My understanding has been that if you know how to set up a problem as a formula then you know how to make it out of toothpicks and rubber bands so you understand it.
This requires a certain ability to manipulate formulas:
2xy/(x+y)=2/(1/x+1/y)=1/(((1/x)+(1/y))/2)
as I’m sure you know.
(but I am 1-in-a-million atypical on this)
So like, the formula for concentrations goes like this:
If we add 1volume of xconcentration to 2volume of yconcentration then:
we are adding 1volume of stuff to 2volume of stuff and getting 1volume+2volume of stuff.
We are adding xconcentration x 1volume to yconcentration x 2volume of special stuff and getting xconcentration x 1volume+yconcentration x 2volume of stuff
so our final concentration is the amount of special stuff divided by the amount of stuff, or
(xconcetration x 1volume+ yconcentration x 2volume)/(1volume+2volume)=zconcentration
So, I mean, that’s all there is. That’s the formula, that’s how the problem works, that’s all that’s at issue.
It’s equivalent to the problem:
You spend 2volume time at yconcentration speed. How much time must you spend at xconcentration speed to attain an average speed of zconcentration?
My understanding has been that if you know how to set up a problem as a formula then you know how to make it out of toothpicks and rubber bands so you understand it.
This requires a certain ability to manipulate formulas:
2xy/(x+y)=2/(1/x+1/y)=1/(((1/x)+(1/y))/2)
as I’m sure you know.
(but I am 1-in-a-million atypical on this)
So like, the formula for concentrations goes like this:
If we add 1volume of xconcentration to 2volume of yconcentration then:
we are adding 1volume of stuff to 2volume of stuff and getting 1volume+2volume of stuff.
We are adding xconcentration x 1volume to yconcentration x 2volume of special stuff and getting xconcentration x 1volume+yconcentration x 2volume of stuff
so our final concentration is the amount of special stuff divided by the amount of stuff, or
(xconcetration x 1volume+ yconcentration x 2volume)/(1volume+2volume)=zconcentration
So, I mean, that’s all there is. That’s the formula, that’s how the problem works, that’s all that’s at issue.
It’s equivalent to the problem:
You spend 2volume time at yconcentration speed. How much time must you spend at xconcentration speed to attain an average speed of zconcentration?