Answer:
D. 4
Step-by-step explanation:
![[(p^2) (q^{-3}) ]^{-2}.[(p)^{-3}(q)^5] ^{-2}\\\\=[(p^2) (q^{-3}) \times(p)^{-3}(q)^5 ]^{-2}\\\\=[(p^{2}) \times(p)^{-3} \times (q^{-3}) \times(q)^5 ]^{-2}\\\\=[(p^{2-3}) \times (q^{5-3}) ]^{-2}\\\\=[(p^{-1}) \times (q^{2}) ]^{-2}\\\\=(p^{-1\times (-2)}) \times (q^{2\times (-2) }) \\\\=p^{2}\times q^{-4} \\\\= \frac{p^2}{q^4}\\\\= \frac{(-2)^2}{(-1)^4}\\\\= \frac{4}{1}\\\\= 4](https://tex.z-dn.net/?f=%20%5B%28p%5E2%29%20%28q%5E%7B-3%7D%29%20%5D%5E%7B-2%7D.%5B%28p%29%5E%7B-3%7D%28q%29%5E5%5D%20%5E%7B-2%7D%5C%5C%5C%5C%3C%2Fp%3E%3Cp%3E%3C%2Fp%3E%3Cp%3E%3D%5B%28p%5E2%29%20%28q%5E%7B-3%7D%29%20%5Ctimes%28p%29%5E%7B-3%7D%28q%29%5E5%20%5D%5E%7B-2%7D%5C%5C%5C%5C%3C%2Fp%3E%3Cp%3E%3C%2Fp%3E%3Cp%3E%3D%5B%28p%5E%7B2%7D%29%20%5Ctimes%28p%29%5E%7B-3%7D%20%5Ctimes%20%28q%5E%7B-3%7D%29%20%5Ctimes%28q%29%5E5%20%5D%5E%7B-2%7D%5C%5C%5C%5C%3C%2Fp%3E%3Cp%3E%3C%2Fp%3E%3Cp%3E%3D%5B%28p%5E%7B2-3%7D%29%20%5Ctimes%20%28q%5E%7B5-3%7D%29%20%5D%5E%7B-2%7D%5C%5C%5C%5C%3C%2Fp%3E%3Cp%3E%3C%2Fp%3E%3Cp%3E%3D%5B%28p%5E%7B-1%7D%29%20%5Ctimes%20%28q%5E%7B2%7D%29%20%5D%5E%7B-2%7D%5C%5C%5C%5C%3C%2Fp%3E%3Cp%3E%3C%2Fp%3E%3Cp%3E%3D%28p%5E%7B-1%5Ctimes%20%28-2%29%7D%29%20%5Ctimes%20%28q%5E%7B2%5Ctimes%20%28-2%29%20%7D%29%20%5C%5C%5C%5C%3C%2Fp%3E%3Cp%3E%3C%2Fp%3E%3Cp%3E%3Dp%5E%7B2%7D%5Ctimes%20q%5E%7B-4%7D%20%5C%5C%5C%5C%3C%2Fp%3E%3Cp%3E%3C%2Fp%3E%3Cp%3E%3D%20%5Cfrac%7Bp%5E2%7D%7Bq%5E4%7D%5C%5C%5C%5C%3C%2Fp%3E%3Cp%3E%3C%2Fp%3E%3Cp%3E%3D%20%5Cfrac%7B%28-2%29%5E2%7D%7B%28-1%29%5E4%7D%5C%5C%5C%5C%3C%2Fp%3E%3Cp%3E%3C%2Fp%3E%3Cp%3E%3D%20%5Cfrac%7B4%7D%7B1%7D%5C%5C%5C%5C%3C%2Fp%3E%3Cp%3E%3C%2Fp%3E%3Cp%3E%3D%204)
Answer:
Well, I'm not sure what you mean but Ptolemy's Theorem gives a relationship between the side lengths and the diagonals of a cyclic quadrilateral; it is the equality case of Ptolemy's Inequality. Ptolemy's Theorem frequently shows up as an intermediate step in problems involving inscribed figures.
Answer:
last oneeee
Step-by-step explanation:
Answer: The required solution is

Step-by-step explanation: We are given to solve the following differential equation :

Let us consider that
be an auxiliary solution of equation (i).
Then, we have

Substituting these values in equation (i), we get
![m^2e^{mt}+10me^{mt}+25e^{mt}=0\\\\\Rightarrow (m^2+10y+25)e^{mt}=0\\\\\Rightarrow m^2+10m+25=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow m^2+2\times m\times5+5^2=0\\\\\Rightarrow (m+5)^2=0\\\\\Rightarrow m=-5,-5.](https://tex.z-dn.net/?f=m%5E2e%5E%7Bmt%7D%2B10me%5E%7Bmt%7D%2B25e%5E%7Bmt%7D%3D0%5C%5C%5C%5C%5CRightarrow%20%28m%5E2%2B10y%2B25%29e%5E%7Bmt%7D%3D0%5C%5C%5C%5C%5CRightarrow%20m%5E2%2B10m%2B25%3D0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~%5B%5Ctextup%7Bsince%20%7De%5E%7Bmt%7D%5Cneq0%5D%5C%5C%5C%5C%5CRightarrow%20m%5E2%2B2%5Ctimes%20m%5Ctimes5%2B5%5E2%3D0%5C%5C%5C%5C%5CRightarrow%20%28m%2B5%29%5E2%3D0%5C%5C%5C%5C%5CRightarrow%20m%3D-5%2C-5.)
So, the general solution of the given equation is

Differentiating with respect to t, we get

According to the given conditions, we have

and

Thus, the required solution is
