How to right a radical in exponential form

Mathematics

1 answer:

Law Incorporation [45]3 years ago

3 0

Answer:

$\sqrt[n]{x}=x^{\frac{1}{n}}$

Step-by-step explanation:

The index of a radical is the denominator of a fractional exponent, and vice versa. If you think about the rules of exponents, you know this must be so.

For example, consider the cube root:

$\sqrt[3]{x}\cdot \sqrt[3]{x}\cdot \sqrt[3]{x}=(\sqrt[3]{x})^3=x\\\\(x^{\frac{1}{3}})^3=x^{\frac{3}{3}}=x^1=x$

That is ...

$\sqrt[3]{x}=x^{\frac{1}{3}} \quad\text{radical index = fraction denominator}$

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Which parent function is represented by the table?

polet [3.4K]

Answer:

x^2

Step-by-step explanation:

So there's very two obvious ones we can elimination immediately, that's f(x) = x, and f(x) = |x|. The reason for this is because we can just look at the first ordered pair of (-2, 4). If it was f(x) = x, then the ordered pair would be (-2, -2), but it isn't. Very similar reasoning for f(x) = |x|, except for this function each ordered pair should have the same magnitude or absolute value. So the f(-2) should output 2, since it's the absolute value, but of course this isn't the case since it's 4.

So let's look at 2^x

$2^{-2} = \frac{1}{2^2} = \frac{1}{4}$