I would say the first choose that the vertical axis should use $0.50 intervals instead of $1.
Because this would make it much easier to read and understand and would it would also be more accurate.
1. 0.6
2. 10
3. -½
the answers are thrre ...
Step-by-step explanation:
2 = 3 - q
1 = (3 - q)×1/2
9 =(3 - q) ×1/2×9
9 = 27 - Q/2
If you use this equation then you say that the ground is h=0 and solve as a quadratic.
The quadratic formula is (-b±<span>√(b^2-4ac))/2a when an equation is in the form ax^2 + bx + c
So the equation you have been given would be -16t^2-15t-151 = 0
This equation has no real roots which leads me to believe it is incorrect.
This is probably where your difficulty is coming from, it's a mistake.
The equation is formed from S=ut+(1/2)at^2+(So) where (So) is the initial height and S is the height that you want to find.
In this case you want S = 0.
If the initial height is +151 and the initial velocity and acceleration are downwards (negative) and the initial velocity (u) is -15 and the initial acceleration is -32 then you get the equation S=-15t-16t^2+151
Solving this using the quadratic formula gives you t = 2.64 or t = -3.58
Obviously -3.58s can't be the answer so you're left with 2.64 seconds.
Hope this makes sense.
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Answer:
Dominos is the better deal.
