# Math 105 Exam 2 Practice (old)

## Formulas:

$$\frac{y_2-y_1}{x_2-x_1}$$

$$y-y_1=m(x-x_1)$$

$$D=RT$$

$$I=Prt$$

$$A+B+C=180$$

## 1. Find the $y=mx+b$ form for the equation of the line through $(3,-7)$
with slope $-\frac{1}{4}$.

:::spoiler
<summary> Solution:</summary>

To find the equation of a line in the form $y = mx + b$ given a point $(3, -7)$ and a slope of $-\frac{1}{4}$, follow these steps:

1. **Start with the slope-intercept form of a line**: $y = mx + b$
2. **Insert the given slope**: Since the slope $m$ is $-\frac{1}{4}$, the equation becomes $y = -\frac{1}{4}x + b$.
3. **Use the given point to find $b$**: Substitute $x = 3$ and $y = -7$ into the equation to solve for $b$.
4. **Solve for $b$**: $-7 = -\frac{1}{4}(3) + b$
5. **Simplify and solve the equation for $b$**:
$-7 = -\frac{3}{4} + b$
$b = -7 + \frac{3}{4}$
$b = -\frac{28}{4} + \frac{3}{4}$
$b = -\frac{25}{4}$
6. **Write the final equation** with both $m$ and $b$ in the $y = mx + b$ format:
$y = -\frac{1}{4}x - \frac{25}{4}$

:::

## 2. Consider the line $2x-7y=8$.

(a) What is its slope?

:::spoiler
<summary> Solution:</summary>

To find the slope of the line represented by the equation $2x - 7y = 8$, follow these steps:

1. **Start with the given equation**: $2x - 7y = 8$
2. **Transform the equation into slope-intercept form ($y = mx + b$)**: This involves solving the equation for $y$ to identify the slope $m$ directly from the equation.
3. **Isolate $y$**:
- Move the term involving $x$ to the other side by subtracting $2x$ from both sides, resulting in $-7y = -2x + 8$.
- Divide the entire equation by $-7$ to solve for $y$, yielding $y = \frac{2}{7}x - \frac{8}{7}$.
4. **Identify the slope**:
- In the slope-intercept form, the coefficient of $x$ is the slope. Therefore, the slope $m$ is $\frac{2}{7}$.

**Conclusion**: The slope of the line given by the equation $2x - 7y = 8$ is $\frac{2}{7}$.

:::

(b) What is the equation of the line perpendicular to this line through $(1,-5)$?

:::spoiler
<summary> Solution:</summary>

To determine the equation of a line that is perpendicular to the line represented by $2x - 7y = 8$ and passes through the point $(1, -5)$, perform the following steps:

1. **Find the slope of the original line**: Convert the given equation $2x - 7y = 8$ into slope-intercept form ($y = mx + b$) to identify its slope ($m$). The equation can be rewritten as $y = \frac{2}{7}x - \frac{8}{7}$, indicating a slope of $\frac{2}{7}$.
2. **Determine the slope of the perpendicular line**: The slope of a line perpendicular to another line is the negative reciprocal of the original line's slope. Therefore, if the original slope is $\frac{2}{7}$, the perpendicular slope will be $-\frac{7}{2}$.
3. **Use the point-slope form to find the equation of the perpendicular line**: With a slope of $-\frac{7}{2}$ and passing through $(1, -5)$, use the point-slope form equation, $y - y_1 = m(x - x_1)$, where $(x_1, y_1)$ is the point the line passes through, and $m$ is the slope.
4. **Substitute the known values and solve for $y$**:

$y+5=-\dfrac{7}{2} (x-1)$

$y+5=-\dfrac{7}{2}x+\dfrac{7}{2}$

$y=-\dfrac{7}{2}x +\dfrac{7}{2}-5$

$y=-\dfrac{7}{2}x +\dfrac{7}{2}-\dfrac{10}{2}$

$y=-\dfrac{7}{2}x-\dfrac{3}{2}$

After substitution and simplification, the equation of the perpendicular line is found to be $y=-\dfrac{7}{2}x-\dfrac{3}{2}$.

