so the investigator found the skid marks were 75 feet long hmmm what speed will that be?
![s=\sqrt{30fd}~~ \begin{cases} f=\stackrel{friction}{factor}\\ d=\stackrel{skid}{feet}\\[-0.5em] \hrulefill\\ f=\stackrel{dry~day}{0.7}\\ d=75 \end{cases}\implies s=\sqrt{30(0.7)(75)}\implies s\approx 39.69~\frac{m}{h}](https://tex.z-dn.net/?f=s%3D%5Csqrt%7B30fd%7D~~%20%5Cbegin%7Bcases%7D%20f%3D%5Cstackrel%7Bfriction%7D%7Bfactor%7D%5C%5C%20d%3D%5Cstackrel%7Bskid%7D%7Bfeet%7D%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20f%3D%5Cstackrel%7Bdry~day%7D%7B0.7%7D%5C%5C%20d%3D75%20%5Cend%7Bcases%7D%5Cimplies%20s%3D%5Csqrt%7B30%280.7%29%2875%29%7D%5Cimplies%20s%5Capprox%2039.69~%5Cfrac%7Bm%7D%7Bh%7D)
nope, the analysis shows that Charlie was going faster than 35 m/h.
now, assuming Charlie was indeed going at 35 m/h, then his skid marks would have been
![s=\sqrt{30fd}~~ \begin{cases} f=\stackrel{friction}{factor}\\ d=\stackrel{skid}{feet}\\[-0.5em] \hrulefill\\ f=\stackrel{dry~day}{0.7}\\ s=35 \end{cases}\implies 35=\sqrt{30(0.7)d} \\\\\\ 35^2=30(0.7)d\implies \cfrac{35^2}{30(0.7)}=d\implies 58~ft\approx d](https://tex.z-dn.net/?f=s%3D%5Csqrt%7B30fd%7D~~%20%5Cbegin%7Bcases%7D%20f%3D%5Cstackrel%7Bfriction%7D%7Bfactor%7D%5C%5C%20d%3D%5Cstackrel%7Bskid%7D%7Bfeet%7D%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20f%3D%5Cstackrel%7Bdry~day%7D%7B0.7%7D%5C%5C%20s%3D35%20%5Cend%7Bcases%7D%5Cimplies%2035%3D%5Csqrt%7B30%280.7%29d%7D%20%5C%5C%5C%5C%5C%5C%2035%5E2%3D30%280.7%29d%5Cimplies%20%5Ccfrac%7B35%5E2%7D%7B30%280.7%29%7D%3Dd%5Cimplies%2058~ft%5Capprox%20d)
18 - (8 - 3 • (2t + 5)) = 0
6t + 25 = 0
3.1 Solve : 6t+25 = 0
Subtract 25 from both sides of the equation :
6t = -25
Divide both sides of the equation by 6:
t = -25/6 = -4.167
Final answer is
t = -25/6 = -4.167
Answer: 2x+3y=18
Step-by-step explanation: Any line parallel to 2x+3y+11=0 is the form 2x+3y =k this can also be written as x/(k/2)+y/k/3 =1 this is the line whose sum of the intercepts is 15, then k/2 +k/3 = 15 after solving we get that indeedk= 18.
