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3 years ago

Not really sure how to do this one (ignore writing above)

Mathematics

1 answer:

vladimir2022 [97]3 years ago

4 0

$\bf \left( x^{4a-3} \right)^2=x^{10}\implies x^{(4a-3)2}=x^{10}
\\\\\\
\textit{since the bases are the same, the exponents must also be the same}
\\\\\\
2(4a-3)=10\implies 4a-3=\cfrac{10}{2}\implies 4a-3=5\implies 4a=8
\\\\\\
a=\cfrac{8}{4}\implies \boxed{a=2}$

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Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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A point of discontinuity refers to a particular point which is included in the simplified function, but which is not included in the original one. In this case, the point which is not included in the unsimplified function is at $x = 3$ . In the simplified version of the function, if we plug in $x = 3$ , we get

$y = \frac{1}{((3) - 9)} = -\frac{1}{6}$

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It's also important to distinguish between specific points of discontinuity and vertical asymptotes. This function also has a vertical asymptote at $x = 9$ (since it causes the denominator to be 0), but the difference in behavior is that in the case of the asymptote, only the denominator becomes 0 for a specific value of $x$

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Match the expressions with their equivalent simplified expressions.

Tasya [4]

Answer:

$\sqrt[4]{\frac{16x^6y^4}{81x^2y^8}}\rightarrow\frac{2x}{3y}\\\sqrt[4]{\frac{81x^2y^{10}}{81x^6y^6}} \rightarrow\frac{3y}{2x}\\\sqrt[3]{\frac{64x^8y^7}{125x^2y^{10}}}\rightarrow\frac{4x^2}{5y}\\\sqrt[5]{\frac{243x^{17}y^{16}}{32x^7y^{21}}}\rightarrow\frac{3x^2}{2y}\\\sqrt[5]{\frac{32x^{12}y^{15}}{243x^7y^{10}}} \rightarrow\frac{2xy}{3}\\\sqrt[4]{\frac{16x^{10}y^{9}}{256x^2y^{17}}}\rightarrow\frac{x}{2y}$

Step-by-step explanation:

$\sqrt[4]{\frac{16x^6y^4}{81x^2y^8}} =\sqrt[4]{\frac{(2^4)(x^{6-2})(y^{4-8})}{(3^4)}} =\sqrt[4]{\frac{2^4x^4y^{-4}}{3^4}} =\frac{2xy^{-1}}{3}=\frac{2x}{3y}$

$\sqrt[4]{\frac{81x^2y^{10}}{81x^6y^6}} =\sqrt[4]{\frac{(3^4)(x^{2-6})(y^{10-6})}{(2^4)}} =\sqrt[4]{\frac{3^4x^{-4}y^{4}}{2^4}} =\frac{3x^{-1}y^1}{3}=\frac{3y}{2x}$

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$\sqrt[5]{\frac{32x^{12}y^{15}}{243x^7y^{10}}} =\sqrt[5]{\frac{(2^5)(x^{12-7})(y^{15-10})}{(3^5)}} =\sqrt[5]{\frac{2^5x^{5}y^{5}}{3^5}} =\frac{2x^1y^{1}}{3}=\frac{2xy}{3}$

$\sqrt[4]{\frac{16x^{10}y^{9}}{256x^2y^{17}}} =\sqrt[4]{\frac{(2^4)(x^{10-2})(y^{9-17})}{(4^4)}} =\sqrt[4]{\frac{2^4x^{8}y^{-8}}{4^4}} =\frac{2x^{1}y^{-1}}{4}=\frac{x}{2y}$

Thus,

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3 years ago