Answer:
After 14.92 seconds, the arrow hits the ground.
Step-by-step explanation:
Given:
Initial velocity of the arrow is, ![u=240\ ft/s or 73.152\ m/s](https://tex.z-dn.net/?f=u%3D240%5C%20ft%2Fs%20or%2073.152%5C%20m%2Fs)
The arrow goes up and come back to the same starting position. So, initial position is same as that of final position.
Therefore, displacement of the arrow is 0 m.
The displacement of the arrow is given as:
(equation 1)
Plug all value in equation 1 and solve for ![t](https://tex.z-dn.net/?f=t)
![0=73.152\timests t-\frac{1}{2} \times9.8 \times t^{2}](https://tex.z-dn.net/?f=0%3D73.152%5Ctimests%20t-%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes9.8%20%5Ctimes%20t%5E%7B2%7D)
![0=73.152\timests t-4.9 \times t^{2}](https://tex.z-dn.net/?f=0%3D73.152%5Ctimests%20t-4.9%20%5Ctimes%20t%5E%7B2%7D)
![0=t(73.152-4.9 \times t)](https://tex.z-dn.net/?f=0%3Dt%2873.152-4.9%20%5Ctimes%20t%29)
where ![t\neq 0](https://tex.z-dn.net/?f=t%5Cneq%200)
![4.9\times t=73.152](https://tex.z-dn.net/?f=4.9%5Ctimes%20t%3D73.152)
![t=\frac{73.152}{4.9}= 14.92\ s](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B73.152%7D%7B4.9%7D%3D%2014.92%5C%20s)
∴ After 14.92 seconds, the arrow hits the ground.
5(x²+x-2x-2)
5(x²-x-2)
5x²-5x-10
you can also simplify to x²-x-2.
Is this multiple choice? I know that a is correct.
Answer:
A. 2(x+y+1). lmk if you need any more awnsers
Answer:
its an exponent
Step-by-step explanation: