Join the IDNLearn.com community and start finding the answers you need today. Ask any question and receive accurate, in-depth responses from our dedicated team of experts.
Sagot :
In this problem, we have the following system of inequalities:
[tex]\begin{gathered} y>x^2-4x-5, \\ y<-x^2-5x+6. \end{gathered}[/tex]We must check which of the given pair of values (x, y) is not a solution of the system. To do that, we replace the values (x, y) in the system, and then we check if the inequality holds.
A) Replacing (x, y) = (1, -5), we have:
[tex]\begin{gathered} -5>1^2-4\cdot1-5=-8\Rightarrow-5>-8\text{ }✓ \\ -5<-1^2-5\cdot1+6=0\Rightarrow-5<0\text{ }✓ \end{gathered}[/tex]B) Replacing (x, y) = (-1, 6), we have:
[tex]\begin{gathered} 6>(-1)^2-4\cdot(-1)-5=0\Rightarrow6>0\text{ }✓ \\ 6<-(-1)^2-5\cdot(-1)+6=10\Rightarrow6<10\text{ }✓ \end{gathered}[/tex]C) Replacing (x, y) = (1, 7), we have:
[tex]\begin{gathered} 7>1^2-4\cdot1-5=-8\Rightarrow7>-8\text{ }✓ \\ 7<-1^2-5\cdot1+6=0\Rightarrow7<0\text{ }✖ \end{gathered}[/tex]We see that the second inequality doesn't hold. So the ordered pair (1, 7) is not a solution of the system.
D) Replacing (x, y) = (1, -7), we have:
[tex]\begin{gathered} -7>1^2-4\cdot1-5=-8\text{ }✓ \\ -7<-1^2-5\cdot1+6=0\text{ }✓ \end{gathered}[/tex]Answer
C) (1, 7)
We value your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. For trustworthy answers, visit IDNLearn.com. Thank you for your visit, and see you next time for more reliable solutions.
