a) For this question we set h=59 and solve for t, in order to do so we use the general formula for second-degree equations:
![\begin{gathered} t=\frac{-124\pm\sqrt[]{124^2-4(-16)(-40)}}{2(-16)} \\ t=\frac{-124\pm113.21}{-32} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20t%3D%5Cfrac%7B-124%5Cpm%5Csqrt%5B%5D%7B124%5E2-4%28-16%29%28-40%29%7D%7D%7B2%28-16%29%7D%20%5C%5C%20t%3D%5Cfrac%7B-124%5Cpm113.21%7D%7B-32%7D%20%5Cend%7Bgathered%7D)
The height of the object will be 59 feet at t=7.41 seconds and t=0.34 seconds.
b) When the object reaches the ground, h=0 therefore:
Solving for t we get:
![\begin{gathered} t=\frac{-124\pm\sqrt[]{124^2-4(-16)(19)}}{2(-16)} \\ t=\frac{-124\pm\sqrt[]{16592}}{-32}=\frac{-124\pm128.81}{-32} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20t%3D%5Cfrac%7B-124%5Cpm%5Csqrt%5B%5D%7B124%5E2-4%28-16%29%2819%29%7D%7D%7B2%28-16%29%7D%20%5C%5C%20t%3D%5Cfrac%7B-124%5Cpm%5Csqrt%5B%5D%7B16592%7D%7D%7B-32%7D%3D%5Cfrac%7B-124%5Cpm128.81%7D%7B-32%7D%20%5Cend%7Bgathered%7D)
Therefore, since t cannot be negative the solution is t=7.9 seconds.
About 1.5 per min but im not sure if it is exactly 1.5
Answer:
Angle 4 and angle 5
Explanation:
Alternate interior angles are angles within two parallel lines that fall at the opposite sides of the transversal.
By carefully observing the figure shown:
Angle 3 and angle 8 are alternate interior angles because they are within the two parallel lines and are at opposite sides of the transversal
Also, angle 4 and angle 5 are alternate interior angles because they are within the two parallel lines and are at opposite sides of the transversal
Out of the given options, only angle 4 and angle 5 are aleternate interior angles
Answer:
$13.42 each
Step-by-step explanation:
80.52/6=$13.42
Answer:
D. 71 units
Step-by-step explanation:
