19. A number $$ n $$ is the algebraic sum of two terms, one of which varies directly as $$ u $$ and the other inversely as $$ u^{2} $$. If $$ n=22 $$ when $$ u=2 $$ and $$ n=56.5 $$ when $$ u=8 $$, calculate the on value of $$ n $$ when $$ u=10 $$.

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Given that the number $$ n $$ is the algebraic sum of two terms, one of which varies directly as $$ u $$ and the other inversely as $$ u^{2} $$, we can express this relationship as:

$$ n = ku + \frac{c}{u^2} $$

where $$ k $$ and $$ c $$ are constants.

We are given two sets of values for $$ n $$ and $$ u $$:
1. When $$ u = 2 $$, $$ n = 22 $$
2. When $$ u = 8 $$, $$ n = 56.5 $$

We can use these values to form a system of equations to solve for the constants $$ k $$ and $$ c $$.

Let's start by substituting the given values into the equation:

1. When $$ u = 2 $$, $$ n = 22 $$:
$$ 22 = k(2) + \frac{c}{2^2} $$
$$ 22 = 2k + \frac{c}{4} $$

2. When $$ u = 8 $$, $$ n = 56.5 $$:
$$ 56.5 = k(8) + \frac{c}{8^2} $$
$$ 56.5 = 8k + \frac{c}{64} $$

Now, we have a system of two equations with two unknowns. We can solve this system to find the values of $$ k $$ and $$ c $$.

Let's solve the system of equations to find the values of $$ k $$ and $$ c $$.
Solve the system of equations by following steps:
- step0: Solve using the substitution method:
  $$\left\{ \begin{array}{l}22=2k+\frac{c}{4}\56.5=8k+\frac{c}{64}\end{array}\right.$$
- step1: Solve the equation:
  $$\left\{ \begin{array}{l}k=\frac{88-c}{8}\56.5=8k+\frac{c}{64}\end{array}\right.$$
- step2: Substitute the value of $$k:$$
  $$56.5=8\times \frac{88-c}{8}+\frac{c}{64}$$
- step3: Multiply the terms:
  $$56.5=88-c+\frac{c}{64}$$
- step4: Swap the sides:
  $$88-c+\frac{c}{64}=56.5$$
- step5: Convert the expressions:
  $$88-c+\frac{c}{64}=\frac{113}{2}$$
- step6: Multiply both sides of the equation by LCD:
  $$\left(88-c+\frac{c}{64}\right)\times 64=\frac{113}{2}\times 64$$
- step7: Simplify the equation:
  $$5632-63c=3616$$
- step8: Move the constant to the right side:
  $$-63c=3616-5632$$
- step9: Subtract the numbers:
  $$-63c=-2016$$
- step10: Change the signs:
  $$63c=2016$$
- step11: Divide both sides:
  $$\frac{63c}{63}=\frac{2016}{63}$$
- step12: Divide the numbers:
  $$c=32$$
- step13: Substitute the value of $$c:$$
  $$k=\frac{88-32}{8}$$
- step14: Calculate:
  $$k=7$$
- step15: Calculate:
  $$\left\{ \begin{array}{l}c=32\k=7\end{array}\right.$$
- step16: Check the solution:
  $$\left\{ \begin{array}{l}c=32\k=7\end{array}\right.$$
- step17: Rewrite:
  $$\left(c,k\right) = \left(32,7\right)$$
The solution to the system of equations is $$ c = 32 $$ and $$ k = 7 $$.

Now that we have found the values of $$ k $$ and $$ c $$, we can substitute them back into the original equation to find the value of $$ n $$ when $$ u = 10 $$:

$$ n = 7u + \frac{32}{u^2} $$

Substitute $$ u = 10 $$ into the equation:

$$ n = 7(10) + \frac{32}{10^2} $$
$$ n = 70 + \frac{32}{100} $$
$$ n = 70 + 0.32 $$
$$ n = 70.32 $$

Therefore, the value of $$ n $$ when $$ u = 10 $$ is $$ 70.32 $$.
When $$ u = 10 $$, $$ n = 70.32 $$.

19. A number \( n \) is the algebraic sum of two terms, one of which varies directly as \( u \) and the other inversely as \( u^{2} \). If \( n=22 \) when \( u=2 \) and \( n=56.5 \) when \( u=8 \), calculate the on value of \( n \) when \( u=10 \).

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