Answer: Choice C) Infinitely many solutions

If you solve the first equation for y, you get
2y = 14 - 2x
2y = 14 + (-2x)
2y = -2x + 14
2y/2 = (-2x)/2 + 14/2 ... divide every term by 2
y = -x + 7

This result of y = -x+7 is identical to the second equation in the system. So the two graphs are going to be the same. We only produce one line. One graph is right on top of the other. The two lines will intersect infinitely many times. Any point on the blue line is a solution to the system.

3 0

3 years ago

Suppose you have a water-balloon launcher. The balloon is 3 ft high when it leaves the launcher. Use the equation 0 = ?16t2 + 38

mario62 [17]

Answer:

t = 2.5 s

Step-by-step explanation:

Suppose you have a water-balloon launcher. The balloon is 3 ft high when it leaves the launcher. Its equation is :

$16t^2 + 38.8t + 3 =0$

The above equation is a quadratic equation. The general equation is :

$ax^2+bx+c=0$

Here, a = 16, b = 38.8 and c = 3

The solution of quadratic equation is given by :

$x=\dfrac{-b\pm \sqrt{b^2-4ac} }{2a}\\\\t=\dfrac{-38.8\pm \sqrt{(38.8)^2-4\times (-16)\times 3} }{2\times (-16)}\\\\t=\dfrac{-38.8+\sqrt{(38.8)^{2}-4\times-16\times3}}{-2\times16}, \dfrac{-38.8-\sqrt{(38.8)^{2}-4\times-16\times3}}{-2\times16}\\\\t=-0.075\ s,2.5\ s$

So, at t = 2.5 s the balloon is in air.

8 0

3 years ago