# Quick help with Year 5 maths homework please? Topic: Quick help with Year 5 maths homework please?
July 16, 2019 / By Urbanna
Question: By the way, the numbers under on the 'Part of pattern' on the right are the 'Number of Counters'. Part of pattern (term) Number of Counters 1 1 2 3 3 6 4 10 5 15 Write the algebraic expression to find the pattern. Thanks. First correcter answer gets 10 points!  Savanna | 1 day ago
x f(x) = Σ i         i=1 f(1) = 1 f(2) = 1 + 2 = 3 f(3) = 1 + 2 + 3 = 6 f(4) = 1 + 2 + 3 + 4 = 10 f(5) = 1 + 2 + 3 + 4 + 5 = 15 f(6) = 1 + 2 + 3 + 4 + 5 + 6 = 21 ... have a look at 1 + 2 + 3 + 4 + 5 + 6 rearange it like this: (1+6) + (2+5) + (3+4) = 7 + 7 + 7 you see that each term has become (x+1), and you take it x/2 times so f(x) = (x + 1) ∙ x / 2 f(1) = (1 + 1) ∙ 1 / 2 = 1 f(2) = (2 + 1) ∙ 2 / 2 = 3 f(3) = (3 + 1) ∙ 3 / 2 = 6 f(4) = (4 + 1) ∙ 4 / 2 = 10 f(5) = (5 + 1) ∙ 5 / 2 = 15 f(6) = (6 + 1) ∙ 6 / 2 = 21 ... PS: it's a bit more difficult to explain with odd numbers, but basicaly it is the same...
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x f(x) = Σ i         i=1 f(1) = 1 f(2) = 1 + 2 = 3 f(3) = 1 + 2 + 3 = 6 f(4) = 1 + 2 + 3 + 4 = 10 f(5) = 1 + 2 + 3 + 4 + 5 = 15 f(6) = 1 + 2 + 3 + 4 + 5 + 6 = 21 ... have a look at 1 + 2 + 3 + 4 + 5 + 6 rearange it like this: (1+6) + (2+5) + (3+4) = 7 + 7 + 7 you see that each term has become (x+1), and you take it x/2 times so f(x) = (x + 1) ∙ x / 2 f(1) = (1 + 1) ∙ 1 / 2 = 1 f(2) = (2 + 1) ∙ 2 / 2 = 3 f(3) = (3 + 1) ∙ 3 / 2 = 6 f(4) = (4 + 1) ∙ 4 / 2 = 10 f(5) = (5 + 1) ∙ 5 / 2 = 15 f(6) = (6 + 1) ∙ 6 / 2 = 21 ... PS: it's a bit more difficult to explain with odd numbers, but basicaly it is the same... Originally Answered: Quick help with Year 5 maths homework please?
I'm not sure what is meant by either "Part of pattern" or "Number of Counters," but the numbers on the right are the sums of the numbers on the left. 1 = 1 1 + 2 = 3 1 + 2 + 3 = 6 etc. If summation notation is allowed, something like y = sum(n), where n goes from 1 to x. If summation notation is not allowed, check the differences between the "Number of Counters" values: 3 - 1 = 2 6 - 3 = 3 10 - 6 = 4 etc. Since these differences are not constant, try the differences between these differences: 3 - 2 = 1 4 - 3 = 1 etc. Here the constant difference is one. Since this constant difference showed up in our second set of subtractions, the function must have an x^2 term. One way to find the specific function is to write quadratic equations for each value and solve for the unknown variables. The general quadratic expression is ax^2 + bx + c. When x = 1 the quadratic expression is a(1)^2 + b(1) + c, or a + b + c. For the second term, the expression is a a(2)^2 +b(2)+ c, or 4a + 2b + c. Putting these in the chart makes it look like this: 3 – 1 = (4a + 2b + c) – (a + b + c) = 3a + b 6 – 3 = (9a + 3b + c) – (4a + 2b + c) = 5a + b 10 – 6 = (16a + 4b +c) – (9a +3b +c) = 7a + b The second order differences look like: (5a + b) – (3a + b) = 2a (7a + b) – (5a + b) = 2a Again, there’s a constant difference. Equating this generic expression difference to the sequence difference gives 2a = 1. Solving, we see a = ½. Substituting into the first differences, we see 3a + b = 2. Since a = ½, b = ½. This holds true for 5a + b = 3 and 7a + b = 4. Substituting into the basic expression, ax^2 + bx + c = 1, gives ½ + ½ + c = 1 when x = 1, and we see c = 0. This holds true when x = 2 -- ½(4) + ½(2) + c = 3, when x = 3 – ½(9) + ½(3) + c = 6, etc. The pattern, therefore, is (½)x^2 + (½)x, or (x^2 + x)/2. Noreen
