fatguy
fatguy
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0.80 * Value > 8400 Value > 8400/0.80 Value > 10500
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0.80 * Value > 8400 Value > 8400/0.80 Value > 10500
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f(x) = 0.18x f(40) = 0.18(40) = $7.20
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I believe that it correct. an "increase [of] 140%" is 100% + 140% of the original value.
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solve for what? y? 24y = 8y*L 3y = L and y =L/3 are the best you can do. have to have at least as many equations as variables in order to "solve"
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solve for what? y? 24y = 8y*L 3y = L and y =L/3 are the best you can do. have to have at least as many equations as variables in order to "solve"
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solve for what? y? 24y = 8y*L 3y = L and y =L/3 are the best you can do. have to have at least as many equations as variables in order to "solve"
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solve for what? y? 24y = 8y*L 3y = L and y =L/3 are the best you can do. have to have at least as many equations as variables in order to "solve"
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xy=8 ; y = 8/x 1/x - 1/(8/x) =1/4 1/x - x/8 =1/4 1 - (1/8)x^2 = x/4 8 - x^2 = 2x x^2 + 2x - 8 = 0 x^2 - 2x + 4x - 8 = 0 x(x - 2) + 4(x - 2) = 0 (x + 4)(x - 2) = 0 x = -4, 2; y = 8/x when x = -4 y = 8/-4 y = -2 (-4, -2) when x=2 …
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xy=8 ; y = 8/x 1/x - 1/(8/x) =1/4 1/x - x/8 =1/4 1 - (1/8)x^2 = x/4 8 - x^2 = 2x x^2 + 2x - 8 = 0 x^2 - 2x + 4x - 8 = 0 x(x - 2) + 4(x - 2) = 0 (x + 4)(x - 2) = 0 x = -4, 2; y = 8/x when x = -4 y = 8/-4 y = -2 (-4, -2) when x=2 …
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xy=8 ; y = 8/x 1/x - 1/(8/x) =1/4 1/x - x/8 =1/4 1 - (1/8)x^2 = x/4 8 - x^2 = 2x x^2 + 2x - 8 = 0 x^2 - 2x + 4x - 8 = 0 x(x - 2) + 4(x - 2) = 0 (x + 4)(x - 2) = 0 x = -4, 2; y = 8/x when x = -4 y = 8/-4 y = -2 (-4, -2) when x=2 …
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solve for what? y? 24y = 8y*L 3y = L and y =L/3 are the best you can do. have to have at least as many equations as variables in order to "solve"
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solve for what? y? 24y = 8y*L 3y = L and y =L/3 are the best you can do. have to have at least as many equations as variables in order to "solve"
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I believe that it correct. an "increase [of] 140%" is 100% + 140% of the original value.
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I believe that it correct. an "increase [of] 140%" is 100% + 140% of the original value.
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I believe that it correct. an "increase [of] 140%" is 100% + 140% of the original value.
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I believe that it correct. an "increase [of] 140%" is 100% + 140% of the original value.
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I believe that it correct. an "increase [of] 140%" is 100% + 140% of the original value.
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f(x) = 0.18x f(40) = 0.18(40) = $7.20
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0.80 * Value > 8400 Value > 8400/0.80 Value > 10500