Multiply 4 with (2x-5)
8x-20+15=11
add 20 both sides
8x+15=31
subtract 15 both sides
8x=16
divide both sides by the x (8)
x=2
Answer:
The maximum height of the particle is 100 ft
A is correct
Step-by-step explanation:
A particle is moving along a projectile path where the initial height is 96 feet with an initial speed of 16 feet per second.
The standard equation of parabolic projectile.
![h(t)=-16t^2+v_0t+h_0](https://tex.z-dn.net/?f=h%28t%29%3D-16t%5E2%2Bv_0t%2Bh_0)
where,
![v_0=16\ ft/s](https://tex.z-dn.net/?f=v_0%3D16%5C%20ft%2Fs)
![h_0=96\ ft](https://tex.z-dn.net/?f=h_0%3D96%5C%20ft)
![h(t)=-16t^2+16t+96](https://tex.z-dn.net/?f=h%28t%29%3D-16t%5E2%2B16t%2B96)
So, we get the parabolic equation of path.
The maximum value at vertex.
![t=-\dfrac{b}{2a}](https://tex.z-dn.net/?f=t%3D-%5Cdfrac%7Bb%7D%7B2a%7D)
![t=-\dfrac{16}{-32}=\dfrac{1}{2}](https://tex.z-dn.net/?f=t%3D-%5Cdfrac%7B16%7D%7B-32%7D%3D%5Cdfrac%7B1%7D%7B2%7D)
Maximum at t=0.5
![h(0.5)=-16(0.5)^2+16(0.5)+96](https://tex.z-dn.net/?f=h%280.5%29%3D-16%280.5%29%5E2%2B16%280.5%29%2B96)
![h(0.5)=100](https://tex.z-dn.net/?f=h%280.5%29%3D100)
Hence, The maximum height of the particle is 100 ft
Answer: its b
Step-by-step explanation: