Welcome to Westonci.ca, the place where your questions are answered by a community of knowledgeable contributors. Discover solutions to your questions from experienced professionals across multiple fields on our comprehensive Q&A platform. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.
Sagot :
### 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].
Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.