(j)* Engineers use two formulae to design electrical circuits: $$ \mathrm{P}=\mathrm{IV} $$ and $$ \mathrm{V}=\mathrm{LR} $$ If $$ \mathrm{P}=100 $$ and $$ \mathrm{R}=4 $$, calculate the positive value of L . (k)* The formulae for the surface area and volume of a cylinder are $$ \mathrm{S}=2 \pi r^{2}+2 \pi r h $$ and $$ \mathrm{V}=\pi r^{2} h $$. Calculate the surface area ( S ) if $$ r=3 $$ and $$ \mathrm{V}=45 \pi $$. (1)* Given the formula $$ v=\sqrt{u^{2}+2 a s} $$. (1) Calculate the value of $$ v $$ if $$ u=8, a=6 $$ and $$ s=3 $$. (2) Calculate the positive value of $$ u $$ if $$ v=6, a=0,5 $$ and $$ s=11 $$.

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Simplify the expression by following steps:
- step0: Solution:
  $$2\pi \times 3^{2}+2\pi \times 3h$$
- step1: Multiply the terms:
  $$18\pi +2\pi \times 3h$$
- step2: Multiply the terms:
  $$18\pi +6\pi h$$
Solve the equation $$ 100=IV $$.
Solve the equation by following steps:
- step0: Solve for $$I$$:
  $$100=IV$$
- step1: Rewrite the expression:
  $$100=VI$$
- step2: Swap the sides:
  $$VI=100$$
- step3: Divide both sides:
  $$\frac{VI}{V}=\frac{100}{V}$$
- step4: Divide the numbers:
  $$I=\frac{100}{V}$$
Solve the equation $$ 6=\sqrt(u^2 + 2*0.5*11) $$.
Solve the equation by following steps:
- step0: Solve for $$u$$:
  $$6=\sqrt{u^{2}+2\times 0.5\times 11}$$
- step1: Multiply the terms:
  $$6=\sqrt{u^{2}+11}$$
- step2: Swap the sides:
  $$\sqrt{u^{2}+11}=6$$
- step3: Raise both sides to the $$2$$-th power$$:$$
  $$\left(\sqrt{u^{2}+11}\right)^{2}=6^{2}$$
- step4: Evaluate the power:
  $$u^{2}+11=36$$
- step5: Move the constant to the right side:
  $$u^{2}=36-11$$
- step6: Subtract the numbers:
  $$u^{2}=25$$
- step7: Simplify the expression:
  $$u=\pm \sqrt{25}$$
- step8: Simplify:
  $$u=\pm 5$$
- step9: Separate into possible cases:
  $$\begin{align}&u=5\&u=-5\end{align}$$
- step10: Check the solution:
  $$\begin{align}&u=5\&u=-5\end{align}$$
- step11: Rewrite:
  $$u_{1}=-5,u_{2}=5$$
Solve the equation $$ V=LR $$.
Solve the equation by following steps:
- step0: Solve for $$L$$:
  $$V=LR$$
- step1: Rewrite the expression:
  $$V=RL$$
- step2: Swap the sides:
  $$RL=V$$
- step3: Divide both sides:
  $$\frac{RL}{R}=\frac{V}{R}$$
- step4: Divide the numbers:
  $$L=\frac{V}{R}$$
Solve the equation $$ V=\pi*3^2*h $$.
Solve the equation by following steps:
- step0: Solve for $$h$$:
  $$V=\pi \times 3^{2}h$$
- step1: Multiply the terms:
  $$V=9\pi h$$
- step2: Swap the sides:
  $$9\pi h=V$$
- step3: Divide both sides:
  $$\frac{9\pi h}{9\pi }=\frac{V}{9\pi }$$
- step4: Divide the numbers:
  $$h=\frac{V}{9\pi }$$
Calculate or simplify the expression $$ \sqrt(8^2 + 2*6*3) $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\sqrt{8^{2}+2\times 6\times 3}$$
- step1: Multiply the terms:
  $$\sqrt{8^{2}+36}$$
- step2: Add the numbers:
  $$\sqrt{100}$$
- step3: Write in exponential form:
  $$\sqrt{10^{2}}$$
- step4: Simplify the root:
  $$10$$
Let's solve each part step by step.

### Part (j)
We have two equations:
1. $$ P = IV $$
2. $$ V = LR $$

Given:
- $$ P = 100 $$
- $$ R = 4 $$

**Step 1: Solve for $$ I $$ using the first equation.**
$$
I = \frac{P}{V} = \frac{100}{V}
$$

**Step 2: Substitute $$ I $$ into the second equation to find $$ L $$.**
$$
V = LR \implies L = \frac{V}{R} = \frac{V}{4}
$$

### Part (k)
We need to calculate the surface area $$ S $$ of a cylinder given:
- $$ r = 3 $$
- $$ V = 45\pi $$

**Step 1: Use the volume formula to find $$ h $$.**
$$
V = \pi r^2 h \implies 45\pi = \pi (3^2) h \implies h = \frac{45\pi}{9\pi} = 5
$$

**Step 2: Substitute $$ r $$ and $$ h $$ into the surface area formula.**
$$
S = 2\pi r^2 + 2\pi rh = 2\pi(3^2) + 2\pi(3)(5) = 18\pi + 30\pi = 48\pi
$$

### Part (1)
We have the formula:
$$
v = \sqrt{u^2 + 2as}
$$

**Step 1: Calculate $$ v $$ with $$ u = 8 $$, $$ a = 6 $$, and $$ s = 3 $$.**
$$
v = \sqrt{8^2 + 2 \cdot 6 \cdot 3} = \sqrt{64 + 36} = \sqrt{100} = 10
$$

**Step 2: Calculate $$ u $$ with $$ v = 6 $$, $$ a = 0.5 $$, and $$ s = 11 $$.**
$$
6 = \sqrt{u^2 + 2 \cdot 0.5 \cdot 11} \implies 6^2 = u^2 + 11 \implies 36 = u^2 + 11 \implies u^2 = 25 \implies u = 5 \text{ (positive value)}
$$

### Summary of Results
- For Part (j): $$ I = \frac{100}{V} $$ and $$ L = \frac{V}{4} $$
- For Part (k): $$ S = 48\pi $$
- For Part (1): 
  - $$ v = 10 $$
  - $$ u = 5 $$ (positive value)
- For Part (j): $$ I = \frac{100}{V} $$ and $$ L = \frac{V}{4} $$ - For Part (k): $$ S = 48\pi $$ - For Part (1): - $$ v = 10 $$ - $$ u = 5 $$

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