Answer:
she got back 63 times
Step-by-step explanation:
Here, we want to know the number of times she is at bat
from the question, we are made to know that she strikes out twice after 9 times
So striking out 14 times, the number of times she has got back will be ;
14/2 * 9 = 63 times
Answer:
![L(f(t)) = \dfrac{6}{S^2+1} [\sqrt{3} \ S +1 ]](https://tex.z-dn.net/?f=L%28f%28t%29%29%20%3D%20%5Cdfrac%7B6%7D%7BS%5E2%2B1%7D%20%5B%5Csqrt%7B3%7D%20%5C%20S%20%2B1%20%5D)
Step-by-step explanation:
Given that:

recall that:
cos (A-B) = cos AcosB + sin A sin B
∴
![f(t) = 12 [cos\ t \ cos \dfrac{\pi}{6}+ sin \ t \ sin \dfrac{\pi}{6}]](https://tex.z-dn.net/?f=f%28t%29%20%3D%2012%20%5Bcos%5C%20%20t%20%5C%20%20cos%20%5Cdfrac%7B%5Cpi%7D%7B6%7D%2B%20sin%20%5C%20t%20%20%5C%20sin%20%5Cdfrac%7B%5Cpi%7D%7B6%7D%5D)
![f(t) = 12 [cos \ t \ \dfrac{3}{2}+ sin \ t \ sin \dfrac{1}{2}]](https://tex.z-dn.net/?f=f%28t%29%20%3D%2012%20%5Bcos%20%5C%20%20t%20%5C%20%5Cdfrac%7B3%7D%7B2%7D%2B%20sin%20%20%5C%20t%20%20%5C%20sin%20%5Cdfrac%7B1%7D%7B2%7D%5D)

![L(f(t)) = L ( 6 \sqrt{3} \ cos \ (t) + 6 \ sin \ (t) ]](https://tex.z-dn.net/?f=L%28f%28t%29%29%20%3D%20L%20%28%206%20%5Csqrt%7B3%7D%20%5C%20cos%20%5C%20%28t%29%20%2B%206%20%5C%20sin%20%5C%20%28t%29%20%5D)
![L(f(t)) = 6 \sqrt{3} \ L [cos \ (t) ] + 6\ L [ sin \ (t) ]](https://tex.z-dn.net/?f=L%28f%28t%29%29%20%3D%206%20%5Csqrt%7B3%7D%20%5C%20L%20%5Bcos%20%5C%20%28t%29%20%5D%20%2B%206%5C%20L%20%5B%20sin%20%5C%20%28t%29%20%5D)



![L(f(t)) = \dfrac{6}{S^2+1} [\sqrt{3} \ S +1 ]](https://tex.z-dn.net/?f=L%28f%28t%29%29%20%3D%20%5Cdfrac%7B6%7D%7BS%5E2%2B1%7D%20%5B%5Csqrt%7B3%7D%20%5C%20S%20%2B1%20%5D)
The answer is C that is X=27
view the picture for step by step solution
Answer:
½ or 11/10
Step-by-step explanation:
3/10 = 8/10 - X
X = 8/10 - 3/10 = 5/10 = 1/2
Or,
X - 8/10 = 3/10
X = 11/10