**Conclusion**: The equation of the line perpendicular to $2x - 7y = 8$ and passing through the point $(1, -5)$ is $y=-\dfrac{7}{2}x-\dfrac{3}{2}$.

:::

## 3. Consider the line $2x-9y=8$.

(a) What is the slope of this line?

:::spoiler
<summary> Solution:</summary>

1. **Given Equation**: The equation of the line is $2x - 9y = 8$.
2. **Slope-Intercept Form**: Convert this equation to the slope-intercept form, $y = mx + b$, where $m$ is the slope.

To do this, isolate $y$:
$2x - 9y = 8$
$-9y = -2x + 8$
$y = \frac{2}{9}x - \frac{8}{9}$

Therefore, the slope $m = \frac{2}{9}$.

:::

(b) What is the equation of the line parallel to this line through $(1,-5)$?

:::spoiler
<summary> Solution:</summary>

1. **Slope of the Parallel Line**: Parallel lines have the same slope. Thus, the slope of the line parallel to $2x - 9y = 8$ is also $m = \frac{2}{9}$.
2. **Point through which the parallel line passes**: $(1, -5)$.
3. **Point-Slope Form**: Use the point-slope form of the equation, $y - y_1 = m(x - x_1)$, where $(x_1, y_1)$ is the point the line passes through.

Substituting the given point and the slope:
$y + 5 = \frac{2}{9}(x - 1)$

:::

## 4. Consider the points $(2,4)$ and $(-8,7)$.

(a) What is the slope through these points?

:::spoiler
<summary> Solution:</summary>

The slope is $\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{7-4}{-8-2}=\dfrac{3}{-10}=-\dfrac{3}{10}$

:::

(b) What is the equation of the line through these points?

:::spoiler
<summary> Solution:</summary>

Since we have $m=-\dfrac{3}{10}$, $x_1=2$ and $y_1=4$, we plug it into the point-slope equation of a line

$y-y_1=m(x-x_1)$

$y-4=-\dfrac{3}{10}(x-2)$

:::

## 5. Find the equation of the line through $(4,-2)$ parallel to the line $x=-4$. (Show work in the graph.)

:::spoiler
<summary> Solution:</summary>

Take note of the equations of horizontal and vertical lines.

Vertical lines: $x=?$

Horizontal Lines: $y=?$

We have a vertical line $x=-4$ and must find an equation of another vertical line ($x=?$) that passes through the point $(4,-2)$.

Graphing this on the grid:


Thus $x=4$.

:::

## 6. Find the equation of the line through $(4,-2)$ perpendicular to the line $x=-4$. (Show work in the graph.)

:::spoiler
<summary> Solution:</summary>

Take note of the equations of horizontal and vertical lines.

Vertical lines: $x=?$

Horizontal Lines: $y=?$

We have a vertical line and must find an equation of a horizontal line ($y=?$) that passes through the point $(4,-2)$.

Graphing this on the grid:


Thus $y=-2$.

:::

## 7. Solve the equation $$2(2x+1)-3(-x+5)=4(3x-1)$$

:::spoiler
<summary> Solution:</summary>

To solve the equation in a detailed and easy-to-understand manner, we will break it down step by step in a table format:

| Step | Operation | Equation |
|------|-----------|----------|
| 1 | **Write the equation** | $$2(2x+1)-3(-x+5)=4(3x-1)$$ |
| 2 | **Distribute the numbers** | $$4x + 2 + 3x - 15 = 12x - 4$$ |
| 3 | **Combine like terms on both sides** | $$7x - 13 = 12x - 4$$ |
| 4 | **Move all terms involving x to one side** | $$7x - 12x = 4 - 13$$ |
| 5 | **Simplify both sides** | $$-5x = -9$$ |
| 6 | **Solve for x** | $$x = \frac{-9}{-5}$$ |
| 7 | **Simplify the fraction** | $$x = \frac{9}{5}$$ |

Thus, by following these steps, we find that the solution to the equation is:

$$x = \frac{9}{5}.$$

:::

