<span>7745.96669241 is the square root...hope it helps :)</span>
Answer:
52
Step-by-step explanation:
2(4^2 + 10) =
We do what's in the parenthesis first.
4^2 = 16 and 16+10 =26.
2(26)
= 52
Given:
Consider the given expression is
To find:
The radical form of given expression.
Solution:
We have,
![(4x^3y^2)^{\frac{3}{10}}=\sqrt[5]{2^3}\sqrt[10]{x^9}\sqrt[5]{y^3}](https://tex.z-dn.net/?f=%284x%5E3y%5E2%29%5E%7B%5Cfrac%7B3%7D%7B10%7D%7D%3D%5Csqrt%5B5%5D%7B2%5E3%7D%5Csqrt%5B10%5D%7Bx%5E9%7D%5Csqrt%5B5%5D%7By%5E3%7D)
![[\because x^{\frac{1}{n}}=\sqrt[n]{x}]](https://tex.z-dn.net/?f=%5B%5Cbecause%20x%5E%7B%5Cfrac%7B1%7D%7Bn%7D%7D%3D%5Csqrt%5Bn%5D%7Bx%7D%5D)
![(4x^3y^2)^{\frac{3}{10}}=\sqrt[5]{8y^3}\sqrt[10]{x^9}](https://tex.z-dn.net/?f=%284x%5E3y%5E2%29%5E%7B%5Cfrac%7B3%7D%7B10%7D%7D%3D%5Csqrt%5B5%5D%7B8y%5E3%7D%5Csqrt%5B10%5D%7Bx%5E9%7D)
Therefore, the required radical form is ![\sqrt[5]{8y^3}\sqrt[10]{x^9}](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7B8y%5E3%7D%5Csqrt%5B10%5D%7Bx%5E9%7D)
.
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.
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