Answer :
### Step-by-Step Solution:
#### 1. Determine the cost of the bicycle:
The equation given to model the amount of money Hugo still needs to pay for the bike is:
[tex]\[ y - 10 = -2(x - 10) \][/tex]
First, we need to rearrange this equation to solve for [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex]:
[tex]\[ y - 10 = -2(x - 10) \][/tex]
Distribute the [tex]\(-2\)[/tex]:
[tex]\[ y - 10 = -2x + 20 \][/tex]
Add 10 to both sides to isolate [tex]\( y \)[/tex]:
[tex]\[ y = -2x + 30 \][/tex]
To find the initial amount Hugo needs to pay (which represents the total cost of the bicycle), we need to determine [tex]\( y \)[/tex] when [tex]\( x = 0 \)[/tex]:
[tex]\[ y = -2(0) + 30 \][/tex]
[tex]\[ y = 30 \][/tex]
So, the bicycle cost:
[tex]\[ \boxed{30} \][/tex]
#### 2. Determine the number of weeks Hugo will finish paying for the bike:
To find the number of weeks it takes for Hugo to finish paying for the bike, we set [tex]\( y \)[/tex] to 0 (the point at which he owes no more money) and solve for [tex]\( x \)[/tex]:
[tex]\[ 0 = -2x + 30 \][/tex]
Rearrange the equation to solve for [tex]\( x \)[/tex]:
[tex]\[ 2x = 30 \][/tex]
[tex]\[ x = 15 \][/tex]
So, Hugo will finish paying for the bike after:
[tex]\[ \boxed{15} \text{ weeks} \][/tex]
#### 3. Populate the table for the graph:
To graph the equation [tex]\( y - 10 = -2(x - 10) \)[/tex], we need points [tex]\( (x, y) \)[/tex] that satisfy the equation. Here are some values:
1. When [tex]\( x = 0 \)[/tex]:
[tex]\[ y = -2(0) + 30 = 30 \][/tex]
2. When [tex]\( x = 5 \)[/tex]:
[tex]\[ y = -2(5) + 30 = 20 \][/tex]
3. When [tex]\( x = 10 \)[/tex]:
[tex]\[ y = -2(10) + 30 = 10 \][/tex]
4. When [tex]\( x = 15 \)[/tex]:
[tex]\[ y = -2(15) + 30 = 0 \][/tex]
So, the table for the graph is:
[tex]\[ \begin{array}{|l|l|} \hline x & y \\ \hline 0 & 30 \\ 5 & 20 \\ 10 & 10 \\ 15 & 0 \\ \hline \end{array} \][/tex]
#### 4. Draw the graph:
Using the table values, you can plot the points [tex]\( (0, 30) \)[/tex], [tex]\( (5, 20) \)[/tex], [tex]\( (10, 10) \)[/tex], and [tex]\( (15, 0) \)[/tex] on the graph. The line that connects these points represents the equation [tex]\( y - 10 = -2(x - 10) \)[/tex].
The graph will be a straight line decreasing from [tex]\( (0, 30) \)[/tex] to [tex]\( (15, 0) \)[/tex].
#### 1. Determine the cost of the bicycle:
The equation given to model the amount of money Hugo still needs to pay for the bike is:
[tex]\[ y - 10 = -2(x - 10) \][/tex]
First, we need to rearrange this equation to solve for [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex]:
[tex]\[ y - 10 = -2(x - 10) \][/tex]
Distribute the [tex]\(-2\)[/tex]:
[tex]\[ y - 10 = -2x + 20 \][/tex]
Add 10 to both sides to isolate [tex]\( y \)[/tex]:
[tex]\[ y = -2x + 30 \][/tex]
To find the initial amount Hugo needs to pay (which represents the total cost of the bicycle), we need to determine [tex]\( y \)[/tex] when [tex]\( x = 0 \)[/tex]:
[tex]\[ y = -2(0) + 30 \][/tex]
[tex]\[ y = 30 \][/tex]
So, the bicycle cost:
[tex]\[ \boxed{30} \][/tex]
#### 2. Determine the number of weeks Hugo will finish paying for the bike:
To find the number of weeks it takes for Hugo to finish paying for the bike, we set [tex]\( y \)[/tex] to 0 (the point at which he owes no more money) and solve for [tex]\( x \)[/tex]:
[tex]\[ 0 = -2x + 30 \][/tex]
Rearrange the equation to solve for [tex]\( x \)[/tex]:
[tex]\[ 2x = 30 \][/tex]
[tex]\[ x = 15 \][/tex]
So, Hugo will finish paying for the bike after:
[tex]\[ \boxed{15} \text{ weeks} \][/tex]
#### 3. Populate the table for the graph:
To graph the equation [tex]\( y - 10 = -2(x - 10) \)[/tex], we need points [tex]\( (x, y) \)[/tex] that satisfy the equation. Here are some values:
1. When [tex]\( x = 0 \)[/tex]:
[tex]\[ y = -2(0) + 30 = 30 \][/tex]
2. When [tex]\( x = 5 \)[/tex]:
[tex]\[ y = -2(5) + 30 = 20 \][/tex]
3. When [tex]\( x = 10 \)[/tex]:
[tex]\[ y = -2(10) + 30 = 10 \][/tex]
4. When [tex]\( x = 15 \)[/tex]:
[tex]\[ y = -2(15) + 30 = 0 \][/tex]
So, the table for the graph is:
[tex]\[ \begin{array}{|l|l|} \hline x & y \\ \hline 0 & 30 \\ 5 & 20 \\ 10 & 10 \\ 15 & 0 \\ \hline \end{array} \][/tex]
#### 4. Draw the graph:
Using the table values, you can plot the points [tex]\( (0, 30) \)[/tex], [tex]\( (5, 20) \)[/tex], [tex]\( (10, 10) \)[/tex], and [tex]\( (15, 0) \)[/tex] on the graph. The line that connects these points represents the equation [tex]\( y - 10 = -2(x - 10) \)[/tex].
The graph will be a straight line decreasing from [tex]\( (0, 30) \)[/tex] to [tex]\( (15, 0) \)[/tex].
