Samhita faces a three-headed dragon. She wins a "Tactical medal" if she manages to defeat exactly one of the three heads.
The battle proceeds head-by-head under the following conditions:
The probability of defeating the first head is $\frac{\mathbf{1}}{\mathbf{3}}$.
After a win: if she defeats a head, the probability of defeating the next head is $\frac{2}{3}$.
After a loss: if she fails to defeat a head, the probability of defeating the next head is $\frac{\mathbf{1}}{\mathbf{4}}$.
What is the probability that Samhita earns the "Tactical medal"?
$\frac{23}{72}$
$\frac{5}{36}$
$\frac{17}{72}$
$\frac{19}{72}$
$$ \int e^{2 x} \cos (5 x+3) d x= $$
$$ \frac{e^{2 x}}{29}[2 \sin (5 x+3)+5 \cos (5 x+3)]+C $$
$$ \frac{e^{2 x}}{29}[2 \sin (5 x+3)+5 \cos (5 x+3)]+C $$
$$ \frac{e^{2 x}}{29}[2 \cos (5 x+3)+5 \sin (5 x+3)]+C $$
$$ \frac{e^{2 x}}{29}[5 \cos (5 x+3)-2 \sin (5 x+3)]+C $$
Given $A=\left(\begin{array}{ll}1 & 2 \\ 2 & 3\end{array}\right)$ and $f(x)=x^2-2 x-3$ then $f(A)$ is:
Identity matrix
Skew symmetric matrix
Null matrix
Symmetric Matrix
The value of $\mathop {\lim }\limits_{x \to 3}\left[\frac{1}{x-3}+\frac{9 x}{27-x^3}\right]$ is:
$\frac{1}{2}$
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