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HELP I NEED THIS FOR TODAY!! 100 points



Two water balloons were launched into the air at different moments and collided. The water balloons were modeled by the quadratic functions: y = −3x2 + 24x + 3 and y = −4x2 + 25x + 15, where y represents the height in meters and x represents the time in seconds after the launch. What is the time, in seconds, that the balloons collided at the highest point?

3.25 seconds
3.50 seconds
3.75 seconds
4.00 seconds



What are the solutions to the following system of equations?

x − y = 6
y = x2 −6

(0, −6) and (7, 1)
(0, −6) and (1, −5)
(0, −6) and (9, 3)
(0, −6) and (−9, 3)

Based on the graphs of the equations y = x + 7 and y = x2 – 3x + 7, the solutions are located at points:

(4, 11) and (0, 7)
(4, 11) and (–7, 0)
(–7, 0) and (1.5, 4.75)
(1.5, 4.75) and (0, 7)

Which system of equations is represented by the graph?

Graph of a quadratic function intersecting a linear function at point 2 comma 0.

y = x2 + 3x + 2
x + y = 2
y = x2 − 3x + 2
x − y = 2
y = x2 − 3x + 2
x − y = 3
y = x2 − 3x + 2
x + y = 3

Respuesta :

The solutions of the given equations are given by equating the

variables to gives;

  • First part; 4.00 seconds
  • Second part; (0, -6) and (1, -5)
  • Third part; (4, 11) and (0, 7)
  • Fourth part: y = x² + 3·x + 2; x + y = 2

How can the (common) solution be found?

First part;

The given functions are;

y = -3·x² + 24·x + 3, and y = -4·x² + 25·x + 15

When the balloons collided, we have;

The height are equal, which gives;

y = y

-3·x² + 24·x + 3 = -4·x² + 25·x + 15

-3·x² + 24·x + 3 - (-4·x² + 25·x + 15) = x² - x - 12 = 0

Which gives;

x² - x - 12 = (x - 4) × (x + 3) = 0

x = 4, and x = -3

The time at which the balloon collide at the highest point is therefore;

  • At the time, x = 4.00 seconds

Second part;

The given system of equations are;

x - y = 6

y = x² - 6

y = x - 6

Which gives;

x - 6 = x² - 6

x = x²

Which gives;

x = 0 or x = 1

Similarly, we have;

x = y + 6

Which gives;

y = (y + 6)² - 6

y² + 11·y + 30 = 0

(y + 5)·(y + 6) = 0

Which gives;

y = -5, or y = -6

When y = -5, we have; x = -5 + 6 = 1

The solution are therefore;

  • (0, -6) and (1, -5)

Third part;

The equations are; y = x + 7 and y = x² - 3·x + 7

At the solution, the u-values are equal, which gives;

y = y

x + 7 = x² - 3·x + 7

x² - 3·x + 7 - (x + 7) = 0

x² - 4·x = 0

x·(x - 4) = 0

x = 4, or x = 0

When x = 4, y = 4 + 7 = 11

When x = 0, y = 0 + 7 = 7

The solutions are located at points;

  • (4, 11) and (0, 7)

Fourth part;

Given that the point of intersection of the line and the quadratic function

is the point (2, 0), we have;

On the linear function, when y = 0, x = 2

The possible linear function functions is therefore;

x + y = 2

From the quadratic function that accompanies the above linear function, we have;

y = x² + 3·x + 2

When x = 0, y = 0² + 3×0 + 2 = 2

Therefore;

  • The correct option is; y = x² + 3·x + 2, and x + y = 2

Learn more about the solutions of equations here:

https://brainly.com/question/24400554

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