Alan

Show us exactly what you typed in!

You do, indeed, add all the like terms, but, unfortunately, your step 2 doesn't match the original expression, Anonymous!

8x + 4y + 4z - 5x + 2z -8y =

3x + 4y + 4z + 2z -8y     combining the x's

3x - 4y + 4z + 2z            combining the y's

3x - 4y + 6z                   combining the z's

We can also attack this by noting that f(x) will be zero when the ball lands as well as when it starts.

-5x² + 40x = 0

x*(-5x + 40) = 0

so x=0 (start)

and -5x + 40 = 0 or 5x = 40 or x = 8 (when it lands).

Perhaps this graph will help.

tan(64°) is given by the distance from radio mast to library, divided by height of balloon (h), or:

tan(64°)=24/h

Rearrange to get h = 24/tan(64°)

$${\mathtt{h}} = {\frac{{\mathtt{24}}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}{\left({\mathtt{64}}^\circ\right)}}} \Rightarrow {\mathtt{h}} = {\mathtt{11.705\: \!582\: \!125\: \!582\: \!365\: \!3}}$$  

h ≈ 11.7 miles

 Edit:  I should have used 63° not 64° in the above!!  

         Thanks for pointing this out Chris.

$${\mathtt{h}} = {\frac{{\mathtt{24}}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}{\left({\mathtt{63}}^\circ\right)}}} \Rightarrow {\mathtt{h}} = {\mathtt{12.228\: \!610\: \!787\: \!867\: \!229\: \!7}}$$

h ≈ 12.2 miles

Plot the ordered pairs on a graph.  See if that suggests an obvious functional relationship (like a polynomial of some sort, for example).

Either change the n to x or solve for n!!!

(2n+5)*(n-3) = 2n²-n-15 = 0

so n = -5 or n = 3

Similarly,

2x²-x-15=0 means x = -5 or x = 3

If, on the calculator on the Home page here, you press 2nd and then the const button, you will get a drop-down list of lots of physical constants that you can just select, instead of typing them in.

My view is:

Probability that 4 non-skiers died (hence 3 skiers survived) is:

$${\frac{{\mathtt{14}}{\mathtt{\,\times\,}}{\mathtt{13}}{\mathtt{\,\times\,}}{\mathtt{12}}{\mathtt{\,\times\,}}{\mathtt{11}}}{\left({\mathtt{17}}{\mathtt{\,\times\,}}{\mathtt{16}}{\mathtt{\,\times\,}}{\mathtt{15}}{\mathtt{\,\times\,}}{\mathtt{14}}\right)}} = {\frac{{\mathtt{143}}}{{\mathtt{340}}}} = {\mathtt{0.420\: \!588\: \!235\: \!294\: \!117\: \!6}}$$

 Oops! Just noticed that this is the same as Bertie's reply (though Bertie did a more complete job by including an explanation).

pH is -log to base 10 of H⁺ concentration, so:

a.  $${\mathtt{pH}} = {\mathtt{\,-\,}}{log}_{10}\left({\mathtt{7.9}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{-{\mathtt{9}}}\right) \Rightarrow {\mathtt{pH}} = {\mathtt{8.102\: \!372\: \!908\: \!709\: \!558}}$$   

or pH≈8.1

b.  $${\mathtt{conc}} = {{\mathtt{10}}}^{-{\mathtt{3.2}}} \Rightarrow {\mathtt{conc}} = {\mathtt{0.000\: \!630\: \!957\: \!344\: \!480\: \!2}}$$

conc of H⁺ ≈ 6.3*10^-4