Question

PLS PLS HELP, I DONT WANT TO FAIL!!

Answer 1

Answer:

2.0 seconds

Step-by-step explanation:

Given quadratic functions:

$\begin{cases}y=-7x^2+26x+3\\y=-6x^2 +23x+5\end{cases}$

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:

$\begin{aligned}-6x^2+23x+5 & = -7x^2+26x+3\\-6x^2+23x+5+7x^2 & = -7x^2+26x+3+7x^2\\x^2+23x+5 & = 26x+3\\x^2+23x+5-26x & = 26x+3-26x\\x^2-3x+5& = 3\\x^2-3x+5-3& = 3-3\\x^2-3x+2& = 0\end{aligned}$

Factor the quadratic:

$\begin{aligned}x^2-3x+2 & = 0\\x^2-2x-x+2 & = 0\\x(x-2)-1(x-2) & = 0\\(x-1)(x-2) & = 0\\\end{aligned}$

Apply the zero-product property to solve for x:

$\implies (x-1)=0 \implies x=1$

$\implies (x-2)=0 \implies x=2$

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:

$f(1)=-6(1)^2 +23(1)+5=22$

$f(2)=-6(2)^2 +23(2)+5=27$

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:

brainly.com/question/27930827

Answer 2

Equate both and solve

• -7x²+26x+3=-6x²+23x+5
• -7x²+6x²+26x-23x+3-5=0
• -x²+3x-2=0
• x²-3x+2=0
• (x-2)(x-1)=0

x=2,1

The collisions happen at 1s and 2s

As only 2s is included in our options option A is correct