Answer:
x = 2root(22)
see image.
Step-by-step explanation:
The triangle shown is one big triangle cut into two more smaller triangles: one medium-sized and one smaller.
ALL THREE TRIANGLES ARE SIMILAR BY AA.
Set the two smaller triangles up so you can see the corresponding sides. x is the short leg in one triangle and it is the long leg in the smallest triangle. Set up a proportion.
22/x = x/4
crossmultiply
x^2 = 22•4
x^2 = 88
square root both sides.
x = sqroot(88)
x = 2sqroot(22)
see image.
225 miles / 45 miles/hr = 5hr
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)
Given
Present investment, P = 22000
APR, r = 0.0525
compounding time = 10 years
Future amount, A
A. compounded annually
n=10*1=10
i=r=0.0525
A=P(1+i)^n
=22000(1+0.0525)^10
=36698.11
B. compounded quarterly
n=10*4=40
i=r/4=0.0525/4
A=P(1+i)^n
=22000*(1+0.0525/4)^40
=37063.29
Therefore, by compounding quarterly, she will get, at the end of 10 years investment, an additional amount of
37063.29-36698.11
=$365.18
