Sorasyn

Well it was a shot. Lol.

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.

Permutation is the act of re-arranging a set of objects (Primarily numbers, but also applicable with about anything), to every feasible sub-set of that particular set. 

{0, 1, 2} = (0, 1, 2), (0, 2, 1), (1, 0, 2), (1, 2, 0), (2, 0, 1), (2, 1, 0);

Number of Permutations = n! where "n" is the length of a particular set.

Pythagorean's Theorem states that, with respect to triangles:

$${{\mathtt{A}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{B}}}^{{\mathtt{2}}} = {{\mathtt{C}}}^{{\mathtt{2}}}$$

Where A, and B, are side lengths, and C is the hypotenuse. Since you know both A, and B, in terms of meters, we're left with one variable. 

$${\mathtt{36}}{\mathtt{\,\small\textbf+\,}}{\mathtt{81}} = {{\mathtt{C}}}^{{\mathtt{2}}}$$

Solving for C gives us: 

$${\mathtt{C}} = {\sqrt{{\mathtt{117}}}}$$

How do you decompose 57.9 times 23?

$${\mathtt{57.9}} = {\mathtt{57}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{9}}}{{\mathtt{10}}}}$$

So, break down 57.9, and distribute 23 across both coefficients.

$${\mathtt{23}}{\mathtt{\,\times\,}}\left({\mathtt{57}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{9}}}{{\mathtt{10}}}}\right) = {\mathtt{X}}$$

Solve for X. :)

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. 

Not sure whether or not this is a troll post, but ":" means ratio. So you have 10:2 or 5:1. Five to one, or 5/1.

I'm feeling lazy today, so I'm not gonna do some grand write-up of how I got the answer. However I'll give you the basics.

$$Grams \Leftrightarrow Mols \Leftrightarrow Atoms$$

Since you know how many atoms of calcium you have, you'll divide that number by Avogadro's Constant (6.022 x 10²³) to deduce the amount of mols of Calcium you have, giving:

$${\mathtt{0.516}}{\mathtt{\,\times\,}}{\mathtt{molCa}}$$

Checking the periodic table shows us that Calcium has a molar weight of 40.08g/mol Ca. Multiply the number of mols, and it's molar weight, giving:

$${\mathtt{0.516}}{\mathtt{\,\times\,}}{\mathtt{molCa}}{\mathtt{\,\times\,}}\left({\frac{\left({\mathtt{40.08}}{\mathtt{\,\times\,}}{\mathtt{gCa}}\right)}{\left({\mathtt{1}}{\mathtt{\,\times\,}}{\mathtt{molCa}}\right)}}\right) = {\mathtt{20.68}}{\mathtt{\,\times\,}}{\mathtt{gCa}}$$

Good luck.

Still got a bee in our bonnet about an incorrect answer I gave back in December, have we? As flattering as it is that you've held on for so long, I'll have to ask you to get over it. Mistakes happen; I'm quite sure that you, yourself, are no angel.

Narcissists are an endearing breed. 

@Melody : Thank you, but I don't spend much time worrying about what other people have to say.

$$pH = -log([H+])$$

According to logarithmic identities, if we raise both sides to the power of the logarithmic base, we can eliminate the log. To do this, we first divide both sides by -1:

$$-pH = log([H+])$$

Then, we raise both sides to the power of 10, thus eliminating the logarithm. Leaving us with:

$$10^-^p^H = [H^+]$$

Hope this helps. 

4.34 x 10^-1

I'll need some Ugg boots, and a Pumpkin Spice latte from Starbucks before I can answer that question. Lol.

1.25 x 10^-2

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:

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