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Champion Juicer Grain Mill Attachment Model G-90 |  |
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US $126.30 | 23d 13h 50m |
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KITCHENAID GRAIN MILL ATTACHMENT BY HOBART MODEL GM-A VINTAGE | |
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US $135.00 | 18d 20h 14m |
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Model G-90 Grain Master Grain Mill Grinder Attachment for Champion Juicer | |
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US $69.99 | 7d 15h 53m |
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Mill Attachment Model
will an older Kitchenaid grain mill attachment, model # GM, work with a newer model mitchenaid mixer?
WILL AN OLDER GRAIN MILL ATTACHMENT (MODEL #GM) WORK WITH A NEW KITCHENAID MIXER?
I have no idea. Write to kitchenaid.com...
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|
Champion Juicer Grain Mill Attachment Model G-90 | |
|
US $126.30 | 23d 13h 50m |
|
KITCHENAID GRAIN MILL ATTACHMENT BY HOBART MODEL GM-A VINTAGE | |
|
US $135.00 | 18d 20h 14m |
|
Model G-90 Grain Master Grain Mill Grinder Attachment for Champion Juicer | |
|
US $69.99 | 7d 15h 53m |
| Powered by phpBay Pro |
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KitchenAid Mixer - grain mill attachment
Calculus question.. HELP?
here's a calculus question that i need to know the answer to. pls if you know it solve it step by step till the final answer, if ur comfortableusing microsoft for this then u can email me the answer as an attachment n mention over here that u have. b4 i rwrite the question... THANX a
MILL who ever helps me out... (its integration/anti derevative by the way)
A bacteria popoulatin N thousand, has a growth rate modelled by the equation dN/dt=4000/(1+0.5t), t>= 0 where t is messured in days. Initialy there are 250 bacteria the popoulation, Find the popoulation size after 10 days.
dN/dt = 4000/(1+0.5t)
N(t) = ∫ 4000/(1+0.5t) dt
Use substitution:
u = 1+0.5t
du = 0.5 dt
dt = 2 du
N(t) = ∫ 4000/(1+0.5t) dt
= ∫ 4000/u * 2 du
= 8000 ∫ (1/u) du
= 8000 ln(u) + C
N(t) = 8000 ln(1+0.5t) + C
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Initially there are 250 bacteria in the population:
N(0) = 250
8000 ln(1+0.5(0)) + C = 250
8000 ln(1) + C = 250
8000*0 + C = 250
C = 250
N(t) = 8000 ln(1+0.5t) + 250
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Fin the population size after 10 days:
N(10) = 8000 ln(1+0.5*10) + 250
N(10) = 8000 ln(6) + 250
N(10) = 14584.0757538
There are 14,584 bacteria after 10 days
