The speed of the moving walkway relative to the airport terminal exists at 1.84 m/s.
<h3>How to estimate the speed of the moving walkway relative to the airport terminal?</h3>
Let x be the speed of the walkway.
(2.8 + x) = speed of child moving in direction of the walkway
(2.8 - x) = speed of child moving against the direction of the walkway
Travel time = distance/speed
Travel time of child moving in direction of walkway = 23/(2.8+x)
Total elapsed time given = 29s
23/(2.8 + x)+ 23 / (2.8-x) = 29
LCD = (2.8 + x)(2.8 - x)

simplifying the equation, we get




Speed of walkway = 1.84 m/s
The speed of the moving walkway relative to the airport terminal exists at 1.84 m/s.
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<h3>
Answer: Choice A</h3>
![x^2\left(\sqrt[4]{x^2}\right)](https://tex.z-dn.net/?f=x%5E2%5Cleft%28%5Csqrt%5B4%5D%7Bx%5E2%7D%5Cright%29)
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Explanation:
The fourth root of x is the same as x^(1/4)
I.e,
![\sqrt[4]{x} = x^{1/4}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7Bx%7D%20%3D%20x%5E%7B1%2F4%7D)
The same applies to x^10 as well
![\sqrt[4]{x^{10}} = \left(x^{10}\right)^{1/4}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7Bx%5E%7B10%7D%7D%20%3D%20%5Cleft%28x%5E%7B10%7D%5Cright%29%5E%7B1%2F4%7D)
Multiply the exponents 10 and 1/4 to get 10/4
![\sqrt[4]{x^{10}} = \left(x^{10}\right)^{1/4} = x^{10*1/4} = x^{10/4}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7Bx%5E%7B10%7D%7D%20%3D%20%5Cleft%28x%5E%7B10%7D%5Cright%29%5E%7B1%2F4%7D%20%3D%20x%5E%7B10%2A1%2F4%7D%20%3D%20x%5E%7B10%2F4%7D)
![\sqrt[4]{x^{10}} = x^{10/4}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7Bx%5E%7B10%7D%7D%20%3D%20x%5E%7B10%2F4%7D)
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If we have an expression in the form x^(m/n), with m > n, then we can simplify it into an equivalent form as shown below
![x^{m/n} = x^a\sqrt[n]{x^b}](https://tex.z-dn.net/?f=x%5E%7Bm%2Fn%7D%20%3D%20x%5Ea%5Csqrt%5Bn%5D%7Bx%5Eb%7D)
The 'a' and 'b' are found through dividing m/n
m/n = a remainder b
'a' is the quotient, b is the remainder
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The general formula can easily be confusing, so let's replace m and n with the proper numbers. In this case, m = 10 and n = 4
m/n = 10/4 = 2 remainder 2
We have a = 2 and b = 2
So
![x^{m/n} = x^a\sqrt[n]{x^b}](https://tex.z-dn.net/?f=x%5E%7Bm%2Fn%7D%20%3D%20x%5Ea%5Csqrt%5Bn%5D%7Bx%5Eb%7D)
turns into
![x^{10/4} = x^2\sqrt[4]{x^2}](https://tex.z-dn.net/?f=x%5E%7B10%2F4%7D%20%3D%20x%5E2%5Csqrt%5B4%5D%7Bx%5E2%7D)
which means
![\sqrt[4]{x^{10}} = {x^2} \sqrt[4]{x^2}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7Bx%5E%7B10%7D%7D%20%3D%20%7Bx%5E2%7D%20%5Csqrt%5B4%5D%7Bx%5E2%7D)
Answer:x=1
Step-by-step explanation:
C^2-64 is a square roots factoring one. you take the square root of c^2 which is c and then the square root of 64 which is 8 or -8 so the answer is (c-8)(c+8). and it is a special product