## 8. Solve the equation $$3(2x+1)-2(x+5)=4(x-3)$$

:::spoiler
<summary> Solution:</summary>

To solve the equation in a detailed and understandable manner, let's break it down step by step in a table format:

| Step | Operation | Equation |
|------|--------------------------------------------|---------------------------------------|
| 1 | **Expand the equation** | $$3(2x+1)-2(x+5)=4(x-3)$$ |
| 2 | **Distribute the numbers** | $$6x + 3 - 2x - 10 = 4x - 12$$ |
| 3 | **Combine like terms on both sides** | $$4x - 7 = 4x - 12$$ |
| 4 | **Move all terms involving x to one side** | Subtract $$4x$$ from both sides: $$-7 = -12$$ |
| 5 | **Simplify both sides** | The equation simplifies to a contradiction. |
| 6 | **Conclusion** | Since $$-7 \neq -12$$, there is no solution. |

After following these steps, we find that the given equation results in a contradiction, indicating that there is no solution to the equation.

:::

## 9. Solve the inequality: $1-2(3x-1)\leq 2x+5$

:::spoiler
<summary> Solution:</summary>

To solve the inequality in a clear and structured manner, let's detail the steps in a table format:

| Step | Operation | Inequality |
|------|-----------------------------------------------|-------------------------------------|
| 1 | **Expand the inequality** | $$1-2(3x-1)\leq 2x+5$$ |
| 2 | **Distribute the numbers** | $$1 - 6x + 2 \leq 2x + 5$$ |
| 3 | **Combine like terms on both sides** | $$3 - 6x \leq 2x + 5$$ |
| 4 | **Subtract 2x from both sides** | $$3 - 6x -2x\leq 5$$ |
| 5 | **Combine like terms on both sides** | $$3 - 8x\leq 5$$ |
| 6 | **Subtract 3 from both sides** | $$ - 8x\leq 5-3$$ |
| 7 | **Simplify both sides** | $$-8x \leq 2$$ |
| 8 | **Solve for x** (Switch inequality) | $$x \geq -\frac{2}{8}$$ |
| 9 | **Simplify the fraction** | $$x \geq -\frac{1}{4}$$ |

Thus, by following these steps, we find that the solution to the inequality is:

$$x \geq -\frac{1}{4}.$$

:::

## 10. Find the domain of $f(x)=\sqrt{5x+9}$.

:::spoiler
<summary> Solution:</summary>

To find the domain of the function

$$f(x) = \sqrt{5x + 9},$$

we need to determine the set of all possible values of \(x\) that make the expression under the square root non-negative (since the square root of a negative number is not a real number). Let's solve this step by step in a table format:

| Step | Operation | Condition |
|------|-----------|-----------|
| 1 | **Set the expression under the square root to be greater than or equal to 0** | $$5x + 9 \geq 0$$ |
| 2 | **Solve the inequality** | Subtract 9 from both sides: $$5x \geq -9$$ |
| 3 | **Divide by the coefficient of x** | $$x \geq -\frac{9}{5}$$ |

Thus, the domain of $f(x) = \sqrt{5x + 9}$ is all real numbers \(x\) such that

$$x \geq -\frac{9}{5}.$$

In interval notation, the domain is expressed as

$$\left[-\frac{9}{5}, \infty\right).$$

:::

## 11. Find the domain of $h(x)=\frac{2}{\sqrt{3x+11}}$.

:::spoiler
<summary> Solution:</summary>

To find the domain of the function

$$h(x) = \frac{2}{\sqrt{3x + 11}},$$

we need to determine the set of all possible values of $x$ that make the denominator non-zero and the expression under the square root non-negative (since the square root of a negative number is not defined in the real numbers, and division by zero is undefined). Let's solve this step by step in a table format:

| Step | Operation | Condition |
|------|-----------|-----------|
| 1 | **Set the expression under the square root greater than 0** | $$3x + 11 > 0$$ |
| 2 | **Solve the inequality for x** | Subtract 11 from both sides: $$3x > -11$$ |
| 3 | **Divide by the coefficient of x** | $$x > -\frac{11}{3}$$ |

Thus, the domain of $h(x) = \frac{2}{\sqrt{3x + 11}}$ is all real numbers $x$ such that

$$x > -\frac{11}{3}.$$

In interval notation, the domain is expressed as

$$\left(-\frac{11}{3}, \infty\right).$$

:::

## 12. A graph of a function $f(x)$ is shown in Figure 1. Using the graph, state the intervals where $f(x)$ is increasing, decreasing, and constant. Also state the relative maximum and minimum values for $f(x)$.


