Jack is doing a test launch of his hovercraft for the upcoming STEM competition.

The answer is missing.

I think there is something wrong with the equation on its own

Step-by-step explanation:

Equation for the hovercraft

F(x)= X²+6x+2

Olivia catapult equation

F(x)= √2x

Differential of the both equation will give the velocity at x time

F(x)= X²+6x+2

DF(x)/Dx= 2x +6

F(x)= √2x

F(x)= (2x)^½

DF(x)/Dx= (2x)^-½

So the differential is the velocity of the both equation.

Let's equate both equation to find the value of x at which they have same velocity.

(2x)^-½=- 2x+6

(2x)^-1= (-2x+6)²

(2x)^-1= 4x² -24x +36

0= 8x³ - 48x² +72x -1

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.

$5.41

step-by-step explanation:

$20 - $14.51 = $5.49.