Answer:
The third option listed: ![\sqrt[3]{2x} -6\sqrt[3]{x}\\](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B2x%7D%20-6%5Csqrt%5B3%5D%7Bx%7D%5C%5C)
Step-by-step explanation:
We start by writing all the numerical factors inside the qubic roots in factor form (and if possible with exponent 3 so as to easily identify what can be extracted from the root):
![7\sqrt[3]{2x} -3\sqrt[3]{16x} -3\sqrt[3]{8x} =\\=7\sqrt[3]{2x} -3\sqrt[3]{2^32x} -3\sqrt[3]{2^3x} =\\=7\sqrt[3]{2x} -3*2\sqrt[3]{2x} -3*2\sqrt[3]{x}=\\=7\sqrt[3]{2x} -6\sqrt[3]{2x} -6\sqrt[3]{x}](https://tex.z-dn.net/?f=7%5Csqrt%5B3%5D%7B2x%7D%20%20-3%5Csqrt%5B3%5D%7B16x%7D%20-3%5Csqrt%5B3%5D%7B8x%7D%20%3D%5C%5C%3D7%5Csqrt%5B3%5D%7B2x%7D%20%20-3%5Csqrt%5B3%5D%7B2%5E32x%7D%20-3%5Csqrt%5B3%5D%7B2%5E3x%7D%20%3D%5C%5C%3D7%5Csqrt%5B3%5D%7B2x%7D%20%20-3%2A2%5Csqrt%5B3%5D%7B2x%7D%20-3%2A2%5Csqrt%5B3%5D%7Bx%7D%3D%5C%5C%3D7%5Csqrt%5B3%5D%7B2x%7D%20%20-6%5Csqrt%5B3%5D%7B2x%7D%20-6%5Csqrt%5B3%5D%7Bx%7D)
And now we combine all like terms (notice that the only two terms we can combine are the first two, which contain the exact same radical form:
![7\sqrt[3]{2x} -6\sqrt[3]{2x} -6\sqrt[3]{x}=\\=\sqrt[3]{2x} -6\sqrt[3]{x}](https://tex.z-dn.net/?f=7%5Csqrt%5B3%5D%7B2x%7D%20%20-6%5Csqrt%5B3%5D%7B2x%7D%20-6%5Csqrt%5B3%5D%7Bx%7D%3D%5C%5C%3D%5Csqrt%5B3%5D%7B2x%7D%20-6%5Csqrt%5B3%5D%7Bx%7D)
Therefore this is the simplified radical expression: ![\sqrt[3]{2x} -6\sqrt[3]{x}\\](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B2x%7D%20-6%5Csqrt%5B3%5D%7Bx%7D%5C%5C)
We are given a watermelon dropped at free fall from a building 320 meters above the sidewalk. Superman is headed down at 30 meters per second. We are asked to determine how fast is the watermelon going when it passes Superman. To solve for the final velocity of the watermelon, we will use one of the kinematic equations (free fall):
vf = vi + a*t
where vf is the final velocity
vi is the initial velocity, zero
a is the acceleration, in this case, gravitational acceleration = 9.8m/s^2
t is time
we also need to set-up another equation using the distance:
d = vf + vi / 2 * t
(1) 320 m = vf * t /2
(2) vf = 9.8 m/s^2 * t
From here, we have two equations and two unknown, thus we can solve the problem by substitution.
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