Answer:
2.0 seconds
Step-by-step explanation:
Given quadratic functions:
[tex]\begin{cases}y=-7x^2+26x+3\\y=-6x^2 +23x+5\end{cases}[/tex]
To find the time, in seconds, that the balloons collided at the highest point, substitute one equation into the other equation and rearrange to equal zero:
[tex]\begin{aligned}-6x^2+23x+5 & = -7x^2+26x+3\\-6x^2+23x+5+7x^2 & = -7x^2+26x+3+7x^2\\x^2+23x+5 & = 26x+3\\x^2+23x+5-26x & = 26x+3-26x\\x^2-3x+5& = 3\\x^2-3x+5-3& = 3-3\\x^2-3x+2& = 0\end{aligned}[/tex]
Factor the quadratic:
[tex]\begin{aligned}x^2-3x+2 & = 0\\x^2-2x-x+2 & = 0\\x(x-2)-1(x-2) & = 0\\(x-1)(x-2) & = 0\\\end{aligned}[/tex]
Apply the zero-product property to solve for x:
[tex]\implies (x-1)=0 \implies x=1[/tex]
[tex]\implies (x-2)=0 \implies x=2[/tex]
Therefore, the balloons collided at 1 second and 2 seconds.
To find at which time the highest point of collision occured, substitute both values of x into one of the functions:
[tex]f(1)=-6(1)^2 +23(1)+5=22[/tex]
[tex]f(2)=-6(2)^2 +23(2)+5=27[/tex]
Therefore, the time, in seconds, that the balloons collided at the highest point is 2.0 seconds.
Learn more about quadratic systems of equations here:
https://brainly.com/question/27930827
Equate both and solve
x=2,1
The collisions happen at 1s and 2s
As only 2s is included in our options option A is correct
