Answer:
28.7092
Step-by-step explanation:
2.5892 + 23.5600 = 26.1492
26.1492 = 28.7092
I assume the first equation is supposed to be

and not

As an augmented matrix, this system is given by
![\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\4&-2&4&12\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D5%26-3%262%2613%5C%5C2%26-1%26-3%261%5C%5C4%26-2%264%2612%5Cend%7Barray%7D%5Cright%5D)
Multiply through row 3 by 1/2:
![\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\2&-1&2&6\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D5%26-3%262%2613%5C%5C2%26-1%26-3%261%5C%5C2%26-1%262%266%5Cend%7Barray%7D%5Cright%5D)
Add -1(row 2) to row 3:
![\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\0&0&5&5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D5%26-3%262%2613%5C%5C2%26-1%26-3%261%5C%5C0%260%265%265%5Cend%7Barray%7D%5Cright%5D)
Multiply through row 3 by 1/5:
![\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\0&0&1&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D5%26-3%262%2613%5C%5C2%26-1%26-3%261%5C%5C0%260%261%261%5Cend%7Barray%7D%5Cright%5D)
Add -2(row 3) to row 1, and add 3(row 3) to row 2:
![\left[\begin{array}{ccc|c}5&-3&0&11\\2&-1&0&4\\0&0&1&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D5%26-3%260%2611%5C%5C2%26-1%260%264%5C%5C0%260%261%261%5Cend%7Barray%7D%5Cright%5D)
Add -3(row 2) to row 1:
![\left[\begin{array}{ccc|c}-1&0&0&-1\\2&-1&0&4\\0&0&1&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D-1%260%260%26-1%5C%5C2%26-1%260%264%5C%5C0%260%261%261%5Cend%7Barray%7D%5Cright%5D)
Multiply through row 1 by -1:
![\left[\begin{array}{ccc|c}1&0&0&1\\2&-1&0&4\\0&0&1&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%260%260%261%5C%5C2%26-1%260%264%5C%5C0%260%261%261%5Cend%7Barray%7D%5Cright%5D)
Add -2(row 1) to row 2:
![\left[\begin{array}{ccc|c}1&0&0&1\\0&-1&0&2\\0&0&1&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%260%260%261%5C%5C0%26-1%260%262%5C%5C0%260%261%261%5Cend%7Barray%7D%5Cright%5D)
Multipy through row 2 by -1:
![\left[\begin{array}{ccc|c}1&0&0&1\\0&1&0&-2\\0&0&1&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%260%260%261%5C%5C0%261%260%26-2%5C%5C0%260%261%261%5Cend%7Barray%7D%5Cright%5D)
The solution to the system is then

Of the first 10 numbers, for example, 40 percent<span> are prime — 2, 3, 5 and 7 — but among 10-digit numbers, only about 4 percent are prime.</span>
Answer: 
Step-by-step explanation:
Given: u and v be are the solutions of
Let
is the quadratic equation and u and v are the zeroes/solutions then
Sum of zeroes; 
Product of zeroes; 
Comparing
to
we get a= 3 , b= 5 and c = 7


Now we have to find
adding and subtracting 2uv in numerator we get

Substituting the values from (i) and (ii) we get

Hence, the value of
is 