iggy-rocko
iggy-rocko
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4:59 - 2:15 = 4 - 2 = 2 hours and 59 - 15 = 44 minutes 1:45 - 8:15 = 13:45 - 8:15 = 13 - 8 = 5 hours and 45 - 15 = 30 minutes 2 hours, 44 minutes + 5 hours, 30 minutes = 7 hours, 74 minutes = 8 hours, 14 minutes Round this up to 8 hours, 15 minut…
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Since the numerator and denominator approach 0, use L'Hopital's rule and take the derivatives of the numerator and denominator. Also, it's pi, not pie. lim (1 - tanx)/(sinx - cosx) = x-> pi/4 lim -sec^2x/(cosx + sinx) = -(√2)^2/(√2/2 + √2/2) …
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Since the numerator and denominator approach 0, use L'Hopital's rule and take the derivatives of the numerator and denominator. Also, it's pi, not pie. lim (1 - tanx)/(sinx - cosx) = x-> pi/4 lim -sec^2x/(cosx + sinx) = -(√2)^2/(√2/2 + √2/2) …
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If 31 students go, 35 total will go. 35 is the output for 31 in the table. If 33 students go, 37 total will go. 37 is the output for 33 in the table.
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If 31 students go, 35 total will go. 35 is the output for 31 in the table. If 33 students go, 37 total will go. 37 is the output for 33 in the table.
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Let u = 2x - 1. u^2 - 9 = (u + 3)(u - 3) = (2x - 1 + 3)(2x - 1 - 3) = (2x + 2)(2x - 4) (x - b)^50 - x + b = (x - b)^50 - (x - b) = (x - b)((x - b)^49 - 1) =
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Let u = 2x - 1. u^2 - 9 = (u + 3)(u - 3) = (2x - 1 + 3)(2x - 1 - 3) = (2x + 2)(2x - 4) (x - b)^50 - x + b = (x - b)^50 - (x - b) = (x - b)((x - b)^49 - 1) =
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Let u = 2x - 1. u^2 - 9 = (u + 3)(u - 3) = (2x - 1 + 3)(2x - 1 - 3) = (2x + 2)(2x - 4) (x - b)^50 - x + b = (x - b)^50 - (x - b) = (x - b)((x - b)^49 - 1) =
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Let u = 2x - 1. u^2 - 9 = (u + 3)(u - 3) = (2x - 1 + 3)(2x - 1 - 3) = (2x + 2)(2x - 4) (x - b)^50 - x + b = (x - b)^50 - (x - b) = (x - b)((x - b)^49 - 1) =
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16b^2 - 25b = 0 b(16b - 25) = 0 b = 0 or 16b - 25 = 0 b = 0 or b = 25/16
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(x, y) = (11cos(3pi/4), 11sin(3pi/4)) = (11 * -√2/2, 11 * √2/2) = (-11√2/2, 11√2/2)
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You've given an expression, not an equation so x can't be solved for. At best you can say (x - 1/3)^2 = x^2 - 2x/3 + 1/9
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x + larger angle = 180 Since the problem says "more than," you use an inequality. larger angle > 10x There is no exact solution. If you meant the larger angle's measure equals ten times the smaller angle's measure, then x + 10x = 1…
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x + larger angle = 180 Since the problem says "more than," you use an inequality. larger angle > 10x There is no exact solution. If you meant the larger angle's measure equals ten times the smaller angle's measure, then x + 10x = 1…
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You've given an expression, not an equation so x can't be solved for. At best you can say (x - 1/3)^2 = x^2 - 2x/3 + 1/9
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(x, y) = (11cos(3pi/4), 11sin(3pi/4)) = (11 * -√2/2, 11 * √2/2) = (-11√2/2, 11√2/2)
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(x, y) = (11cos(3pi/4), 11sin(3pi/4)) = (11 * -√2/2, 11 * √2/2) = (-11√2/2, 11√2/2)
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16b^2 - 25b = 0 b(16b - 25) = 0 b = 0 or 16b - 25 = 0 b = 0 or b = 25/16
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Let u = 2x - 1. u^2 - 9 = (u + 3)(u - 3) = (2x - 1 + 3)(2x - 1 - 3) = (2x + 2)(2x - 4) (x - b)^50 - x + b = (x - b)^50 - (x - b) = (x - b)((x - b)^49 - 1) =
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If 31 students go, 35 total will go. 35 is the output for 31 in the table. If 33 students go, 37 total will go. 37 is the output for 33 in the table.
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If 31 students go, 35 total will go. 35 is the output for 31 in the table. If 33 students go, 37 total will go. 37 is the output for 33 in the table.
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Since the numerator and denominator approach 0, use L'Hopital's rule and take the derivatives of the numerator and denominator. Also, it's pi, not pie. lim (1 - tanx)/(sinx - cosx) = x-> pi/4 lim -sec^2x/(cosx + sinx) = -(√2)^2/(√2/2 + √2/2) …