Answer:
Step-by-step explanation:
![\ln \dfrac{4y^5}{x^2}\\\\=\ln(4y^5) - \ln(x^2)~~~~~~~~~~~;\left[ \log_b\left( \dfrac mn \right) = \log_b m - \llog_b n \right]\\\\=\ln 4 + \ln y^5 - 2\ln x~~~~~~~~~~~~;[\log_b m^n = n \log_b m ~\text{and}~\log_b(mn) = \log_b m + \log_b n ]\\\\=\ln 4 + 5 \ln y -2 \ln x\\\\=\ln 4 -2 \ln x +5 \ln y](https://tex.z-dn.net/?f=%5Cln%20%5Cdfrac%7B4y%5E5%7D%7Bx%5E2%7D%5C%5C%5C%5C%3D%5Cln%284y%5E5%29%20-%20%5Cln%28x%5E2%29~~~~~~~~~~~%3B%5Cleft%5B%20%5Clog_b%5Cleft%28%20%5Cdfrac%20mn%20%5Cright%29%20%20%3D%20%5Clog_b%20m%20-%20%5Cllog_b%20n%20%5Cright%5D%5C%5C%5C%5C%3D%5Cln%204%20%2B%20%5Cln%20y%5E5%20-%202%5Cln%20x~~~~~~~~~~~~%3B%5B%5Clog_b%20m%5En%20%3D%20n%20%5Clog_b%20m%20~%5Ctext%7Band%7D~%5Clog_b%28mn%29%20%3D%20%5Clog_b%20m%20%2B%20%5Clog_b%20n%20%5D%5C%5C%5C%5C%3D%5Cln%204%20%2B%205%20%5Cln%20y%20-2%20%5Cln%20x%5C%5C%5C%5C%3D%5Cln%204%20-2%20%5Cln%20x%20%2B5%20%5Cln%20y)
Let
b-----------> the length side of the square box
h------------> the height of the box
SA---------> surface area of the box
we know that
[volume of the box]=b²*h
volume=256 in³
b²*h=256-------> h=256/b²-----> equation 1
surface area of the box=area of the base+perimeter of base*height
area of the base=b²
perimeter of the base=4*b
surface area=b²+(4*b)*h------> SA=b²+4*b*h-----> equation 2
substitute equation 1 in equation 2
SA=b²+4*b*[256/b²]-----> SA=b²+1024/b-----> SA=(b³+1024)/b
the answer is
the formula of the volume of the box is V=b²*h-----> 256=b²*h
the formula of the surface area of the box are
SA=b²+4*b*h
SA=(b³+1024)/b
Average speed = (distance covered) / (time to cover the distance)
= (500 miles) / (6.7 hours)
= (500 / 6.7) (mile/hour)
= 74.63 miles per hour.
I imagine there were quite a number of pit stops included in that.
This is faster than I expected when I first read your question.
That's really gettin' with it for cars in 1911 !
