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2 years ago

10

4 ^-3 rewrite without an exponent

2 answers:

Sergeeva-Olga [200]2 years ago

6 0

Answer:

$\frac{1}{64}$

Step-by-step explanation:

Negative exponents means you bring the base to the denominator, which becomes a fraction.

$4^{-3} \\=\frac{1}{4^{3}}$

The question wants us to write without an exponent.

The $4^{3} = 4 * 4* 4=64$

So, it becomes:

$\frac{1}{64}$

hichkok12 [17]2 years ago

5 0

Answer:

(1/4)³

Step-by-step explanation:

Hope it helps you

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1 year ago

Read 2 more answers

D<br> Evaluate<br> arcsin<br> (6)]<br> at x = 4.<br> dx

sineoko [7]

Answer:

$\frac{1}{2\sqrt{5} }$

Step-by-step explanation:

Let, $\text{sin}^{-1}(\frac{x}{6})$ = y

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$\frac{1}{6}=\text{cos}(y)\frac{dy}{dx}$

$\frac{dy}{dx}=\frac{1}{6\text{cos(y)}}$

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= $\sqrt{1-(\frac{x}{6})^2}$

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Therefore, from equation (1),

$\frac{dy}{dx}=\frac{1}{6\sqrt{1-\frac{x^2}{36}}}$

Or $\frac{d}{dx}[\text{sin}^{-1}(\frac{x}{6})]=\frac{1}{6\sqrt{1-\frac{x^2}{36}}}$

At x = 4,

$\frac{d}{dx}[\text{sin}^{-1}(\frac{4}{6})]=\frac{1}{6\sqrt{1-\frac{4^2}{36}}}$

$\frac{d}{dx}[\text{sin}^{-1}(\frac{2}{3})]=\frac{1}{6\sqrt{1-\frac{16}{36}}}$

$=\frac{1}{6\sqrt{\frac{36-16}{36}}}$

$=\frac{1}{6\sqrt{\frac{20}{36} }}$

$=\frac{1}{\sqrt{20}}$

$=\frac{1}{2\sqrt{5}}$

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