Answer:
7 miles
Step-by-step explanation:
Distance = Rate times Time
Distance #1: (15 mph)(12/60 hr) = 3 miles
Distance #2: (12 mph)(20/60 hr) = 4 miles
Total distance ridden = 7 miles
It depends, but most of the time it's<span> a round plane figure whose boundary (the circumference) consists of points equidistant from a fixed point.</span>
Check the picture below.

so the object hits the ground when h(x) = 0, hmmm how long did it take to hit the ground the first time anyway?

now, we know the 2nd time around it hit the ground, h(x) = 0, but it took less time, it took 0.5 or 1/2 second less, well, the first time it took 3/2, if we subtract 1/2 from it, we get 3/2 - 1/2 = 2/2 = 1, so it took only 1 second this time then, meaning x = 1.
![\bf ~~~~~~\textit{initial velocity in feet} \\\\ h(x) = -16x^2+v_ox+h_o \quad \begin{cases} v_o=\textit{initial velocity}&0\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&\\ \qquad \textit{of the object}\\ h=\textit{object's height}&0\\ \qquad \textit{at "t" seconds}\\ x=\textit{seconds}&1 \end{cases} \\\\\\ 0=-16(1)^2+0x+h_o\implies 0=-16+h_o\implies 16=h_o \\\\[-0.35em] ~\dotfill\\\\ ~\hfill h(x) = -16x^2+16~\hfill](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~%5Ctextit%7Binitial%20velocity%20in%20feet%7D%20%5C%5C%5C%5C%20h%28x%29%20%3D%20-16x%5E2%2Bv_ox%2Bh_o%20%5Cquad%20%5Cbegin%7Bcases%7D%20v_o%3D%5Ctextit%7Binitial%20velocity%7D%260%5C%5C%20%5Cqquad%20%5Ctextit%7Bof%20the%20object%7D%5C%5C%20h_o%3D%5Ctextit%7Binitial%20height%7D%26%5C%5C%20%5Cqquad%20%5Ctextit%7Bof%20the%20object%7D%5C%5C%20h%3D%5Ctextit%7Bobject%27s%20height%7D%260%5C%5C%20%5Cqquad%20%5Ctextit%7Bat%20%22t%22%20seconds%7D%5C%5C%20x%3D%5Ctextit%7Bseconds%7D%261%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%200%3D-16%281%29%5E2%2B0x%2Bh_o%5Cimplies%200%3D-16%2Bh_o%5Cimplies%2016%3Dh_o%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20~%5Chfill%20h%28x%29%20%3D%20-16x%5E2%2B16~%5Chfill)
quick info:
in case you're wondering what's that pesky -16x² doing there, is gravity's pull in ft/s².
16/155 = n
explanation
A linear equation is an equation of a straight line. To solve linear equations, it is important to keep the following key concepts in mind:
Maintain balance of the equation by applying the same operations to both sides of an equation
Isolate for the variable by collecting like terms
Use inverse operations to rearrange an equation
to show your work
-8/ 5 = -31/2 • n
- 8/5 = -31n/2
2(-8/5) = 2 • -31/n
-16/5 = -31n
-16/5 / -31 = -31n/31
16/155 = n