Simon opened a savings account last year using the money he received for graduation. He made additional deposits each month, and his savings grew at a constant rate. The table below shows how much money he had in his account after 4, 6, 12, and 18 months.

To determine how much money Simon deposited when he opened his account, we first need to establish the linear relationship represented by the values in the table. We can represent the total amount in his savings account as a linear equation of the form:

\[ S = mt + b \]

Here, \(S\) is the total money saved, \(t\) is the time in months, \(m\) is the constant rate of saving (slope), and \(b\) is the initial deposit (y-intercept).

Using the data points, we can calculate the slope \(m\) between two points. Let’s choose the points for \(t = 4\) and \(t = 6\):

1. For \(t = 4\), \(S = 780\)
2. For \(t = 6\), \(S = 896\)

Calculating the slope (\(m\)) using these two points:

\[
m = \frac{S_2 - S_1}{t_2 - t_1} = \frac{896 - 780}{6 - 4} = \frac{116}{2} = 58
\]

So, the constant saving rate is \(m = 58\) dollars/month.

Now we can use one of the points to find \(b\). Let's use the point for \(t = 4\), where \(S = 780\):

\[
780 = 58(4) + b
\]
\[
780 = 232 + b
\]
\[
b = 780 - 232 = 548
\]

Thus, Simon's initial deposit was **$548**.

Now we can write the complete equation to represent the situation:

\[
S = 58t + 548
\]

In summary:

The amount of money Simon deposited when he opened his account is **$548**.

The equation to represent the situation is:

\(S = 58t + 548\)

A hot air balloon rose at a constant rate. After 3 minutes, the balloon was 44 meters above the ground. Then, 2 minutes later, the hot air balloon had risen to 70 meters above the ground.

How much did the hot air balloon rise each minute?
___

How far above the ground was the hot air balloon at the start of the ride?
___

Complete the equation that describes the relationship between the altitude of the hot air balloon in meters, A, and the elapsed time in minutes, t.
Write your answer using whole numbers or decimals rounded to the nearest tenth.
A= __ t+ ___

To determine how much the hot air balloon rose each minute and its initial altitude, we can set up a linear equation based on the given information.

### Given Information:
1. After 3 minutes, the balloon was at 44 meters: \((t_1 = 3, A_1 = 44)\).
2. After 5 minutes (which is 2 minutes after the 3-minute mark), the balloon was at 70 meters: \((t_2 = 5, A_2 = 70)\).

### Step 1: Calculate the rate of rise
First, we can find the rise in altitude between the two times:

\[
\text{Change in altitude} = A_2 - A_1 = 70 - 44 = 26 \text{ meters}
\]

Next, we calculate the time difference:

\[
\text{Change in time} = t_2 - t_1 = 5 - 3 = 2 \text{ minutes}
\]

Now, we can find the rate of rise (slope, \(m\)):

\[
m = \frac{\text{Change in altitude}}{\text{Change in time}} = \frac{26}{2} = 13 \text{ meters per minute}
\]

### Step 2: Calculate the altitude at the start
To find out how high the balloon was at the start (when \(t = 0\)), we can use the point at \(t = 3\) minutes, \(A = 44\) meters, in our linear equation format of \(A = mt + b\):

The equation is structured as:

\[
A = 13t + b
\]

Plugging in the values we know:

\[
44 = 13(3) + b
\]
\[
44 = 39 + b
\]

Solving for \(b\):

\[
b = 44 - 39 = 5
\]

Thus, the hot air balloon was **5 meters** above the ground at the start of the ride.

### Step 3: Write the complete equation
Now we can express the relationship between the altitude \(A\) and the elapsed time \(t\):

\[
A = 13t + 5
\]

### Summary:
- The hot air balloon rose **13 meters** each minute.
- The hot air balloon was **5 meters** above the ground at the start of the ride.
- The equation describing the relationship is \(A = 13t + 5\).

A line passes through the points (–8,–6) and (7,4). What is an equation of the line in point-slope form?

