IDNLearn.com makes it easy to find the right answers to your questions. Find the solutions you need quickly and accurately with help from our knowledgeable community.
Sagot :
To determine which tables represent viable solutions for the equation [tex]\( y = 5x \)[/tex], where [tex]\( x \)[/tex] is the number of tickets sold for the school play and [tex]\( y \)[/tex] is the amount of money collected, we will check each table to see if it satisfies the equation.
We do this by examining each pair [tex]\((x, y)\)[/tex] in the table to verify if [tex]\( y = 5x \)[/tex].
### Table 1:
[tex]\[ \begin{array}{|c|c|} \hline \text{Tickets } (x) & \text{Money Collected } (y) \\ \hline-100 & -500 \\ \hline-25 & -125 \\ \hline 40 & 250 \\ \hline 600 & 3,000 \\ \hline \end{array} \][/tex]
- For [tex]\( x = -100 \)[/tex], [tex]\( y \)[/tex] should be [tex]\( 5 \times -100 = -500 \)[/tex]. (True)
- For [tex]\( x = -25 \)[/tex], [tex]\( y \)[/tex] should be [tex]\( 5 \times -25 = -125 \)[/tex]. (True)
- For [tex]\( x = 40 \)[/tex], [tex]\( y \)[/tex] should be [tex]\( 5 \times 40 = 250 \)[/tex]. (True)
- For [tex]\( x = 600 \)[/tex], [tex]\( y \)[/tex] should be [tex]\( 5 \times 600 = 3000 \)[/tex]. (True)
All values in Table 1 satisfy the equation [tex]\( y = 5x \)[/tex].
### Table 2:
[tex]\[ \begin{array}{|c|c|} \hline \text{Tickets } (x) & \text{Money Collected } (y) \\ \hline-20 & -100 \\ \hline 20 & 100 \\ \hline 100 & 500 \\ \hline 109 & 545 \\ \hline \end{array} \][/tex]
- For [tex]\( x = -20 \)[/tex], [tex]\( y \)[/tex] should be [tex]\( 5 \times -20 = -100 \)[/tex]. (True)
- For [tex]\( x = 20 \)[/tex], [tex]\( y \)[/tex] should be [tex]\( 5 \times 20 = 100 \)[/tex]. (True)
- For [tex]\( x = 100 \)[/tex], [tex]\( y \)[/tex] should be [tex]\( 5 \times 100 = 500 \)[/tex]. (True)
- For [tex]\( x = 109 \)[/tex], [tex]\( y \)[/tex] should be [tex]\( 5 \times 109 = 545 \)[/tex]. (True)
All values in Table 2 satisfy the equation [tex]\( y = 5x \)[/tex].
### Table 3:
[tex]\[ \begin{array}{|c|c|} \hline \text{Tickets } ( x ) & \text{Money Collected } ( y ) \\ \hline 0 & 0 \\ \hline 10 & 50 \\ \hline 51 & 255 \\ \hline 400 & 2000 \\ \hline \end{array} \][/tex]
- For [tex]\( x = 0 \)[/tex], [tex]\( y \)[/tex] should be [tex]\( 5 \times 0 = 0 \)[/tex]. (True)
- For [tex]\( x = 10 \)[/tex], [tex]\( y \)[/tex] should be [tex]\( 5 \times 10 = 50 \)[/tex]. (True)
- For [tex]\( x = 51 \)[/tex], [tex]\( y \)[/tex] should be [tex]\( 5 \times 51 = 255 \)[/tex]. (True)
- For [tex]\( x = 400 \)[/tex], [tex]\( y \)[/tex] should be [tex]\( 5 \times 400 = 2000 \)[/tex]. (True)
All values in Table 3 satisfy the equation [tex]\( y = 5x \)[/tex].
### Table 4:
[tex]\[ \begin{array}{|c|c|} \hline \text{Tickets } (x) & \text{Money Collected } (y) \\ \hline 5 & 25 \\ \hline 65 & 350 \\ \hline \end{array} \][/tex]
- For [tex]\( x = 5 \)[/tex], [tex]\( y \)[/tex] should be [tex]\( 5 \times 5 = 25 \)[/tex]. (True)
- For [tex]\( x = 65 \)[/tex], [tex]\( y \)[/tex] should be [tex]\( 5 \times 65 = 325 \)[/tex]. (False)
In Table 4, only the first pair satisfies the equation [tex]\( y = 5x \)[/tex]. The second pair does not.
