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3 years ago

Can you help me with this problem, please

Mathematics

1 answer:

DanielleElmas [232]3 years ago

5 0

Answer:

$\dfrac{2x}{3y}\sqrt{3y}$

Step-by-step explanation:

When we simplify a radical, we want to remove any squares from under the radical. Often, we also want to eliminate any fractions under the radical.

Here, that means we need to multiply by the denominator so as to make it a perfect square, then remove the square from under the radical.

$\sqrt{\dfrac{4x^2}{3y}}=\sqrt{\dfrac{(2x)^2(3y)}{(3y)(3y)}}=\sqrt{\dfrac{(2x)^2(3y)}{(3y)^2}}\\\\=\dfrac{2x}{3y}\sqrt{3y}$

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The domain and the range of a function are the set of input and output values, the function can take.

The domain and the range of $g(x) = 9x^2 - 2$ is $(-\infty, \infty)$ .
The parent function $f(x) =x^2$ is vertically compressed by 9, then shifted down by 5 units to get $g(x) = 9x^2 - 2$

Given

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Domain and range

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This means that the domain and the range are $(-\infty, \infty)$

$(-\infty, \infty)$ is in interval notation

The corresponding set notation is: $- \infty < x < \infty$

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The rule of this transformation is:

$(x,y) \to (x,9y)$

So, we have:

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Next, the function is shifted down by 5 units.

So, we have:

$g(x) = f'(x) - 5$

$g(x) = 9x^2 - 5$

Read more about functions at:

brainly.com/question/21027387

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