nineScherlenzi
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Assuming x > 0, which of these expressions is equivalent to 11 times the square root of 245 x to the third plus 9 times the square root of 45 x to the third? ?

5 x times the square root of 104 x

20 times the square root of 290 x to the sixth

20 x times the square root of 290 x

104 x times the square root of 5 x

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JcAlmighty
The answer is 104 x times the square root of 5 x 

11 times the square root of 245 x to the third plus 9 times the square root of 45 x to the third is:
[tex]11 \sqrt{245 x^{3} } +9 \sqrt{45 x^{3} } =11 \sqrt{245} *\sqrt{ x^{3} } +9 \sqrt{45} *\sqrt{ x^{3} }= \\ \\ =11 \sqrt{5*49} * \sqrt{ x^{2} *x} +9 \sqrt{5*9} * \sqrt{ x^{2} *x}= \\ \\ =11* \sqrt{5} * \sqrt{49} * \sqrt{ x^{2}} * \sqrt{x} +9* \sqrt{5} * \sqrt{9} * \sqrt{ x^{2} }* \sqrt{x} = \\ \\ =11* \sqrt{5} *7*x* \sqrt{x} +9* \sqrt{5} *3*x* \sqrt{x} = \\ \\ =77x* \sqrt{5*x} +27x* \sqrt{5*x} = \\ \\ =77x \sqrt{5x} +27x \sqrt{5x} = \\ \\ =104x \sqrt{5x} [/tex]
scme1702
Writing the expression mathematically:
11√(245x³)  +  9√(45x³)
11√(7² * 5 * x² * x) + 9√(3² * 5 * x² * x)
77x√(5x) + 27x√(5x)
104x√(5x)

104x times the square root of 5x
The last option is correct.

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