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.
</span><span>
</span>
<span>commutative property of multiplication</span>
Answer:
Step-by-step explanation:
<u>Simplfy in Algebra
</u>
We have the following expression
Simplifying like factors in the denominator and numerator
All the factors are perfect squares except 
, thus we rewrite:
Taking the square root of all the perfect square factors:
Answer:
Step-by-step explanation:
Divide both sides of the equation by 2
2x/2 = 0/2
x = 0
(4+612+x)=2(3x+8)
4+612+x=6x+16
4+612-16=6x-x
600=5x
Divide both sides of the equation by 5
600/5 = 5x
120 = x
Answer:
-1/5
Step-by-step explanation:
: )
