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Answers 24


Alright, so I was curious to give this question the old college try, and here's what I came back with. Lol. 


Starting off, we know that John's vegetable garden is 12' by 24'. We also know the volume of the mulch bags he'll be using are 1.53ft, and that they cost $4/bag.


Garden Area = 12ft * 24ft = 288ft2

Volume of Mulch Bag = 1.53ft = 3.375ft3


To find out how much area one mulch bag covers, spread at a height of 3'', we'll permutate our cube's dimensions proportionately. We know that the volume does not change, the height is 3'', and that the width and length of our cube are equal to one another. 


Width = x

Length = x

Height = 0.25' (Converted from inches to feet)


$$Volume = L * W * H = 3.375ft^3$$



Replacing what we know in the Volume equation gives us: 


$$3.375 = 0.25 * X^2$$


Solving for X gives us:


$$X^2 = 13.5ft^2$$

$$X = sqrt(13.5) \approx 3.674ft$$


Now, we know that one bag of mulch covers 13.5ft2 at a height of three inches. We divide the area of John's garden, by the area of one bag of mulch at 3" tall.


$${\frac{{\mathtt{288}}\left[{{ft}}^{{\mathtt{2}}}\right]}{\left({\mathtt{13.5}}{\mathtt{\,\times\,}}\left({\frac{{{\mathtt{ft}}}^{{\mathtt{2}}}}{{\mathtt{bag}}}}\right)\right)}} = {\mathtt{21.33}}{\mathtt{\,\times\,}}{\mathtt{bags}}$$


Multiply this by John's cost per bag, giving us:


$$\left({\mathtt{21.33}}{\mathtt{\,\times\,}}{\mathtt{bags}}\right){\mathtt{\,\times\,}}{\mathtt{4}}{\mathtt{\,\times\,}}\left({\frac{{\mathtt{dollar}}}{{\mathtt{bag}}}}\right) = {\mathtt{85.33}}{dollar}$$


It'll cost $85.33.

Feb 20, 2015

I posted this response to someone else who was converting 4.87*10^22 atoms of Carbon to Grams. It should serve as a good example of how to convert between the three. Note that all units being converted are cancelled out. 


 Here's the link to the question:


Assuming that you're speaking in units of atoms, you can convert them like so:


$$Grams \Leftrightarrow Mols \Leftrightarrow Atoms$$


Using Avogadro's Constant, we know that 1 mol = 6.022 x 10^23 atoms, allowing us to convert from atoms to mols:


$$\left({\mathtt{4.87}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{22}}}{\mathtt{\,\times\,}}{\mathtt{atomsC}}\right){\mathtt{\,\times\,}}\left({\frac{\left({\mathtt{1}}{\mathtt{\,\times\,}}{\mathtt{molC}}\right)}{\left({\mathtt{6.022}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{23}}}{\mathtt{\,\times\,}}{\mathtt{atomsC}}\right)}}\right) = {\mathtt{0.808\: \!7}}{\mathtt{\,\times\,}}{\mathtt{molsC}}$$


To get grams, we convert again only from mols, to grams. Knowing that 1 molC = 12.00gC we get:


$${\mathtt{0.808\: \!7}}{\mathtt{\,\times\,}}{\mathtt{molsC}}{\mathtt{\,\times\,}}\left({\frac{\left({\mathtt{12}}{\mathtt{\,\times\,}}{\mathtt{gC}}\right)}{\left({\mathtt{1}}{\mathtt{\,\times\,}}{\mathtt{molC}}\right)}}\right) = {\mathtt{9.077}}{\mathtt{\,\times\,}}{\mathtt{gC}}$$

Feb 20, 2015
Dec 15, 2014
Dec 11, 2014