Answer:
8 seconds.
Step-by-step explanation:
To solve this question, we need to find the time,
, when
, that is, when the height of the object above the ground is 0 feet.
Substituting
feet per second and
gives the quadratic:
![0 = 128t - 16t^{2}](https://tex.z-dn.net/?f=0%20%3D%20128t%20-%2016t%5E%7B2%7D)
Since 128 is divisible by 16, it can be reduced to
.
We must now solve for
.
We can easily see that one answer to the equation is 0,
(we need not concern ourselves with the 16 outside of the parenthesis as in the equation above, since 16 multiplied by 0 is 0). However this is the time the object is released into the air.
The second answer,
is also easy to see by inspection:
.
Therefore the object lands 8 seconds after it is thrown.
Answer:
vertex=(1,-3)
y-intercept=(0,-2)
x-intercept=(1,+\sqrt{3},0)
axis of symmetry= 1
The two point formula :
![y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)](https://tex.z-dn.net/?f=y-y_1%3D%5Cfrac%7By_2-y_1%7D%7Bx_2-x_1%7D%28x-x_1%29)
The points are (-2, 1) and (1, 10)
Using the formula above :
![\begin{gathered} y-1=\frac{10-1}{1-(-2)}(x+2) \\ y-1=\frac{9}{3}(x+2) \\ y-1=3(x+2) \\ y-1=3x+6 \\ -3x+y=6+1 \\ -3x+y=7 \\ 3x-y=-7 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20y-1%3D%5Cfrac%7B10-1%7D%7B1-%28-2%29%7D%28x%2B2%29%20%5C%5C%20y-1%3D%5Cfrac%7B9%7D%7B3%7D%28x%2B2%29%20%5C%5C%20y-1%3D3%28x%2B2%29%20%5C%5C%20y-1%3D3x%2B6%20%5C%5C%20-3x%2By%3D6%2B1%20%5C%5C%20-3x%2By%3D7%20%5C%5C%203x-y%3D-7%20%5Cend%7Bgathered%7D)
The answer is 3x - y = -7
Hi again!
I used an online calculator to help me graph each of the functions.
The correct answer is option D
Let me know if you have any questions about the answer!