Answer:
D.) If soccer balls are sold for $2.05 or $14.61 each, the store will break even but will not make a profit.
Step-by-step explanation:
Let us assume x = selling price of each soccer ball
y = daily profit earned from selling of soccer balls
Given that
Y= [tex]-6x^2+100x-180[/tex]
where,
a = -6
b = 100
c = -180
Now we have to applied the formula which is as follows
x [tex]= \frac{-b\pm\sqrt{b^2-4ac}}{2a}$[/tex]
[tex]= \frac{-100\pm\sqrt{100^2-4\times -6\times -180}}{2\times -6}$[/tex]
[tex]= \frac{-100\pm\sqrt{10,000 - 4,320}}{-12}$[/tex]
[tex]= \frac{-100\pm\sqrt{5.680}}{-12}$[/tex]
[tex]= \frac{-100 + 75.3658}{-12}$[/tex]
[tex]= \frac{24.6342}{-12}[/tex]
x^1 = -2.05285
Now
x^2 [tex]= \frac{-100 - 75.3658}{-12}$[/tex]
[tex]= \frac{- 175.3658}{-12}[/tex]
x^2 = 14.6138
Based on this the option D is most appropriate as per the given situation
Answer:
If soccer balls are sold for $2.05 or $14.61 each, the store will break even but will not make a profit.
