Answer: [tex]By\ 0.6\ seconds[/tex]
Step-by-step explanation:
[tex]Rachel: \\h=-16t^2+36t+160\\Amber: \\h=-16t^2+50t+160[/tex]
Make each equation equal to 0 and then apply the the Quadratic formula [tex]x=\frac{-b\±\sqrt{b^2-4ac} }{2a}[/tex]
Time it takes Rachel's textbook to reach the ground (in seconds)
Having the equation:
[tex]0=-16t^2+36t+160[/tex]
We can identify that:
[tex]a=-16\\b=36\\c=160[/tex]
Substituting values into the Quadratic formula, we get:
[tex]x=\frac{-36\±\sqrt{(36)^2-4(-16)(160)} }{2(-16)}\\\\x_1=4.48\\\\x_2=-2.23[/tex]
Time it takes Amber's textbook to reach the ground (in seconds)
Having the equation:
[tex]0=-16t^2+50t+160[/tex]
We can identify that:
[tex]a=-16\\b=50\\c=160[/tex]
Substituting values into the Quadratic formula, we get:
[tex]x=\frac{-50\±\sqrt{(50)^2-4(-16)(160)} }{2(-16)}\\\\x_1=5.08\\\\x_2=-1.96[/tex]
The difference of the positive values is:
[tex]5.08\ seconds-4.48\ seconds=0.6\ seconds[/tex]
Therefore, Rachel's textbook beat Amber's to the ground by [tex]0.6\ seconds[/tex]
