Answer: C
Step-by-step explanation: Hope this help :D
Answer:
Aahhahaha
Step-by-step explanation:
your welcome
Answer:
Step-by-step explanation:
x^9 + 3x^8 +2x^7 +4=0
Hey,I'm not sure enough...i guess if the highest power is 9 then we get 9 roots..sorryvif I'm wrong:')
Answer:
the answer is c
Step-by-step explanation:
Answer:
It takes 1 second for the tape to reach the ground.
Equation to use: ![y(t)=16-\frac{1}{2} g\,t^2[/tex with acceleration due to gravity "g" = 32ft/s^2]Step-by-step explanation:This is an object moving vertically under the action of the acceleration of gravity (32 ft/s^2), with a starting position of 16 feet, and with NO initial velocity (drops from the roof).The equation that describes the"y" position of the object as a function of time (t) will be written as:[tex]y(t) = y_0+ v_0*t-\frac{1}{2} g\,t^2](https://tex.z-dn.net/?f=y%28t%29%3D16-%5Cfrac%7B1%7D%7B2%7D%20g%5C%2Ct%5E2%5B%2Ftex%20with%20acceleration%20due%20to%20gravity%20%22g%22%20%3D%2032ft%2Fs%5E2%5D%3C%2Fp%3E%3Cp%3E%3Cstrong%3EStep-by-step%20explanation%3A%3C%2Fstrong%3E%3C%2Fp%3E%3Cp%3EThis%20is%20an%20object%20moving%20vertically%20under%20the%20action%20of%20the%20acceleration%20of%20gravity%20%2832%20ft%2Fs%5E2%29%2C%20with%20a%20starting%20position%20of%2016%20feet%2C%20and%20with%20NO%20initial%20velocity%20%28drops%20from%20the%20roof%29.%3C%2Fp%3E%3Cp%3EThe%20equation%20that%20describes%20the%22y%22%20%20position%20of%20the%20object%20as%20a%20function%20of%20time%20%28t%29%20will%20be%20written%20as%3A%3C%2Fp%3E%3Cp%3E%5Btex%5Dy%28t%29%20%3D%20y_0%2B%20v_0%2At-%5Cfrac%7B1%7D%7B2%7D%20g%5C%2Ct%5E2)
As explain above, the initial position
s 16 ft, there is no initial velocity, so
, and the acceleration of gravity is 32 ft/s^2, and should be considered negative [as pointing down in the y-direction], so the equation simplifies to:
![y(t) = y_0+ v_0*t-\frac{1}{2} g\,t^2\\y(t)=16-\frac{1}{2} g\,t^2](https://tex.z-dn.net/?f=y%28t%29%20%3D%20y_0%2B%20v_0%2At-%5Cfrac%7B1%7D%7B2%7D%20g%5C%2Ct%5E2%5C%5Cy%28t%29%3D16-%5Cfrac%7B1%7D%7B2%7D%20g%5C%2Ct%5E2)
In order to find the tima it takes it to hit the ground, we simple solve the equation for t when y(t) = 0 (the tape has reached the ground (zero height in the y-direction):
![y(t)=16-\frac{1}{2} g\,t^2\\0=16-\frac{1}{2} g\,t^2\\-16=-\frac{1}{2} 32\,t^2\\-16=-16\,t^2\\t^2=1\\t=+/- 1 \,second](https://tex.z-dn.net/?f=y%28t%29%3D16-%5Cfrac%7B1%7D%7B2%7D%20g%5C%2Ct%5E2%5C%5C0%3D16-%5Cfrac%7B1%7D%7B2%7D%20g%5C%2Ct%5E2%5C%5C-16%3D-%5Cfrac%7B1%7D%7B2%7D%2032%5C%2Ct%5E2%5C%5C-16%3D-16%5C%2Ct%5E2%5C%5Ct%5E2%3D1%5C%5Ct%3D%2B%2F-%201%20%5C%2Csecond)
We select the positive time (+1 second) which is what makes physical sense, since a negative value in time would mean time before the tape was dropped.
So the answer is: It takes 1 second for the tape to reach the ground.