To find the equation of the line in point-slope form that passes through the points \((-8, -6)\) and \((7, 4)\), we first need to calculate the slope of the line.

### Step 1: Calculate the Slope

The slope (\(m\)) of a line through two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by the formula:

\[
m = \frac{y_2 - y_1}{x_2 - x_1}
\]

Here, we can assign:
- \((x_1, y_1) = (-8, -6)\)
- \((x_2, y_2) = (7, 4)\)

Now substituting these values into the slope formula:

\[
m = \frac{4 - (-6)}{7 - (-8)} = \frac{4 + 6}{7 + 8} = \frac{10}{15} = \frac{2}{3}
\]

### Step 2: Use Point-Slope Form

The point-slope form of a line is expressed as:

\[
y - y_1 = m(x - x_1)
\]

We can use either of the two points to write the equation. Let's use the point \((-8, -6)\).

Substituting the slope \(m = \frac{2}{3}\) and the point \((-8, -6)\):

\[
y - (-6) = \frac{2}{3}(x - (-8))
\]

This simplifies to:

\[
y + 6 = \frac{2}{3}(x + 8)
\]

### Final Equation

Thus, the equation of the line in point-slope form is:

\[
y + 6 = \frac{2}{3}(x + 8)
\]

Which statements about the function f(x)=2x–5 are true? Select all that apply.

-Statements:
The domain is (-5)
The range is (infinity)<y<(infinity)
The domain is all whole numbers
the range is x ≥ -5

To analyze the function \( f(x) = 2x - 5 \), we need to determine its domain and range.

### 1. Finding the Domain
The domain of a function is the set of all possible input values (x-values). For the function \( f(x) = 2x - 5 \), which is a linear function, there are no restrictions on the values that \( x \) can take. Therefore:

- The domain is all real numbers: \((-∞, ∞)\).

### 2. Finding the Range
The range of a function is the set of all possible output values (y-values). The linear function \( f(x) = 2x - 5 \) can take any real number value as \( x \) varies over all real numbers. As \( x \) approaches positive or negative infinity, \( f(x) \) also approaches positive or negative infinity. Therefore:

- The range is all real numbers: \((-∞, ∞)\).

### Evaluation of the Statements
Now, let's evaluate the given statements:

1. **The domain is (-5)**
- **False**: The domain is all real numbers, not limited to just \(-5\).

2. **The range is (infinity)<y<(infinity)**
- **False**: The notation used is incorrect. The correct range is all real numbers, which is better expressed as \((-∞, ∞)\).

3. **The domain is all whole numbers**
- **False**: The domain includes all real numbers, not just whole numbers.

4. **The range is \( x ≥ -5 \)**
- **False**: The range is not constrained in this way; the range is all real numbers.

### Conclusion
None of the provided statements about the function \( f(x) = 2x - 5 \) are true. Hence, the correct domain and range are both all real numbers.

My mistake, I meant (-infinity)<y<(infinity).

No problem! Let’s reevaluate the statements with the corrected information.

Given the function \( f(x) = 2x - 5 \):

### Domain and Range Recap
1. **Domain**: The domain of \( f(x) = 2x - 5 \) is all real numbers: \((-∞, ∞)\).

2. **Range**: The range of \( f(x) = 2x - 5 \) is also all real numbers: \((-∞, ∞)\).

### Evaluation of the Statements
Now, let's evaluate the statements based on your correction:

1. **The domain is (-5)**
- **False**: The domain is all real numbers, not limited to just \(-5\).

2. **The range is \((-∞, ∞)\)**
- **True**: The range is indeed all real numbers, which can be expressed as \((-∞, ∞)\).

3. **The domain is all whole numbers**
- **False**: The domain includes all real numbers, not just whole numbers.

4. **The range is \(x ≥ -5\)**
- **False**: The range is not constrained in this way; it is all real numbers, so it is not limited to values greater than or equal to \(-5\).

### Conclusion
Based on this evaluation:

- The only true statement is: **The range is \((-∞, ∞)\)**.

The other statements are false.