TV
A frog is sitting on the ground when he is scared by a big dog. His movement is modeled by
the equation h = -6t2 + 6t where h is the frog's height at any given time, t. How many
seconds until the frog is back on the ground?
2 seconds
1.5 seconds
3 seconds
1 second
0.5 seconds

Step-by-step explanation:

Model of the frog jump is

h = -6t² + 6t

When will the frog be on the ground? The frog will be on the ground at h=0

Therefore,

0 = -6t² + 6t

Divide through by 6

0 = -t² + t

0 = t(-t + 1)

So, t = 0 OR. 0 = -t + 1

So, t = 0. OR. t = 1

So, we should note that, at t = 0, was when he wanted to lift of from the ground... So, t = 0 will be discarded

So, at t = 1 is when the frog is back to the ground.

It is like it starts from the ground at the 0 seconds and get backs to the ground at 1 seconds

I hope this helps!!!

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'.

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step-by-step explanation: