 → GCD→ greatest common factor. Step 2 Taj is correct because the greatest common factor of 3 and –2s is 1 and factoring out 1 does not change the expression. Question 2 How many permutations of the letters ABCDEFG contain the string BCD? Answer 1 Definitions Definition permutation (order is important): 𝑃 ( 𝑛 , 𝑟 ) = 𝑛 ! ( 𝑛 − 𝑟 ) ! P(n,r)= (n−r)! n! ​ Definition combination (order is not important): 𝐶 ( 𝑛 , 𝑟 ) = ( 𝑛 𝑟 ) = 𝑛 ! 𝑟 ! ( 𝑛 − 𝑟 ) ! C(n,r)=( r n ​ )= r!(n−r)! n! ​ with 𝑛 ! = 𝑛 ⋅ ( 𝑛 − 1 ) ⋯ 2 ⋅ 1 n!=n⋅(n−1)⋯2⋅1. We treat the mentioned string as one single letter, thus we have then the possible letters A, BCD, E, F, G. We then have 5 letters of which we want to select 5 letters. 𝑛 = 5 n=5 𝑟 = 5 r=5 Evaluate the definition of a permutation: 𝑃 ( 5 , 5 ) = 5 ! ( 5 − 5 ) ! = 5 ! 0 ! = 5 ! = 120 P(5,5)= (5−5)! 5! ​ = 0! 5! ​ =5!=120 Answer 2 Step 1: Count the total number of permutations of the letters ABCDEFG. The number of permutations of a set of n distinct objects is given by 𝑛 ! n!. Step 2: Count the number of permutations that contain the string BCD. To count the number of permutations that contain the string BCD, we can treat BCD as a single unit. So, we have 4 units: BCD, A, E, F, G. The number of permutations of these 4 units is 4!. Step 3: Count the number of permutations where BCD can appear in any position. Since BCD can appear in any position within the 4 units, we need to multiply the number of permutations in Step 2 by the number of possible positions for BCD. In this case, there are 5 possible positions: at the beginning, after A, after E, after F, and after G. Therefore, we multiply 4! by 5. Step 4: Calculate the final result. To find the number of permutations of the letters ABCDEFG that contain the string BCD, we multiply the result from Step 3 by the total number of permutations in Step 1. Putting it all together: Total number of permutations = Step 1 × × Step 3 = 7 ! × 4 ! × 5 = ( 7 × 6 × 5 × 4 × 3 × 2 × 1 ) × ( 4 × 3 × 2 × 1 ) × 5 7!×4!×5=(7×6×5×4×3×2×1)×(4×3×2×1)×5 ( 4 × 3 × 2 × 1 ) × 5 = 5040 × 24 × 5 = 604 , 800 (4×3×2×1)×5=5040×24×5=604,800 Therefore, there are 604,800 permutations of the letters ABCDEFG that contain the string BCD. Answer 3 To solve the problem of finding how many permutations of the letters ABCDEFG contain the string BCD, we can use the concept of permutations with restrictions. Let's denote the string BCD as a single entity, say X. Now, we need to find the number of permutations of the letters ABCDEFG where X appears together. To find the total number of permutations, we consider X as a single letter, which means we have 6 distinct entities (A, X, E, F, G) to arrange. The total number of permutations of these entities is 6 ! 6!. However, within the entity X, there are 3 distinct letters (B, C, D), which can be arranged in 3 ! 3! ways. Therefore, the total number of permutations of the letters ABCDEFG containing the string BCD is 6 ! × 3 ! 6!×3!. Answer 4 7 ! − ( 5 ! × 3 ! ) = 7 × 6 × 5 × 4 × 3 × 2 × 1 − 5 × 4 × 3 × 2 × 1 × 3 × 2 × 1 7!−(5!×3!)=7×6×5×4×3×2×1−5×4×3×2×1×3×2×1 Explanation: First, let’s determine the total number of permutations of the letters ABCDEFG. Since we have 7 distinct letters, the number of permutations is given by 7 ! 7!. 7 ! = 7 × 6 × 5 × 4 × 3 × 2 × 1 7!=7×6×5×4×3×2×1 Next, we need to consider the number of permutations that contain the string BCD. To do this, we can treat the string BCD as a single unit. Now, we have 5 distinct units: BCD, A, E, F, G. The number of permutations of these units is given by 5 ! 5!. Question 3 A piece of wire 10 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle. How should the wire be cut so that the total area enclosed is (a) a maximum? (b) A minimum? Answer 1 Let the amount cut for the square be 𝑥 x Then the amount left for the equilateral triangle will be 10 − 𝑥 10−x Side length of the square is 𝑥 4 4 x ​ Area of the square = ( 𝑥 4 ) 2 = 𝑥 2 16 Area of the square=( 4 x ​ ) 2 = 16 x 2 ​ Area of an equilateral triangle with side length 𝑎 a is 3 4 𝑎 2 4 3 ​ ​ a 2 Side length of the equilateral triangle is 10 − 𝑥 3 3 10−x ​ Area of the equilateral triangle is 3 4 ( 10 − 𝑥 3 ) 2 = 3 ( 10 − 𝑥 ) 2 36 4 3 ​ ​ ( 3 10−x ​ ) 2 = 36 3 ​ (10−x) 2 ​ Total area = Area of square + Area of equilateral triangle 𝐴 ( 𝑥 ) = 𝑥 2 16 + 3 ( 10 − 𝑥 ) 2 36 A(x)= 16 x 2 ​ + 36 3 ​ (10−x) 2 ​ Differentiate 𝐴 ′ ( 𝑥 ) = 0 A ′ (x)=0 𝑥 8 = 3 ( 10 − 𝑥 ) 18 8 x ​ = 18 3 ​ (10−x) ​ 18x = 80√3 (10 - x) Question 4 Quentin has 18 unit cubes. How many different rectangular prisms can he build if he uses all of the cubes? Answer 1 Consider the factors of 18. 1,2,3,6,9,18 Quentin could build 3 different rectangular prisms with the following arrangements of unit cubes: 18 x 1 x 1 9 x 2 x 1 6 x 3 x 1 Result: 3 different rectangular prisms Question 5 How to factor 𝑥 2 − 2 𝑥 x 2 −2x Answer 1 Yes, the solution is 𝑥 ( 𝑥 − 2 ) x(x−2). The reason is that in both