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Maksim231197 [3]
3 years ago
9

How is the process of this operation?

sqrt%7Bz%2B5%7D%20%7D%20" id="TexFormula1" title=" \frac{ \sqrt[6]{z+5} }{ \sqrt{z+5} } " alt=" \frac{ \sqrt[6]{z+5} }{ \sqrt{z+5} } " align="absmiddle" class="latex-formula">
Mathematics
1 answer:
Yuki888 [10]3 years ago
4 0
a^\frac{n}{m}=\sqrt[m]{a^n}\\---------------\\\\ \dfrac{ \sqrt[6]{z+5} }{ \sqrt{z+5} } =(z+5)^\frac{1}{6}:(z+5)^\frac{1}{2}\\\\/use:a^n\cdot a^m=a^{n+m}/\\\\=(z+5)^{\frac{1}{6}+\frac{1}{2}}=(z+5)^{\frac{1}{6}+\frac{3}{6}}=(z+5)^\frac{4}{6}=(z+5)^\frac{2}{3}\\\\=\sqrt[3]{(z+5)^2}=\sqrt[3]{z^2+2z\cdot5+5^2}=\sqrt[3]{z^2+10z+25}
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Answer:

The height of the rocket at 3 seconds is 270 feet

Step-by-step explanation:

To calculate the height of the rocket at 3 seconds,

First, we will determine the height the rocket has traveled in 3 seconds

The height traveled by the rocket is given by

Height (h) = Velocity (v) × Time (t)

From the question,

The rocket travels 85 feet per second, that is the velocity of the rocket is 85 feet per second,

Now,

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Hence,

Height (h) = 85 feet /second × 3 seconds

Height = 255 feet

∴ the rocket would have traveled 255 feet in 3 seconds,

But, the rocket starts on a launch pad 15 feet of the ground,

Then, the height of the rocket at 3 seconds is

255 feet + 15 feet = 270 feet.

Hence, the height of the rocket at 3 seconds is 270 feet

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Step-by-step explanation:

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Step-by-step explanation:

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