For this case we can propose a rule of three according to the Shontaro base performance.
4 hits ------------> 12 times at bat
x -------------------> 36 times at bat
Where "x" represents the number of hits he will have the following week, based on his previous performance.
So, Shintaro will have 12 hits in 36 times at bat
Answer:
Shintaro will have 12 hits in 36 times at bat
OPtion B
Explanation is in the file
tinyurl.com/wpazsebu
well, if the diameter is 5, thus its radius must be half that, or 2.5, and therefore, the radius of the one four times as much will be (4)(2.5).
Let's simply get their difference, since that'd be how much more is needed from the smaller to larger sphere.
![~\hfill \stackrel{\textit{surface area of a sphere}}{SA=4\pi r^2}\qquad \qquad r=radius~\hfill \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{\large difference of their areas}}{\stackrel{\textit{radius of (4)(2.5)}}{4\pi (4)(2.5)^2}~~ - ~~\stackrel{\textit{radius of 2.5}}{4\pi (2.5)^2}}\implies 100\pi -25\pi \implies 75\pi ~~ \approx ~~235.62~ft^2](https://tex.z-dn.net/?f=~%5Chfill%20%5Cstackrel%7B%5Ctextit%7Bsurface%20area%20of%20a%20sphere%7D%7D%7BSA%3D4%5Cpi%20r%5E2%7D%5Cqquad%20%5Cqquad%20r%3Dradius~%5Chfill%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7B%5Clarge%20difference%20of%20their%20areas%7D%7D%7B%5Cstackrel%7B%5Ctextit%7Bradius%20of%20%284%29%282.5%29%7D%7D%7B4%5Cpi%20%284%29%282.5%29%5E2%7D~~%20-%20~~%5Cstackrel%7B%5Ctextit%7Bradius%20of%202.5%7D%7D%7B4%5Cpi%20%282.5%29%5E2%7D%7D%5Cimplies%20100%5Cpi%20-25%5Cpi%20%5Cimplies%2075%5Cpi%20~~%20%5Capprox%20~~235.62~ft%5E2)
We know that x > 2 ( or : x ≠ +/- 2 )
We have to factorize the numerator and the denominator:
x² - x - 6 = x² - 3 x + 2 x - 6 = x ( x - 3 ) + 2 ( x - 3 ) = ( x - 3 ) ( x + 2 )
x ² - 4 = ( x - 2 ) ( x + 2 )
