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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I'm pretty sure the answer is .32
The answer would be 19,728.19
Equation of the line in point slope form is y + 3 = -1/2(x - 7)
Step-by-step explanation:
- Step 1: Given slope of the line is -1/2 and passes through the point (x1, y1) = (7, -3). Equation in point slope form is y - y1 = m(x - x1)
⇒ y - (-3) = -1/2(x - 7)
⇒ y + 3 = -1/2(x - 7)
Answer:
from looking at this we can see
1*4=4
2*4=8
3*4=12
6*4=24
a
Hope This Helps!!!
