<h3>
Answer: E) x^5</h3>
![\sqrt{x^{10}} = x^5](https://tex.z-dn.net/?f=%5Csqrt%7Bx%5E%7B10%7D%7D%20%3D%20x%5E5)
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Explanation:
We simply take half of the exponent 10 to get 5. This applies to square roots only.
So the rule is ![\sqrt{a^b} = a^{b/2}](https://tex.z-dn.net/?f=%5Csqrt%7Ba%5Eb%7D%20%3D%20a%5E%7Bb%2F2%7D)
A more general rule is
![\sqrt[n]{a^b} = a^{b/n}](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Ba%5Eb%7D%20%3D%20a%5E%7Bb%2Fn%7D)
If n = 2, then we're dealing with square roots like with this problem. In this case, a = x and b = 10.
Answer:
A=πr²=π·3²≈28.27433
To find the area of a circle, use the equation πr²
The image is a bit blurry so from what I saw r (radius) is 3 otherwise you can use this equation if r is not 3.
If π was replaced with 3.14 the answer is 28.26 ft².
Answer:
<h2>
x₁ = - 2 + √2 , x₂ = - 2 - √2</h2>
Step-by-step explanation:
![3x^2 + 12x + 6 = 0\\\\a=3\,,\ b=12\,,\ c=6\\\\\\ x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\ x=\dfrac{-12\pm\sqrt{12^2-4\cdot3\cdot6}}{2\cdot3}=\dfrac{-12\pm\sqrt{144-72}}{2\cdot3}=\dfrac{-12\pm\sqrt{72}}{2\cdot3}=\dfrac{-12\pm6\sqrt{2}}{6}\\\\x_1=\dfrac{6(-2+\sqrt{2})}{6}=-2+\sqrt2\\\\x_2=\dfrac{6(-2-\sqrt{2})}{6}=-2-\sqrt2](https://tex.z-dn.net/?f=3x%5E2%20%2B%2012x%20%2B%206%20%3D%200%5C%5C%5C%5Ca%3D3%5C%2C%2C%5C%20b%3D12%5C%2C%2C%5C%20c%3D6%5C%5C%5C%5C%5C%5C%20x%3D%5Cdfrac%7B-b%5Cpm%5Csqrt%7Bb%5E2-4ac%7D%7D%7B2a%7D%5C%5C%5C%5C%20x%3D%5Cdfrac%7B-12%5Cpm%5Csqrt%7B12%5E2-4%5Ccdot3%5Ccdot6%7D%7D%7B2%5Ccdot3%7D%3D%5Cdfrac%7B-12%5Cpm%5Csqrt%7B144-72%7D%7D%7B2%5Ccdot3%7D%3D%5Cdfrac%7B-12%5Cpm%5Csqrt%7B72%7D%7D%7B2%5Ccdot3%7D%3D%5Cdfrac%7B-12%5Cpm6%5Csqrt%7B2%7D%7D%7B6%7D%5C%5C%5C%5Cx_1%3D%5Cdfrac%7B6%28-2%2B%5Csqrt%7B2%7D%29%7D%7B6%7D%3D-2%2B%5Csqrt2%5C%5C%5C%5Cx_2%3D%5Cdfrac%7B6%28-2-%5Csqrt%7B2%7D%29%7D%7B6%7D%3D-2-%5Csqrt2)
√18 ÷ √12 = 1.22474487139159