pranil
pranil
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consider flowerpots a, b, c, d you can put a in 4 positions either of 4 you can put b in 3 positions either of remaining 3 you can put c in 2 positions either of remaining 2 you can put d in 1 position remaining 1 total number of ways = 4 × 3 …
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(2 – x)/4 + 3 = 1 (2 – x)/4 = 1 – 3 (2 – x)/4 = – 2 (2 – x) = – 8 – x = – 8 – 2 – x = – 10 x = 10 3 + z/4 + 5 = 19 z/4 + 8 = 19 z/4 = 19 + 8 z/4 = 27 z = 27 Ã 4 = 108 or is it (3 + z)/4 + 5 = 19 (3 + z)/4 = 19 – 5 (3 + z)…
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Sum of 1 to n = n/2[2a + (n – 1)d for 1 to 10 sum = 10/2(a + 9d) 5a + 45d = 495 a + 9d = 99 -------- (1) sum of 1 to 20 = 20/2[2a + 19d] 10a + 190d = 1690 a + 19d = 169 --------(2) (2) – (1) 10d = 70 d = 7 a + 9d = 99 ------…
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(6√0.9)^7 = (√36×0.9)^7 = (√32.4)^7 = (32.4)^7/2 ------
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angle L = angle H ------- 35° angle M = angle I -------- 65° = 2x + 10 angle N = angle J 2x + 10 = 65 2x = 55 x = 27.5° ----
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2 / 9 = 0.222... this is recurring decimals answer is written as bar or dot on the digit which is repeating so answer is 0.2 (bar or dot on 2)
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(2x-1)^2 - 9 use a² – b² = (a + b)(a – b) = (2x-1) - 3)(2x-1) + 3) = (2x - 4)(2x + 2) = 2(x - 2)(2x + 2) = 4(x - 2)(x + 1) (x-b)^50 - x + b = (x – b)^50 - (x – b) = (x – b)[(x – b)^49 – 1] = (x – b)[(x – b)^49/2 – 1][(x – b)^49/2 + 1] …
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(2x-1)^2 - 9 use a² – b² = (a + b)(a – b) = (2x-1) - 3)(2x-1) + 3) = (2x - 4)(2x + 2) = 2(x - 2)(2x + 2) = 4(x - 2)(x + 1) (x-b)^50 - x + b = (x – b)^50 - (x – b) = (x – b)[(x – b)^49 – 1] = (x – b)[(x – b)^49/2 – 1][(x – b)^49/2 + 1] …
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(2x-1)^2 - 9 use a² – b² = (a + b)(a – b) = (2x-1) - 3)(2x-1) + 3) = (2x - 4)(2x + 2) = 2(x - 2)(2x + 2) = 4(x - 2)(x + 1) (x-b)^50 - x + b = (x – b)^50 - (x – b) = (x – b)[(x – b)^49 – 1] = (x – b)[(x – b)^49/2 – 1][(x – b)^49/2 + 1] …
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(2x-1)^2 - 9 use a² – b² = (a + b)(a – b) = (2x-1) - 3)(2x-1) + 3) = (2x - 4)(2x + 2) = 2(x - 2)(2x + 2) = 4(x - 2)(x + 1) (x-b)^50 - x + b = (x – b)^50 - (x – b) = (x – b)[(x – b)^49 – 1] = (x – b)[(x – b)^49/2 – 1][(x – b)^49/2 + 1] …
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volume of new box = l × b × h as it is a square volume = l × l × h l² = v/h = 48/6 = 8 l = √8 = 2√2 original length = 6 + 2√2 + 6 (6' from both sides added) = 12 + 2√2 -----
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volume of new box = l × b × h as it is a square volume = l × l × h l² = v/h = 48/6 = 8 l = √8 = 2√2 original length = 6 + 2√2 + 6 (6' from both sides added) = 12 + 2√2 -----
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150 mi in 3 hr While writing ratio the units must be same here units are minutes and hours. Either convert minutes into hours or hours into minutes Converting minutes into hours 150 min = 150/60 hr. = 2.5 hours So ratio = 2.5 / 3= 25 / 30 = 0.…
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150 mi in 3 hr While writing ratio the units must be same here units are minutes and hours. Either convert minutes into hours or hours into minutes Converting minutes into hours 150 min = 150/60 hr. = 2.5 hours So ratio = 2.5 / 3= 25 / 30 = 0.…
