2.2 Change the subject of $$ S=\frac{1}{2} g t^{2} $$ to $$ t \ $$. Solve for $$ x $$ and illustrate your answer on a number line if $$ x \in \mathrm{~N}_{0} $$ $$ 3 x-\frac{4 x+4}{16}\frac{3(x-1)}{4} $$

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2.2 Change the subject of $$ S=\frac{1}{2} g t^{2} $$ to $$ t \ $$. Solve for $$ x $$ and illustrate your answer on a number line if $$ x \in \mathrm{~N}_{0} $$ $$ 3 x-\frac{4 x+4}{16}>\frac{3(x-1)}{4} $$

UpStudy AI · Accepted Answer

**Solution for the first part**  
We start with  
$$
S = \frac{1}{2} g t^{2}.
$$  
To solve for $$ t $$, multiply both sides by $$ 2 $$ to isolate $$ g t^2 $$:
$$
2S = g t^{2}.
$$  
Now, divide both sides by $$ g $$:
$$
t^{2} = \frac{2S}{g}.
$$  
Taking the square root of both sides (and noting that $$ t $$ is typically nonnegative) gives:
$$
t = \sqrt{\frac{2S}{g}}.
$$

---

**Solution for the second part**  
We need to solve the inequality for $$ x $$:
$$
3x - \frac{4x + 4}{16} > \frac{3(x-1)}{4}.
$$

1. **Clear the fractions:** Multiply every term by $$ 16 $$ (the least common denominator):
   $$
   16 \cdot 3x - (4x + 4) > 16 \cdot \frac{3(x-1)}{4}.
   $$
   
2. **Simplify each term:**  
   The left-hand side becomes:
   $$
   48x - (4x + 4) = 48x - 4x - 4 = 44x - 4.
   $$
   The right-hand side simplifies as:
   $$
   16 \cdot \frac{3(x-1)}{4} = 4 \cdot 3(x-1) = 12(x-1) = 12x - 12.
   $$
   So, the inequality now is:
   $$
   44x - 4 > 12x - 12.
   $$

3. **Solve for $$ x $$:**  
   Subtract $$ 12x $$ from both sides:
   $$
   44x - 12x - 4 > -12,
   $$
   which simplifies to:
   $$
   32x - 4 > -12.
   $$
   Add $$ 4 $$ to each side:
   $$
   32x > -8.
   $$
   Divide by $$ 32 $$:
   $$
   x > -\frac{8}{32}.
   $$
   Simplify the fraction:
   $$
   x > -\frac{1}{4}.
   $$

4. **Interpretation for $$ x \in \mathbb{N}_0 $$:**  
   Since $$ \mathbb{N}_0 $$ denotes the set of nonnegative integers ($$ 0, 1, 2, \ldots $$), and every nonnegative integer is greater than $$-\frac{1}{4}$$, the inequality is true for:
   $$
   x \in \{0, 1, 2, 3, \dots\}.
   $$

5. **Illustration on a number line:**

$$
   \text{Number line:} \quad 0 \quad 1 \quad 2 \quad 3 \quad \dots
   $$
   Marked as:
   
   $$
   \begin{array}{cccccccccc}
   \dots & \quad -1 & \quad 0 & \quad 1 & \quad 2 & \quad 3 & \quad \dots \
   & \quad \text{(unshaded)} & \quad \bullet & \quad \bullet & \quad \bullet & \quad \bullet & \quad \dots \
   \end{array}
   $$
   All points starting from $$ 0 $$ (including $$ 0 $$) are part of the solution.

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**Final Answers:**

1. For $$ S = \frac{1}{2} g t^{2} $$, solving for $$ t $$ gives:
   $$
   t = \sqrt{\frac{2S}{g}}.
   $$

2. The inequality
   $$
   3x - \frac{4x+4}{16} > \frac{3(x-1)}{4}
   $$
   holds for all nonnegative integers:
   $$
   x \in \{0, 1, 2, 3, \dots\}.
   $$
For the equation $$ S = \frac{1}{2} g t^{2} $$, solving for $$ t $$ gives $$ t = \sqrt{\frac{2S}{g}} $$. For the inequality $$ 3x - \frac{4x+4}{16} > \frac{3(x-1)}{4} $$, the solution is all nonnegative integers $$ x \geq 0 $$.

2.2 Change the subject of \( S=\frac{1}{2} g t^{2} \) to \( t \ \). Solve for \( x \) and illustrate your answer on a number line if \( x \in \mathrm{~N}_{0} \) \( 3 x-\frac{4 x+4}{16}>\frac{3(x-1)}{4} \)

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