Increasing:

:::spoiler
<summary> Solution:</summary>

$f(x)$ is increasing when $-2<x<0$ and when $x>1$.
Thus the function is increasing on the interval:
$(-2,0) \cup (1,\infty)$

:::

Decreasing:

:::spoiler
<summary> Solution:</summary>

$f(x)$ is decreasing when $x<-2$ and when $0<x<1$.
Thus the function is increasing on the interval:
$(-\infty,-2) \cup (0,1)$

:::

Relative maximum:

:::spoiler
<summary> Solution:</summary>

$f(x)$ has a relative maximum at $x=0$ and $y=3$

:::

Relative minimum:

:::spoiler
<summary> Solution:</summary>

$f(x)$ has two relative minimums.

The first relative minimum is at $x=-2$ and $y=1$, while the second relative minimum is at $x=1$ and $y=-1$.

:::

## 13. Let

$$ f(x) = \begin{cases}
\frac{1}{3}x-2 & \text{if } x\leq 0 \\
x^2 & \text{if } 0<x<3 \\
2x+3 & \text{if } x\geq 3
\end{cases}
$$

Find (a) $f(5)$

:::spoiler
<summary> Solution:</summary>

To find $f(5)$ for the given piecewise function

$$
f(x) = \begin{cases}
\frac{1}{3}x-2 & x\leq 0 \\
x^2 & 0<x<3 \\
2x+3 & x\geq 3
\end{cases},
$$

we need to determine which condition the input $x = 5$ satisfies, and then use the corresponding expression to calculate $f(5)$. Let's solve this step by step in a table format:

| Step | Condition Check | Applicable Case |
|------|-----------------------|----------------------------|
| 1 | Is $x \leq 0$? | No, since $5 > 0$. |
| 2 | Is $0 < x < 3$? | No, since $5$ is not less than $3$. |
| 3 | Is $x \geq 3$? | Yes, since $5$ is greater than $3$. |

Given that $x = 5$ falls under the condition $x \geq 3$, we use the corresponding expression for $f(x)$:

$$f(5) = 2(5) + 3 = 10 + 3 = 13.$$

Therefore, $f(5) = 13$.

:::

(b) $f(2)$

:::spoiler
<summary> Solution:</summary>

To find $f(2)$ for the given piecewise function

$$
f(x) = \begin{cases}
\frac{1}{3}x-2 & x\leq 0 \\
x^2 & 0<x<3 \\
2x+3 & x\geq 3
\end{cases},
$$

we need to determine which condition the input $x = 2$ satisfies, and then use the corresponding expression to calculate $f(2)$. Let's solve this step by step in a table format:

| Step | Condition Check | Applicable Case |
|------|-----------------------|----------------------------|
| 1 | Is $x \leq 0$? | No, since $2 > 0$. |
| 2 | Is $0 < x < 3$? | Yes, since $2$ is greater than $0$ and less than $3$. |
| 3 | Is $x \geq 3$? | No, since $2$ is less than $3$. |

Given that $x = 2$ falls under the condition $0 < x < 3$, we use the corresponding expression for $f(x)$:

$$f(2) = (2)^2 = 4.$$

Therefore, $f(2) = 4$.

:::

(c) $f(-4)$.

:::spoiler
<summary> Solution:</summary>

To find $f(-4)$ for the given piecewise function

$$
f(x) = \begin{cases}
\frac{1}{3}x-2 & x\leq 0 \\
x^2 & 0<x<3 \\
2x+3 & x\geq 3
\end{cases},
$$

we need to determine which condition the input $x = -4$ satisfies, and then use the corresponding expression to calculate $f(-4)$. Let's solve this step by step in a table format:

| Step | Condition Check | Applicable Case |
|------|-----------------------|----------------------------|
| 1 | Is $x \leq 0$? | Yes, since $-4 < 0$. |
| 2 | Is $0 < x < 3$? | No, since $-4$ is less than $0$. |
| 3 | Is $x \geq 3$? | No, since $-4$ is less than $3$. |

Given that $x = -4$ falls under the condition $x \leq 0$, we use the corresponding expression for $f(x)$:

$$f(-4) = \frac{1}{3}(-4)-2 = -\frac{4}{3}-2 = -\frac{4}{3}-\frac{6}{3} = -\frac{10}{3}.$$

Therefore, $f(-4)=-\dfrac{10}{3}$.

