seamus-o
seamus-o
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Let number of qt 25% solution = x and let number of qt 75% solution = y 0.25x + 0.75y = 0.4 x 50 0.25x + 0.75y = 20 ... [eqn 1] x + y = 50 y = 50 - x ... [eqn 2] Subs [eqn 2] into [eqn 1] → 0.25x + 0.75(50 - x) = 20 0.25x + 37…
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y = -0.005x² + x + 5 y is its height above ground level so when it hits the ground y = 0 so -0.005x² + x + 5 = 0 Now use the quadratic formula x = [-b ± â(b² - 4ac)] / 2a to find x (the distance the ball has traveled horizontally) …
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Yes ... you've chosen the correct formula to use Air resistance is being ignored ... so the time taken by the glasses and by the pen to drop the 32 m to the ground is the same So first find the time it takes the glasses to fall to the ground…
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Yes ... you've chosen the correct formula to use Air resistance is being ignored ... so the time taken by the glasses and by the pen to drop the 32 m to the ground is the same So first find the time it takes the glasses to fall to the ground…
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Yes ... you've chosen the correct formula to use Air resistance is being ignored ... so the time taken by the glasses and by the pen to drop the 32 m to the ground is the same So first find the time it takes the glasses to fall to the ground…
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y = -0.005x² + x + 5 y is its height above ground level so when it hits the ground y = 0 so -0.005x² + x + 5 = 0 Now use the quadratic formula x = [-b ± â(b² - 4ac)] / 2a to find x (the distance the ball has traveled horizontally) …
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Let number of qt 25% solution = x and let number of qt 75% solution = y 0.25x + 0.75y = 0.4 x 50 0.25x + 0.75y = 20 ... [eqn 1] x + y = 50 y = 50 - x ... [eqn 2] Subs [eqn 2] into [eqn 1] → 0.25x + 0.75(50 - x) = 20 0.25x + 37…