Please Help Out With This - Romance - Nairaland
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| Please Help Out With This by senatepresido(op): 4:14pm On Oct 27, 2022 |
Mathematicians in the house should please help attend to these questions and do well to show their workings MATHEMATICS INSTRUCTIONS: ANSWER ALL THEORY Work out 1. 9×(3/4 + 1/3) 2. 6÷(34/9 - 1 7/9) 3. (4 1/3 + 9 2/3)÷(2/5 + 3/10) 4. Arrange 7/11, 4/13, 5/6, 3/7 in descending order. 5. Find the missing numbers or letters 2600/7800 = 26/t = u/39 = 1/v Section B Evaluate in base 2 1. (111 + 101) x (111 - 101) 2. 1 111 011 ÷ 101 001 3.Simplify (a+3b)/3 - (5a-3b)/6 4. Find the product of 2 1/2 and -4, and divide the result by -1 1/3 5. Translate (3×15-9)/6 into words |
| Re: Please Help Out With This by kushme: 11:02am On Oct 28, 2022*. Modified: 2:43pm On Oct 28, 2022 |
2. 6÷(34/9 - 1 7/9)= Change 1 7/9 to improper fraction= 16/9 6÷(34/9 - 16/9) = Remove bracket (LCM=9) 9÷9×34=34, also 9÷9×16=16 (34-16)/9= 18/9 6 ÷ 18/9 = 6÷2 = 3 3. ( 4 1/3 + 9 2/3) ÷ ( 2/5 + 3/10) = 20 First thing first, change the mixed fractions to improper fractions, then remove the brackets. 4 1/3= 13/3, 9 2/3= 29/3 (13/3 + 29/3) ÷ ( 2/5 + 3/10 ) = Cut the LCM ÷ and working short, First bracket ÷ second bracket 42/3 ÷ 7/10 Operational sign changes from ÷ to ×,thereby making the 7/10 to change position(10 going up, 7 coming down). Like this: 42/3 × 10/7 = 14/1 × 10/7 = 2/1 × 10/1 = 20 |
| Re: Please Help Out With This by kushme: 11:03am On Oct 28, 2022*. Modified: 12:12pm On Oct 28, 2022 |
senatepresido:I no be mathematician but I pass math for O'level ![]() let me try! Section A Solution 1. 9×(3/4+ 1/3) firstly, solve to remove bracket. the LCM of the denominators 4 and 3 is 12 12÷4×3=9 and 12÷3×1=4 Now, we have 9+4/12= 13/12 Therefore 9×13/12= 3×13/4 = 39/4 = 9whole number 3/4 or 9.75 |
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