Answer:
(2, - 5)
Explanation:
The line y = -1 is horizontal: parallel to the y-axis.
The point B (2,3) is above the line y = - 1 at a distance equal to the y-coordinate of B (y = 3) less - 1:
Δy = 3 - ( - 1) = 4
When the point B (2,3) is reflected over the line y = -1, the x-coordinate does not change. The image will be 4 units down the line y = -1, thus the y-coordinate will be - 1 - 4 = - 5.
Hence, the coordinates of the vertex B' are x = 2, y = - 5: (2, - 5).
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:
Step-by-step explanation:
Answer is 28 centimeters
Explanation:
I have no clue download Photomath and ask it lol