Answer:
5,435,820,000 is your answer
Answer:
google maybe
Step-by-step explanation:
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Answer:
the answer is 2% decreasing
Step-by-step explanation:
The given expression is :
![(\sqrt[3]{128} )^x](https://tex.z-dn.net/?f=%28%5Csqrt%5B3%5D%7B128%7D%20%29%5Ex)
We need to solve the above expression.
According to one of the law of indices,
![(\sqrt[a]{m} )^b=(\sqrt{m})^{b/a}](https://tex.z-dn.net/?f=%28%5Csqrt%5Ba%5D%7Bm%7D%20%29%5Eb%3D%28%5Csqrt%7Bm%7D%29%5E%7Bb%2Fa%7D)
Using above property,
![(\sqrt[3]{128} )^x=(128)^{x/3}\\\\=(64\times 2)^{x/3}\\\\=4(2)^{x/3}\ (\because\ \sqrt{64}=4) \\\\=4(\sqrt[3]{2})^x\\\\=4(2^{1/3})^x](https://tex.z-dn.net/?f=%28%5Csqrt%5B3%5D%7B128%7D%20%29%5Ex%3D%28128%29%5E%7Bx%2F3%7D%5C%5C%5C%5C%3D%2864%5Ctimes%202%29%5E%7Bx%2F3%7D%5C%5C%5C%5C%3D4%282%29%5E%7Bx%2F3%7D%5C%20%28%5Cbecause%5C%20%5Csqrt%7B64%7D%3D4%29%20%5C%5C%5C%5C%3D4%28%5Csqrt%5B3%5D%7B2%7D%29%5Ex%5C%5C%5C%5C%3D4%282%5E%7B1%2F3%7D%29%5Ex)
Hence, this is the required solution.
First, an introduction: If the equation of the circle were x^2 + y^2 = 144, then the center would be at (0,0) and the radius would be 12. Note that the distance from the center to P(10,10) is 10sqrt(2), or 14.14. Thus, in this example, P would be OUTSIDE the circle (since 14.14 is greater than the radius 12).
Now let's focus on <span>(x-1)^2 + (y-2)^2 =144. Let x = 12 as an example; find the corresponding y: 9^2 + (y-2)^2 = 144, or (y-2)^2 = 63, and so y-2 is approx. -8 or +8. Then y (for x = 12) is either approx. -10 or 6: (12,-10) or (12,6). Are these inside the circle or outside?
A better way to address this would be as follows:
Find the distance from the center (1, 2) to the point P(10,10). If this distance is less than 12, the point P is inside C; if greater than 12, P is outside C.
This distance is sqrt( (10-2)^2 + (10-1)^2 ), or sqrt (64+81) = sqrt(145).
This is LARGER than sqrt(144). Thus, P is OUTSIDE the circle C.</span>
