Since he only makes commission on sales over $1000, he makes commission on $1500 (we find this by doing 2500-1000.) His commission is 15%, so find 15% of $1500:
So we know how makes $225 in commission.
Now we need to add this to his fixed pay to get his total pay:
So his total pay is $525, which is answer 'a'
Hope I helped, and let me know if you have any questions :)
Factor the following:
2 x^3 + x^2 - 18 x - 9
Factor terms by grouping. 2 x^3 + x^2 - 18 x - 9 = (2 x^3 + x^2) + (-18 x - 9) = x^2 (2 x + 1) - 9 (2 x + 1):
x^2 (2 x + 1) - 9 (2 x + 1)
Factor 2 x + 1 from x^2 (2 x + 1) - 9 (2 x + 1):
(2 x + 1) (x^2 - 9)
x^2 - 9 = x^2 - 3^2:
(2 x + 1) (x^2 - 3^2)
Factor the difference of two squares. x^2 - 3^2 = (x - 3) (x + 3):
Answer: (x - 3) (x + 3) (2 x + 1) thus the Answer is C.
That is a question about perimeter of rectangle.
The perimeter of a geometric shape is the sum of the value of the all sides.
In a rectangle, the opposite sides are cogruentes (equals). For example, look at picture below. In that rectangle, the sides AB and DC are equals, and the sides AD and BC are equals.
So, the sides of that rectangular table are: 1.52m, 1.52m, 0.75m and 0.75.
Therefore, the perimeter of that rectangle is 
m.
But that is not the final answer because that is the perimeter in meters and the question want the answer in centimeters.
1 meter has 100 centimeter, so we need to multiplicate the perimeter by 100.
Thus, the perimeter of that rectangular table is 
cm.
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)
