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kumpel [21]
1 year ago
10

Rewrite in simplest rational exponent form

exFormula1" title="\sqrt{x} * \sqrt[4]{x}" alt="\sqrt{x} * \sqrt[4]{x}" align="absmiddle" class="latex-formula"> Show each step of your process.
Mathematics
1 answer:
Shkiper50 [21]1 year ago
5 0

The given expression written in the simplest rational exponent form is x^{\frac{3}{4} }

<h3>Writing an Expression in Exponent form</h3>

From the question, we are to write the given expression in exponent form

The given expression is

\sqrt{x}  \times \sqrt[4]{x}

In exponent form,

\sqrt{x}  = x^{\frac{1}{2} }

and

\sqrt[4]{x} = x^{\frac{1}{4} }

Thus,  \sqrt{x}  \times \sqrt[4]{x} becomes

x^{\frac{1}{2} } \times x^{\frac{1}{4} }

Applying the multiplication law of indices, we get

x^{\frac{1}{2} + \frac{1}{4} }

= x^{\frac{3}{4} }

Hence, the given expression written in the simplest rational exponent form is x^{\frac{3}{4} }

Learn more on Writing an expression in exponent form here: brainly.com/question/4421494

#SPJ1

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myrzilka [38]

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y intercept ( 0, 0.4)

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4 0
3 years ago
A sine function had an amplitude of 3, period of 6pi, horizontal shift of 3pi/2, &amp; vertical shift of -1.
Simora [160]

Answer: \bold{y=\dfrac{1}{2}}

<u>Step-by-step explanation:</u>

f(x) = A sin (Bx - C) + D

  • amplitude = |A|
  • period =\dfrac{2\pi}{B}
  • phase shift =\dfrac{C}{B}
  • vertical shift = D

<u>A</u>

amplitude of 3 is given so  3 = |A| → A = ± 3, since it is stated that this is a positive function, then A = 3

<u>B</u>

period of 6π is given so 6\pi=\dfrac{2\pi}{B}\quad \rightarrow \quad B=\dfrac{2\pi}{6\pi}\quad \rightarrow \quad B=\dfrac{1}{3}

<u>C</u>

\text{phase shift is given as}\ \dfrac{3\pi}{2}\ \text{so}\ \dfrac{3\pi}{2}=\dfrac{C}{\frac{1}{3}}\quad \rightarrow\quad \dfrac{(\frac{1}{3})3\pi}{2}=C\quad \rightarrow\quad \dfrac{\pi}{2}=C

<u>D</u>

vertical shift of -1 is given so -1 = D


Now, substitute the values of A, B, C, and D into the formula (above):

f(x) = 3\ sin \bigg(\dfrac{1}{3}x - \dfrac{\pi}{2}\bigg) - 1


Next, solve when x = 2π

f(2\pi) = 3\ sin \bigg(\dfrac{1}{3}(2\pi) - \dfrac{\pi}{2}\bigg) - 1

        = 3\ sin \bigg(\dfrac{2\pi}{3} - \dfrac{\pi}{2}\bigg) - 1

        = 3\ sin \bigg(\dfrac{4\pi}{6} - \dfrac{3\pi}{6}\bigg) - 1

        = 3\ sin \bigg(\dfrac{\pi}{6}\bigg) - 1

        = 3\ \bigg(\dfrac{1}{2}\bigg) - 1

        =\dfrac{3}{2}-\dfrac{2}{2}

        =\dfrac{1}{2}

6 0
3 years ago
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