:::

## 15. Let $f(x)=x^2+2$ and $g(x)=3x-2$. Calculate $(f+g)(2)$ and $(fg)(-1)$.

:::spoiler
<summary> Solution:</summary>

### Given Functions:
| Function | Expression |
|----------|--------------|
| $f(x)$ | $x^2 + 2$ |
| $g(x)$ | $3x - 2$ |

### Calculation:

1. **$(f+g)(2)$**:

To find $(f+g)(2)$, we need to first find $f(2)$ and $g(2)$, and then add them.

| Function | Expression | Evaluation | Result |
|----------|--------------|------------|--------|
| $f(x)$ | $x^2 + 2$ | $f(2)$ | $2^2 + 2 = 6$ |
| $g(x)$ | $3x - 2$ | $g(2)$ | $3 \cdot 2 - 2 = 4$ |

Now, $(f+g)(2) = f(2) + g(2) = 6 + 4 = 10$.

2. **$(fg)(-1)$**:

To find $(fg)(-1)$, we need to first find $f(-1)$ and $g(-1)$, and then multiply them.

| Function | Expression | Evaluation | Result |
|----------|--------------|------------|--------|
| $f(x)$ | $x^2 + 2$ | $f(-1)$ | $(-1)^2 + 2 = 3$ |
| $g(x)$ | $3x - 2$ | $g(-1)$ | $3 \cdot (-1) - 2 = -5$ |

Now, $(fg)(-1) = f(-1) \cdot g(-1) = 3 \cdot (-5) = -15$.

### Results:
1. $(f+g)(2) = 10$
2. $(fg)(-1) = -15$

:::

## 16. Let $f(x)=x^2+2$ and $g(x)=3x-2$. Calculate $(f-g)(1)$ and $(f/g)(2)$

:::spoiler
<summary> Solution:</summary>

### Given Functions:
| Function | Expression |
|----------|--------------|
| $f(x)$ | $x^2 + 2$ |
| $g(x)$ | $3x - 2$ |

### Calculation:

1. **$(f-g)(1)$**:

To find $(f-g)(1)$, we need to first find $f(1)$ and $g(1)$, and then subtract $g(1)$ from $f(1)$.

| Function | Expression | Evaluation | Result |
|----------|--------------|------------|--------|
| $f(x)$ | $x^2 + 2$ | $f(1)$ | $1^2 + 2 = 3$ |
| $g(x)$ | $3x - 2$ | $g(1)$ | $3 \cdot 1 - 2 = 1$ |

Now, $(f-g)(1) = f(1) - g(1) = 3 - 1 = 2$.

2. **$(f/g)(2)$**:

To find $(f/g)(2)$, we need to first find $f(2)$ and $g(2)$, and then divide $f(2)$ by $g(2)$.

| Function | Expression | Evaluation | Result |
|----------|--------------|------------|--------|
| $f(x)$ | $x^2 + 2$ | $f(2)$ | $2^2 + 2 = 6$ |
| $g(x)$ | $3x - 2$ | $g(2)$ | $3 \cdot 2 - 2 = 4$ |

Now, $(f/g)(2) = \frac{f(2)}{g(2)} = \frac{6}{4}=\frac{3}{2}$.

### Results:
1. $(f-g)(1) = 2$
2. $(f/g)(2) = \frac{3}{2}$

:::

## 17. Let $f(x)=x-5$ and $g(x)=3x+7$. Determine $(f+g)(x)$ and $(fg)(x)$.

:::spoiler
<summary> Solution:</summary>

### Given Functions:
| Function | Expression |
|----------|--------------|
| $f(x)$ | $x - 5$ |
| $g(x)$ | $3x + 7$ |

### Calculation:

1. **$(f+g)(x)$**:

To find $(f+g)(x)$, we need to add the expressions of $f(x)$ and $g(x)$.