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h – w + l = 32 h + w – l = 28 2h = 60 h = 30 cm to calculate some more data is required as ratio of L : w then only l and w can be calculated. -----
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h – w + l = 32 h + w – l = 28 2h = 60 h = 30 cm to calculate some more data is required as ratio of L : w then only l and w can be calculated. -----
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h – w + l = 32 h + w – l = 28 2h = 60 h = 30 cm to calculate some more data is required as ratio of L : w then only l and w can be calculated. -----
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150 mi in 3 hr While writing ratio the units must be same here units are minutes and hours. Either convert minutes into hours or hours into minutes Converting minutes into hours 150 min = 150/60 hr. = 2.5 hours So ratio = 2.5 / 3= 25 / 30 = 0.…
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volume of new box = l × b × h as it is a square volume = l × l × h l² = v/h = 48/6 = 8 l = √8 = 2√2 original length = 6 + 2√2 + 6 (6' from both sides added) = 12 + 2√2 -----
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volume of new box = l × b × h as it is a square volume = l × l × h l² = v/h = 48/6 = 8 l = √8 = 2√2 original length = 6 + 2√2 + 6 (6' from both sides added) = 12 + 2√2 -----
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volume of new box = l × b × h as it is a square volume = l × l × h l² = v/h = 48/6 = 8 l = √8 = 2√2 original length = 6 + 2√2 + 6 (6' from both sides added) = 12 + 2√2 -----
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(2x-1)^2 - 9 use a² – b² = (a + b)(a – b) = (2x-1) - 3)(2x-1) + 3) = (2x - 4)(2x + 2) = 2(x - 2)(2x + 2) = 4(x - 2)(x + 1) (x-b)^50 - x + b = (x – b)^50 - (x – b) = (x – b)[(x – b)^49 – 1] = (x – b)[(x – b)^49/2 – 1][(x – b)^49/2 + 1] …
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2 / 9 = 0.222... this is recurring decimals answer is written as bar or dot on the digit which is repeating so answer is 0.2 (bar or dot on 2)
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angle L = angle H ------- 35° angle M = angle I -------- 65° = 2x + 10 angle N = angle J 2x + 10 = 65 2x = 55 x = 27.5° ----
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angle L = angle H ------- 35° angle M = angle I -------- 65° = 2x + 10 angle N = angle J 2x + 10 = 65 2x = 55 x = 27.5° ----
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angle L = angle H ------- 35° angle M = angle I -------- 65° = 2x + 10 angle N = angle J 2x + 10 = 65 2x = 55 x = 27.5° ----
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angle L = angle H ------- 35° angle M = angle I -------- 65° = 2x + 10 angle N = angle J 2x + 10 = 65 2x = 55 x = 27.5° ----
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Sum of 1 to n = n/2[2a + (n – 1)d for 1 to 10 sum = 10/2(a + 9d) 5a + 45d = 495 a + 9d = 99 -------- (1) sum of 1 to 20 = 20/2[2a + 19d] 10a + 190d = 1690 a + 19d = 169 --------(2) (2) – (1) 10d = 70 d = 7 a + 9d = 99 ------…
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(2 – x)/4 + 3 = 1 (2 – x)/4 = 1 – 3 (2 – x)/4 = – 2 (2 – x) = – 8 – x = – 8 – 2 – x = – 10 x = 10 3 + z/4 + 5 = 19 z/4 + 8 = 19 z/4 = 19 + 8 z/4 = 27 z = 27 Ã 4 = 108 or is it (3 + z)/4 + 5 = 19 (3 + z)/4 = 19 – 5 (3 + z)…
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consider flowerpots a, b, c, d you can put a in 4 positions either of 4 you can put b in 3 positions either of remaining 3 you can put c in 2 positions either of remaining 2 you can put d in 1 position remaining 1 total number of ways = 4 × 3 …