I'm not sure what is meant by either "Part of pattern" or "Number of Counters," but the numbers on the right are the sums of the numbers on the left. 1 = 1 1 + 2 = 3 1 + 2 + 3 = 6 etc. If summation notation is allowed, something like y = sum(n), where n goes from 1 to x. If summation notation is not allowed, check the differences between the "Number of Counters" values: 3 - 1 = 2 6 - 3 = 3 10 - 6 = 4 etc. Since these differences are not constant, try the differences between these differences: 3 - 2 = 1 4 - 3 = 1 etc. Here the constant difference is one. Since this constant difference showed up in our second set of subtractions, the function must have an x^2 term. One way to find the specific function is to write quadratic equations for each value and solve for the unknown variables. The general quadratic expression is ax^2 + bx + c. When x = 1 the quadratic expression is a(1)^2 + b(1) + c, or a + b + c. For the second term, the expression is a a(2)^2 +b(2)+ c, or 4a + 2b + c. Putting these in the chart makes it look like this: 3 – 1 = (4a + 2b + c) – (a + b + c) = 3a + b 6 – 3 = (9a + 3b + c) – (4a + 2b + c) = 5a + b 10 – 6 = (16a + 4b +c) – (9a +3b +c) = 7a + b The second order differences look like: (5a + b) – (3a + b) = 2a (7a + b) – (5a + b) = 2a Again, there’s a constant difference. Equating this generic expression difference to the sequence difference gives 2a = 1. Solving, we see a = ½. Substituting into the first differences, we see 3a + b = 2. Since a = ½, b = ½. This holds true for 5a + b = 3 and 7a + b = 4. Substituting into the basic expression, ax^2 + bx + c = 1, gives ½ + ½ + c = 1 when x = 1, and we see c = 0. This holds true when x = 2 -- ½(4) + ½(2) + c = 3, when x = 3 – ½(9) + ½(3) + c = 6, etc. The pattern, therefore, is (½)x^2 + (½)x, or (x^2 + x)/2.
👍 120 | 👎 -7 Mable
Q2. some packets of smarties are emptied onto a table. Ness takes away 0.5 of the pile. Marley then takes one- 0.33 of what's left. Adele takes 2 and eats them. 0.5 of those left are purple and there are 8 purple smarties. what proportion smarties have been there initially? Ans2. permit x be the type of unique smarties Ness takes away 0.5 of the pile i.e x/2 so what's left is x-x/2 = x/2 Marley then takes one- 0.33 of what's left i.e (x/2)/ 3 = x/6 Adele takes 2 total of what's left after Marley and Adele take their smarties = ( x/2) - (x/6) - 2= (3x/6 - x/6) - 2 = 2x/6 -2 = x/3 - 2 0.5 of those left are purple i,e (x/3 - 2)/2 =x/6 -a million and there are 8 purple smarties. i.e x/6 -a million = 8 including one to the two factors x/6 = 9 multiplying the two factors via 6 x= fifty 4 i.e sort of unique smartires are fifty 4
👍 120 | 👎 -15 Originally Answered: maths homework pleeeeeez help?
Haha i did this in maths today.. ok you need to find the midpoint for each thing.. eg 25 for the first one. And then multiply by the frequency which i guess is 5 ? I think you then add up all these values, and then divide by the total frequency.. (5+9+13+8+7+8) I think this is right =S sorry if its not x x x

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