### Conclusion:
The tables that represent viable solutions for [tex]\( y = 5x \)[/tex] are:
- [1]
- [2]
- [3]
Therefore, Table 1, Table 2, and Table 3 are the viable solutions for the given equation.
We do this by examining each pair [tex]\((x, y)\)[/tex] in the table to verify if [tex]\( y = 5x \)[/tex].
### Table 1:
[tex]\[ \begin{array}{|c|c|} \hline \text{Tickets } (x) & \text{Money Collected } (y) \\ \hline-100 & -500 \\ \hline-25 & -125 \\ \hline 40 & 250 \\ \hline 600 & 3,000 \\ \hline \end{array} \][/tex]
- For [tex]\( x = -100 \)[/tex], [tex]\( y \)[/tex] should be [tex]\( 5 \times -100 = -500 \)[/tex]. (True)
- For [tex]\( x = -25 \)[/tex], [tex]\( y \)[/tex] should be [tex]\( 5 \times -25 = -125 \)[/tex]. (True)
- For [tex]\( x = 40 \)[/tex], [tex]\( y \)[/tex] should be [tex]\( 5 \times 40 = 250 \)[/tex]. (True)
- For [tex]\( x = 600 \)[/tex], [tex]\( y \)[/tex] should be [tex]\( 5 \times 600 = 3000 \)[/tex]. (True)
All values in Table 1 satisfy the equation [tex]\( y = 5x \)[/tex].
### Table 2:
[tex]\[ \begin{array}{|c|c|} \hline \text{Tickets } (x) & \text{Money Collected } (y) \\ \hline-20 & -100 \\ \hline 20 & 100 \\ \hline 100 & 500 \\ \hline 109 & 545 \\ \hline \end{array} \][/tex]
- For [tex]\( x = -20 \)[/tex], [tex]\( y \)[/tex] should be [tex]\( 5 \times -20 = -100 \)[/tex]. (True)
- For [tex]\( x = 20 \)[/tex], [tex]\( y \)[/tex] should be [tex]\( 5 \times 20 = 100 \)[/tex]. (True)
- For [tex]\( x = 100 \)[/tex], [tex]\( y \)[/tex] should be [tex]\( 5 \times 100 = 500 \)[/tex]. (True)
- For [tex]\( x = 109 \)[/tex], [tex]\( y \)[/tex] should be [tex]\( 5 \times 109 = 545 \)[/tex]. (True)
All values in Table 2 satisfy the equation [tex]\( y = 5x \)[/tex].
### Table 3:
[tex]\[ \begin{array}{|c|c|} \hline \text{Tickets } ( x ) & \text{Money Collected } ( y ) \\ \hline 0 & 0 \\ \hline 10 & 50 \\ \hline 51 & 255 \\ \hline 400 & 2000 \\ \hline \end{array} \][/tex]
- For [tex]\( x = 0 \)[/tex], [tex]\( y \)[/tex] should be [tex]\( 5 \times 0 = 0 \)[/tex]. (True)
- For [tex]\( x = 10 \)[/tex], [tex]\( y \)[/tex] should be [tex]\( 5 \times 10 = 50 \)[/tex]. (True)
- For [tex]\( x = 51 \)[/tex], [tex]\( y \)[/tex] should be [tex]\( 5 \times 51 = 255 \)[/tex]. (True)
- For [tex]\( x = 400 \)[/tex], [tex]\( y \)[/tex] should be [tex]\( 5 \times 400 = 2000 \)[/tex]. (True)
All values in Table 3 satisfy the equation [tex]\( y = 5x \)[/tex].
### Table 4:
[tex]\[ \begin{array}{|c|c|} \hline \text{Tickets } (x) & \text{Money Collected } (y) \\ \hline 5 & 25 \\ \hline 65 & 350 \\ \hline \end{array} \][/tex]
- For [tex]\( x = 5 \)[/tex], [tex]\( y \)[/tex] should be [tex]\( 5 \times 5 = 25 \)[/tex]. (True)
- For [tex]\( x = 65 \)[/tex], [tex]\( y \)[/tex] should be [tex]\( 5 \times 65 = 325 \)[/tex]. (False)
In Table 4, only the first pair satisfies the equation [tex]\( y = 5x \)[/tex]. The second pair does not.
### Conclusion:
The tables that represent viable solutions for [tex]\( y = 5x \)[/tex] are:
- [1]
- [2]
- [3]
Therefore, Table 1, Table 2, and Table 3 are the viable solutions for the given equation.
Thank you for being part of this discussion. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Your questions deserve accurate answers. Thank you for visiting IDNLearn.com, and see you again for more solutions.
