Answer: ![\sqrt[3]{125x^3}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B125x%5E3%7D)
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Step-by-step explanation: Given radical expression
![\sqrt[3]{5x} \times \sqrt[3]{25x^2}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B5x%7D%20%5Ctimes%20%5Csqrt%5B3%5D%7B25x%5E2%7D)
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According to the product property of roots.
![\sqrt[n]{a} \times \sqrt[n]{b} = \sqrt[n]{a \times b}](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Ba%7D%20%5Ctimes%20%5Csqrt%5Bn%5D%7Bb%7D%20%3D%20%5Csqrt%5Bn%5D%7Ba%20%5Ctimes%20b%7D)
On applying above rule, we get
![\sqrt[3]{5x} \times \sqrt[3]{25x^2} = \sqrt[3]{5x \times 25x^2}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B5x%7D%20%5Ctimes%20%5Csqrt%5B3%5D%7B25x%5E2%7D%20%3D%20%5Csqrt%5B3%5D%7B5x%20%5Ctimes%2025x%5E2%7D)
5 × 25 = 125 and
Therefore,
![\sqrt[3]{5x \times 25x^2}= \sqrt[3]{125x^3}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B5x%20%5Ctimes%2025x%5E2%7D%3D%20%5Csqrt%5B3%5D%7B125x%5E3%7D)
<h3>So, the correct option would be second option
.</h3>
Oasis- fertile spot in a desert where water is found.
Answer:
t≈8.0927
Step-by-step explanation:
h(t) = -16t^2 + 128t +12
We want to find when h(t) is zero ( or when it hits the ground)
0 = -16t^2 + 128t +12
Completing the square
Subtract 12 from each side
-12 = -16t^2 + 128t
Divide each side by -16
-12/-16 = -16/-16t^2 + 128/-16t
3/4 = t^2 -8t
Take the coefficient of t and divide it by 8
-8/2 = -4
Then square it
(-4) ^2 = 16
Add 16 to each side
16+3/4 = t^2 -8t+16
64/4 + 3/4= (t-4)^2
67/4 = (t-4)^2
Take the square root of each side
±sqrt(67/4) =sqrt( (t-4)^2)
±1/2sqrt(67) = (t-4)
Add 4 to each side
4 ±1/2sqrt(67) = t
The approximate values for t are
t≈-0.092676
t≈8.0927
The first is before the rocket is launched so the only valid answer is the second one
Answer:
A. 2CO + O2 ---> 2CO2
Step-by-step explanation:
2CO + O2 ---> 2CO2
