5(4x-3)^2 - 3
5(16x^2 - 2(3)(4x) + 9) -3
5(16x^2 - 24x + 9) -3
80x^2 - 120x + 45 - 3
80x^2 - 120x + 42
80(0)^2 - 120(0) + 42
42
Answer:
Step-by-step explanation:
With a factor of (t - 1) we know that zero (ground level) is reached at 1 second from an initial height of (0 - 1)(0 - 1)(0 - 11)(0 - 13)/3 = -1•-1•-11•-13 / 3 = 47⅔ meters at t = 0
As we have <em>two </em>factors of (t - 1) we know the track does not go underground at t = 1, but rises again.
At t = 11 seconds, the car has again returned to ground level, but as we only have a single factor of (t - 11) the car plunges below ground level and returns to above ground level at t = 13 seconds due to the single factor of (t - 13)
we can estimate that the car is the deepest below ground level halfway between 11 and 13 s, so at t = 12. At that time, the depth will be about (12 - 1)(12 - 1)(12 - 11)(12 - 13) / 3 = -(11²/3) = - 40⅓ m.
we can estimate that the car is the highest above ground level halfway between 1 and 11 s, so at t = 6s. At that time, the height will be about (6 - 1)(6 - 1)(6 - 11)(6 - 13) / 3 = 5²•-5•-7 / 3 = 291⅔ m.
It's obvious that the roller coaster car had significant initial velocity at t = 0 to achieve that altitude from an initial height of 47⅔ m
Answer:
4 letters XD
Step-by-step explanation:
you didn't attach a picture.
Answer:
r=8
Step-by-step explanation:
Using the formula they gave us you could plug in the area (64) and divide it by pi which cancels out the pi so taking the square root of 64 gives you 8FT
Hope this helps :)