$(f+g)(x) = f(x) + g(x) = (x - 5) + (3x + 7)$
$= x - 5 + 3x + 7$
$= 4x + 2$.

So, $(f+g)(x) = 4x + 2$.

2. **$(fg)(x)$**:

To find $(fg)(x)$, we need to multiply the expressions of $f(x)$ and $g(x)$.

$(fg)(x) = f(x) \cdot g(x) = (x - 5)(3x + 7)$
$= x(3x + 7) - 5(3x + 7)$
$= 3x^2 + 7x - 15x - 35$
$= 3x^2 - 8x - 35$.

So, $(fg)(x) = 3x^2 - 8x - 35$.

### Results:
1. $(f+g)(x) = 4x + 2$
2. $(fg)(x) = 3x^2 - 8x - 35$

:::

## 18. Let $f(x)=x^3$ and $g(x)=2x-5$. Determine $(f-g)(x)$ and $(fg)(x)$.

:::spoiler
<summary> Solution:</summary>

### Given Functions:
| Function | Expression |
|----------|--------------|
| $f(x)$ | $x^3$ |
| $g(x)$ | $2x - 5$ |

### Calculation:

1. **$(f-g)(x)$**:

To find $(f-g)(x)$, we need to subtract $g(x)$ from $f(x)$.

$(f-g)(x) = f(x) - g(x) = x^3 - (2x - 5)$
$= x^3 - 2x + 5$.

So, $(f-g)(x) = x^3 - 2x + 5$.

2. **$(fg)(x)$**:

To find $(fg)(x)$, we need to multiply $f(x)$ and $g(x)$.

$(fg)(x) = f(x) \cdot g(x) = x^3 \cdot (2x - 5)$
$= 2x^4 - 5x^3$.

So, $(fg)(x) = 2x^4 - 5x^3$.

### Results:
1. $(f-g)(x) = x^3 - 2x + 5$
2. $(fg)(x) = 2x^4 - 5x^3$

:::

## 19. Let $f(x)=2x^2$ and $g(x)=x+3$. Find $(f\circ g)(x)$ and $(g\circ f)(x)$.

:::spoiler
<summary> Solution:</summary>

### Given Functions:
| Function | Expression |
|----------|--------------|
| $f(x)$ | $2x^2$ |
| $g(x)$ | $x + 3$ |

### Calculation:

1. **$(f \circ g)(x)$**:

To find $(f \circ g)(x)$, we need to first find $g(x)$ and then substitute it into $f(x)$.

$g(x) = x + 3$.

Now, substitute $g(x)$ into $f(x)$:

$(f \circ g)(x) = f(g(x)) = f(x + 3) = 2(x + 3)^2$.

Simplifying further:

$(f \circ g)(x) = 2(x^2 + 6x + 9)$
$= 2x^2 + 12x + 18$.

So, $(f \circ g)(x) = 2x^2 + 12x + 18$.

2. **$(g \circ f)(x)$**:

To find $(g \circ f)(x)$, we need to first find $f(x)$ and then substitute it into $g(x)$.

$f(x) = 2x^2$.

Now, substitute $f(x)$ into $g(x)$:

$(g \circ f)(x) = g(f(x)) = g(2x^2) = 2x^2 + 3$.

So, $(g \circ f)(x) = 2x^2 + 3$.

### Results:
1. $(f \circ g)(x) = 2x^2 + 12x + 18$
2. $(g \circ f)(x) = 2x^2 + 3$

:::

## 20. (Won't be on Exam 2) You leave on a car trip driving at $40$ mph. Your friend leaves $2$ hours later on the same trip at $55$ mph. How long does it take your friend to catch up to you?

:::spoiler
<summary> Solution:</summary>

To find out when your friend catches up to you, we can set up an equation where the distance traveled by both of you is the same when your friend catches up.

