The answer to this question is 0.4
Answer:
![4x^{3} y^{2} (\sqrt[3]{4 x y})](https://tex.z-dn.net/?f=4x%5E%7B3%7D%20y%5E%7B2%7D%20%28%5Csqrt%5B3%5D%7B4%20x%20y%7D%29)
Step-by-step explanation:
Another complex expression, let's simplify it step by step...
We'll start by re-writing 256 as 4^4
![\sqrt[3]{256 x^{10} y^{7} } = \sqrt[3]{4^{4} x^{10} y^{7} }](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B256%20x%5E%7B10%7D%20y%5E%7B7%7D%20%7D%20%3D%20%5Csqrt%5B3%5D%7B4%5E%7B4%7D%20x%5E%7B10%7D%20y%5E%7B7%7D%20%7D)
Then we'll extract the 4 from the cubic root. We will then subtract 3 from the exponent (4) to get to a simple 4 inside, and a 4 outside.
![\sqrt[3]{4^{4} x^{10} y^{7} } = 4 \sqrt[3]{4 x^{10} y^{7} }](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B4%5E%7B4%7D%20x%5E%7B10%7D%20y%5E%7B7%7D%20%7D%20%3D%204%20%5Csqrt%5B3%5D%7B4%20x%5E%7B10%7D%20y%5E%7B7%7D%20%7D)
Now, we have x^10, so if we divide the exponent by the root factor, we get 10/3 = 3 1/3, which means we will extract x^9 that will become x^3 outside and x will remain inside.
![4 \sqrt[3]{4 x^{10} y^{7} } = 4x^{3} \sqrt[3]{4 x y^{7} }](https://tex.z-dn.net/?f=4%20%5Csqrt%5B3%5D%7B4%20x%5E%7B10%7D%20y%5E%7B7%7D%20%7D%20%3D%204x%5E%7B3%7D%20%5Csqrt%5B3%5D%7B4%20x%20y%5E%7B7%7D%20%7D)
For the y's we have y^7 inside the cubic root, that means the true exponent is y^(7/3)... so we can extract y^2 and 1 y will remain inside.
![4x^{3} \sqrt[3]{4 x y^{7} } = 4x^{3} y^{2} \sqrt[3]{4 x y}](https://tex.z-dn.net/?f=4x%5E%7B3%7D%20%5Csqrt%5B3%5D%7B4%20x%20y%5E%7B7%7D%20%7D%20%3D%204x%5E%7B3%7D%20y%5E%7B2%7D%20%5Csqrt%5B3%5D%7B4%20x%20y%7D)
The answer is then:
![4x^{3} y^{2} \sqrt[3]{4 x y} = 4x^{3} y^{2} (\sqrt[3]{4 x y})](https://tex.z-dn.net/?f=4x%5E%7B3%7D%20y%5E%7B2%7D%20%5Csqrt%5B3%5D%7B4%20x%20y%7D%20%3D%204x%5E%7B3%7D%20y%5E%7B2%7D%20%28%5Csqrt%5B3%5D%7B4%20x%20y%7D%29)
We will use demonstration of recurrences<span>1) for n=1, 10= 5*1(1+1)=5*2=10, it is just
2) assume that the equation </span>10 + 30 + 60 + ... + 10n = 5n(n + 1) is true, <span>for all positive integers n>=1
</span>3) let's show that the equation<span> is also true for n+1, n>=1
</span><span>10 + 30 + 60 + ... + 10(n+1) = 5(n+1)(n + 2)
</span>let be N=n+1, N is integer because of n+1, so we have
<span>10 + 30 + 60 + ... + 10N = 5N(N+1), it is true according 2)
</span>so the equation<span> is also true for n+1,
</span>finally, 10 + 30 + 60 + ... + 10n = 5n(n + 1) is always true for all positive integers n.
<span>
</span>
If the radius of the circle is 4 than the area would be 50.27.
If you are saying that 4 is the diameter, than divide that by 2 to get the radius which is 2. The area for this would be 12.57.
I don’t know which one you mean but hope this helps :))
Answer:
600
Step-by-step explanation:
Given,
Numbers are 825 and 213.
If the number are equal or greater than 50 then one number will be rounded up. if the number is less then it will rounded down.
Like in this case 25 and 13 are less than 50 so, they will not rounded up.
825 will be rounded to 800.
213 will be rounded to 200.
Subtraction of number
= 800 - 200 = 600
