<span>(2a)^3 =8a^3
hope it helps</span>
Number 10 is D. and 11 is 7 different ways. and nice legs XD
Answer:
Easy peasy, hope it can help you!
Answer:
Kayla, learn this, :D more math is based on it. Just learn it, just memorize the following
Step-by-step explanation:
m = slope
m = rise / run
m = y2 - y1 / x2 - x1 ( this is still rise over run but in point form )
they give us two points to plug into the above formula
P1 = (-4, 8 ) in the form of ( x1, y 1 )
P2 = (2, -1) in the form of (x2, y2 )
now plug those into the above formula
m = ( -1 - 8 ) / ( 2-(-4) )
m = -9 / ( 2 + 4 )
m = -9 / 6
m = - 3/2
since we know the slope now , we can use the point slope equation of
y - y1 = m ( x - x1)
use the m we found and either point , P1 or P2
y - 8 = -3/2(x -(-4) )
y-8 = -3/2(x + 4)
y-8 = -3/2x - 6
y = -3/2x -6 + 8
y = -3/2x + 2
there you go :)
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