For you:
- Rate: $40$ mph
- Time: $t + 2$ hours
- Distance traveled: $40(t + 2)$ miles

For your friend:
- Rate: $55$ mph
- Time: $t$ hours
- Distance traveled: $55t$ miles

| | Rate | Time | Distance |
|---------------------|-------------|-----------------|----------------|
| You | $40$ mph | $t+2$ hours | $40(t+2)$ miles |
| Your Friend | $55$ mph | $t$ hours | $55t$ miles|

Since both of you travel the same distance when your friend catches up to you, we can set up the equation:

$$ 40(t + 2) = 55t $$

Now, let's solve for $t$:

$$ 40t + 80 = 55t $$

Subtract $40t$ from both sides:

$$ 80 = 15t $$

Divide both sides by $15$:

$$ t = \frac{80}{15} = \frac{16}{3} $$

So, it takes your friend $\frac{16}{3}$ hours to catch up to you.

Now, let's convert $\frac{16}{3}$ hours into hours and minutes:

$$ \frac{16}{3} \text{ hours} = 5 \frac{1}{3} \text{ hours} $$

So, your friend catches up to you in $5 \frac{1}{3}$ hours, or $5$ hours and $20$ minutes.

:::

## 21. (Won't be on Exam 2) In triangle $ABC$, angle $B$ has measure $10$ degrees less than that of $A$, and $C$ has four times the measure of $A$. What is the measure of $A$?

:::spoiler
<summary> Solution:</summary>

To solve this problem, let's denote the measure of angle $A$ as $x$ degrees.

Given:
- Angle $B$ has a measure $10$ degrees less than that of angle $A$, so the measure of angle $B$ is $x - 10$ degrees.
- Angle $C$ has four times the measure of angle $A$, so the measure of angle $C$ is $4x$ degrees.

Since the sum of the angles in a triangle is $180$ degrees, we can set up an equation:

Angle sum equation:

$A+B+C=180^\circ$

$x + (x - 10) + 4x = 180^\circ$

Now, let's solve for $x$:

$6x - 10 = 180^\circ$

Add $10$ to both sides:

$6x = 190^\circ$

Divide both sides by $6$:

$x = \frac{190^\circ}{6} =\dfrac{95}{3}= 31.67^\circ$

So, the measure of angle $A$ is $31.67$ degrees.

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## 22. (Maybe on Exam 2?) We invest $1000 into two accounts, one at 4% and the other at 1%. If the total interest in the two accounts after two years is $56, how much is in the account initially at 4\% (i.e., the principal)?

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<summary> Solution:</summary>

To solve this problem, let's denote the initial amount invested at 4% as $x$ dollars.

Given:
- Amount invested at 4%: $x$ dollars
- Amount invested at 1%: $1000 - x$ dollars
- Interest rate for 4% account: 4% or 0.04
- Interest rate for 1% account: 1% or 0.01
- Time: 2 years

Using the formula for simple interest $\text{Interest} = \text{Principal} \times \text{Rate} \times \text{Time} $, we can calculate the interest earned in each account.

For the 4% account:
- Principal: $x$ dollars
- Rate: 0.04
- Time: 2 years
- Interest: $0.04x$ dollars

For the 1% account:
- Principal: $1000 - x$ dollars
- Rate: 0.01
- Time: 2 years
- Interest: $0.02(1000 - x)$ dollars

| | Principal | Rate | Time | Interest |
|---------------------|-------------|-----------------|----------------|----------------|
| 4% | $x$ dollars | $0.04$ | 2 years | $0.08x$ dollars |
| 1% | $1000-x$ dollars | $0.01$ | 2 years| $0.02(1000-x)$ dollars|

The total interest earned is given as $56$ dollars.

So, we can set up the equation:

$$0.08x + 0.02(1000 - x) = 56$$

Now, let's solve for $x$:

$$0.08x + 20 - 0.02x = 56$$

Combine like terms:

$$0.06x + 20 = 56$$

Subtract $20$ from both sides:

$$0.06x = 36$$

Divide both sides by $0.06$:

$$x = \frac{36}{0.06} = \dfrac{3600}{6}=600$$

So, the initial amount invested at 4% is $600$ dollars.

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