helmut
helmut
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R(x) = -20x^2 + 1000x R(x) = -20(x^2 - 50x) R(x) = -20(x^2 - 50x + 25^2 - 25^2) R(x) = -20(x - 25)^2 + 12,500 a. R increases for x < 25 R decreases for x > 25 b. R is maximum at x = 25 c. $12,500
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R(x) = -20x^2 + 1000x R(x) = -20(x^2 - 50x) R(x) = -20(x^2 - 50x + 25^2 - 25^2) R(x) = -20(x - 25)^2 + 12,500 a. R increases for x < 25 R decreases for x > 25 b. R is maximum at x = 25 c. $12,500
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f = 299,792,458/4.4 ≈ 68.1 MHz
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In order to keep the amount invested in the growth fund to a minimum, nothing can be invested in CDs, as they have the lowest return. Thus you have g + b = 50,000 0.146g + 0.087b = 5,000 0.087g + 0.087b = 4,350 0.059g = 650 g = $11,016.95 b =…
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b = c = d = f = g > a > e if the acceleration at a is just before the ball is released. a = b = c = d = f = g > e if the acceleration at a is when or just after the ball is released.
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1. (4/3)ÏR^3 = 234,000 R^3 = 234,000/((4/3)Ï) R = (234,000/((4/3)Ï))^(1/3) C = 2Ï(234,000/((4/3)Ï))^(1/3) C = 240.19 cm 2. Because salt water is more dense than fresh water a floating object doesn't sink as far down before its weight i…
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1. (4/3)ÏR^3 = 234,000 R^3 = 234,000/((4/3)Ï) R = (234,000/((4/3)Ï))^(1/3) C = 2Ï(234,000/((4/3)Ï))^(1/3) C = 240.19 cm 2. Because salt water is more dense than fresh water a floating object doesn't sink as far down before its weight i…
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1. (4/3)ÏR^3 = 234,000 R^3 = 234,000/((4/3)Ï) R = (234,000/((4/3)Ï))^(1/3) C = 2Ï(234,000/((4/3)Ï))^(1/3) C = 240.19 cm 2. Because salt water is more dense than fresh water a floating object doesn't sink as far down before its weight i…
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b = c = d = f = g > a > e if the acceleration at a is just before the ball is released. a = b = c = d = f = g > e if the acceleration at a is when or just after the ball is released.
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In order to keep the amount invested in the growth fund to a minimum, nothing can be invested in CDs, as they have the lowest return. Thus you have g + b = 50,000 0.146g + 0.087b = 5,000 0.087g + 0.087b = 4,350 0.059g = 650 g = $11,016.95 b =…
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f = 299,792,458/4.4 ≈ 68.1 MHz
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f = 299,792,458/4.4 ≈ 68.1 MHz
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R(x) = -20x^2 + 1000x R(x) = -20(x^2 - 50x) R(x) = -20(x^2 - 50x + 25^2 - 25^2) R(x) = -20(x - 25)^2 + 12,500 a. R increases for x < 25 R decreases for x > 25 b. R is maximum at x = 25 c. $12,500