Answer:
check image below
Step-by-step explanation:
Answer:
![14878.04878miles/hours^2](https://tex.z-dn.net/?f=14878.04878miles%2Fhours%5E2)
Step-by-step explanation:
Let's find a solution by understanding the following:
The acceleration rate is defined as the change of velocity within a time interval, which can be written as:
where:
A=acceleration rate
Vf=final velocity
Vi=initial velocity
T=time required for passing from Vi to Vf.
Using the problem's data we have:
Vf=65miles/hour
Vi=6miles/hour
T=14.8seconds
Using the acceleration rate equation we have:
, but look that velocities use 'hours' unit while 'T' uses 'seconds'.
So we need to transform 14.8seconds into Xhours, as follows:
![X=(14.8seconds)*(1hours/60minutes)*(1minute/60seconds)](https://tex.z-dn.net/?f=X%3D%2814.8seconds%29%2A%281hours%2F60minutes%29%2A%281minute%2F60seconds%29)
![X=0.0041hours](https://tex.z-dn.net/?f=X%3D0.0041hours)
Using X=0.0041hours in the previous equation instead of 14.8seconds we have:
![A=(65miles/hour - 6miles/hour)/0.0041hours](https://tex.z-dn.net/?f=A%3D%2865miles%2Fhour%20-%206miles%2Fhour%29%2F0.0041hours)
![A=(61miles/hour)/0.0041hours](https://tex.z-dn.net/?f=A%3D%2861miles%2Fhour%29%2F0.0041hours)
![A=(61miles)/(hour*0.0041hours)](https://tex.z-dn.net/?f=A%3D%2861miles%29%2F%28hour%2A0.0041hours%29)
![A=61miles/0.0041hours^2](https://tex.z-dn.net/?f=A%3D61miles%2F0.0041hours%5E2)
![A=14878.04878miles/hours^2](https://tex.z-dn.net/?f=A%3D14878.04878miles%2Fhours%5E2)
In conclusion, the acceleration rate is ![14878.04878miles/hours^2](https://tex.z-dn.net/?f=14878.04878miles%2Fhours%5E2)
Answer:
$1272.008264
Step-by-step explanation:
If the mark-up was 21%, then the final price is 121% of the original price. Simply divide $1,539.13 by 1.21 to get the original price of $1272.008264
![\bf \begin{cases} x=1\implies &x-1=0\\ x=1\implies &x-1=0\\ x=-\frac{1}{2}\implies 2x=-1\implies &2x+1=0\\ x=2+i\implies &x-2-i=0\\ x=2-i\implies &x-2+i=0 \end{cases} \\\\\\ (x-1)(x-1)(2x+1)(x-2-i)(x-2+i)=\stackrel{original~polynomial}{0} \\\\\\ (x-1)^2(2x+1)~\stackrel{\textit{difference of squares}}{[(x-2)-(i)][(x-2)+(i)]}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Bcases%7D%0Ax%3D1%5Cimplies%20%26x-1%3D0%5C%5C%0Ax%3D1%5Cimplies%20%26x-1%3D0%5C%5C%0Ax%3D-%5Cfrac%7B1%7D%7B2%7D%5Cimplies%202x%3D-1%5Cimplies%20%262x%2B1%3D0%5C%5C%0Ax%3D2%2Bi%5Cimplies%20%26x-2-i%3D0%5C%5C%0Ax%3D2-i%5Cimplies%20%26x-2%2Bi%3D0%0A%5Cend%7Bcases%7D%0A%5C%5C%5C%5C%5C%5C%0A%28x-1%29%28x-1%29%282x%2B1%29%28x-2-i%29%28x-2%2Bi%29%3D%5Cstackrel%7Boriginal~polynomial%7D%7B0%7D%0A%5C%5C%5C%5C%5C%5C%0A%28x-1%29%5E2%282x%2B1%29~%5Cstackrel%7B%5Ctextit%7Bdifference%20of%20squares%7D%7D%7B%5B%28x-2%29-%28i%29%5D%5B%28x-2%29%2B%28i%29%5D%7D)
![\bf (x^2-2x+1)(2x+1)~[(x-2)^2-(i)^2] \\\\\\ (x^2-2x+1)(2x+1)~[(x^2-4x+4)-(-1)] \\\\\\ (x^2-2x+1)(2x+1)~[(x^2-4x+4)+1] \\\\\\ (x^2-2x+1)(2x+1)~[x^2-4x+5] \\\\\\ (x^2-2x+1)(2x+1)(x^2-4x+5)](https://tex.z-dn.net/?f=%5Cbf%20%28x%5E2-2x%2B1%29%282x%2B1%29~%5B%28x-2%29%5E2-%28i%29%5E2%5D%0A%5C%5C%5C%5C%5C%5C%0A%28x%5E2-2x%2B1%29%282x%2B1%29~%5B%28x%5E2-4x%2B4%29-%28-1%29%5D%0A%5C%5C%5C%5C%5C%5C%0A%28x%5E2-2x%2B1%29%282x%2B1%29~%5B%28x%5E2-4x%2B4%29%2B1%5D%0A%5C%5C%5C%5C%5C%5C%0A%28x%5E2-2x%2B1%29%282x%2B1%29~%5Bx%5E2-4x%2B5%5D%0A%5C%5C%5C%5C%5C%5C%0A%28x%5E2-2x%2B1%29%282x%2B1%29%28x%5E2-4x%2B5%29)
of course, you can always use (x-1)(x-1)(2x+1)(x²-4x+5) as well.
Again, just simple addition and subtraction: $1,034.52