Answer:
8 futballs
basketballs 18
beisballs 36
softballs 24
Step-by-step explanation:
dos mas del doble (2) de balones de fut = 8x2 = 16, 16+2 = 18
cuatro menos de 5 veces las pelotas de fut = 8x5 = 40, 40-4 = 36
seis mas de la mitad de beis = 36/2 = 18, 18+6 = 24
LISTO 
 
        
             
        
        
        
 <span>Don't forget S is measured in thousands of units so you are solving for : 
100 < 74.5 + 43.75Sin(πt/6) 
25.5 < 43.75Sin(πt/6) 
Sin(πt/6) >25.5/43.75 = 0.582857 
ASrcSin(πt/6) > 0.62224 radians 
πt/6 > 0.62224 
t > 6 x 0.62224/π = 1.1884 (4dp) 
This initial value occurs when the sine value is increasing and it will reach its maximum value of 1 when Sin(πt/6) = Sinπ/2, that is when t = 3. 
Consequently, monthly sales exceed 100,000 during the period between t = 1.1884 and 4.8116 
[3 - 1.1884 = 1.8116 so the other extreme occurs at 3 + 1.8116] 
Note : on the basis of these calculations, January is 0 ≤ t < 1 : February is 1 ≤ t < 2 :....May is 4 ≤ t < 5 
So the period when sales exceed 100,000 occurs between Feb 6 and May 25 and annually thereafter.</span>
        
             
        
        
        
1,770,000 
find the first part of the equation (2200000) and then the second part (430000) and subtract
        
             
        
        
        
Answer:
B. price
Step-by-step explanation:
The equation is linear and looks like this:
C(x) = 5.99x
where C(x) is the cost of x number of movies.  The cost is the dependent variable, since it is dependent upon how many movies you rent at 5.99 each.
 
        
                    
             
        
        
        
let's notice something, we have a circle with a radius of 12 and one 90° sector is cut off, so only three 90° sectors of the circle are left shaded, so namely the cone will be using 3/4 of that circle.
think of it as, this shaded area is some piece of paper, and you need to pull it upwards and have the cutoff edges meet, and when that happens, you'll end up with a cone-shaped paper cup, and pour in some punch.
now, once we have pulled up the center of the circle to make our paper cup, there will be a circular base, its diameter not going to be 24, it'll be less, but whatever that base is, we know that is going to have the same circumference as those in the shaded area.  Well, what is the circumference of that shaded area?
![\bf \textit{circumference of a circle}\\\\ C=2\pi r~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=12 \end{cases}\implies C=2\pi 12\implies C=24\pi \implies \stackrel{\textit{three quarters of it}}{24\pi \cdot \cfrac{3}{4}} \\\\\\ 6\pi \cdot 3\implies 18\pi](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bcircumference%20of%20a%20circle%7D%5C%5C%5C%5C%20C%3D2%5Cpi%20r~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20r%3D12%20%5Cend%7Bcases%7D%5Cimplies%20C%3D2%5Cpi%2012%5Cimplies%20C%3D24%5Cpi%20%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bthree%20quarters%20of%20it%7D%7D%7B24%5Cpi%20%5Ccdot%20%5Ccfrac%7B3%7D%7B4%7D%7D%20%5C%5C%5C%5C%5C%5C%206%5Cpi%20%5Ccdot%203%5Cimplies%2018%5Cpi)
well then, the circumference of that circle at the bottom will be 18π, so, what is the diameter of a circle with a circumferenc of 18π?
![\bf \textit{circumference of a circle}\\\\ C=2\pi r~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ C=18\pi \end{cases}\implies 18\pi =2\pi r\implies \cfrac{18\pi }{2\pi }=r\implies 9=r \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \stackrel{\textit{diameter is twice the radius}}{d=18}~\hfill](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bcircumference%20of%20a%20circle%7D%5C%5C%5C%5C%20C%3D2%5Cpi%20r~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20C%3D18%5Cpi%20%5Cend%7Bcases%7D%5Cimplies%2018%5Cpi%20%3D2%5Cpi%20r%5Cimplies%20%5Ccfrac%7B18%5Cpi%20%7D%7B2%5Cpi%20%7D%3Dr%5Cimplies%209%3Dr%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20~%5Chfill%20%5Cstackrel%7B%5Ctextit%7Bdiameter%20is%20twice%20the%20radius%7D%7D%7Bd%3D18%7D~%5Chfill)