If a random variable X follows the Binomial distribution $B(10, \quad p)$ such that $5 \mathrm{P}(\mathrm{X}=0)=\mathrm{P}(\mathrm{X}=1)$, then the value of $\frac{\mathrm{P}(\mathrm{X}=5)}{\mathrm{P}(\mathrm{X}=6)}$ is equal to

A particular solution of $\frac{\mathrm{d} y}{\mathrm{~d} x}=(x+9 y)^2$, when $x=0, y=\frac{1}{27}$ is

$$ \mathop {\lim }\limits_{x \to \infty } \frac{\mathrm{e}^{x^4}-1}{\mathrm{e}^{x^4}+1}= $$

The shortest distance between the lines $\bar{r}=(4 \hat{i}-\hat{j})+\lambda(\hat{i}+2 \hat{j}-3 \hat{k})$ and $\bar{r}=(\hat{i}-\hat{j}+2 \hat{k})+\mu(2 \hat{i}+4 \hat{j}-5 \hat{k})$ is