What is the simplified form of (cube root of)16

1 answer:

5 0

Remember that $\sqrt[n]{x^m}= x^{ \frac{n}{m} }$ so

∛16=16^(1/3)
we know that
16=2^3 so
(2^3)^(1/3)
remember another rule
(x^m)^n=x^(mn) so
(2^3)^(1/3)=2^(3/3)=2^1=2

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Solve. 2a + 3b = 5 6= a -5 a = 4 b = -1 1 a = 6 b=1 a=6 6 = -1 a=4 6 = 1<br>

slamgirl [31]

Answer:Solve. 2a + 3b = 5 6= a -5 a = 4 b = -1 1 a = 6 b=1 a=6 6 = -1 a=4 6 = 1

Step-by-step explanation:

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3 years ago

What is -5/6 + -5/6? (its ment to be a fraction)

maks197457 [2]

The answer is: " -5/3 " ; or, write as: " -1 ⅔ " .
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Explanation:
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(-5/6) + (-5/6) = (-5/6) <span>− (5/6) ;

------> {since: "adding a negative" is the same a "subtracting a positive"} ;

------> </span> (-5/6) − (5/6) = (-5 − 5) / 6 ;

= -10/6 = (-10/2) / (6/2) ;
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= " -5/3 " ; or, write as: " -1 ⅔ " .
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4 years ago

Find derivative problem<br> Find B’(6)

dalvyx [7]

Answer:

$B^\prime(6) \approx -28.17$

Step-by-step explanation:

We have:

$\displaystyle B(t)=24.6\sin(\frac{\pi t}{10})(8-t)$

And we want to find B’(6).

So, we will need to find B(t) first. To do so, we will take the derivative of both sides with respect to x. Hence:

$\displaystyle B^\prime(t)=\frac{d}{dt}[24.6\sin(\frac{\pi t}{10})(8-t)]$

We can move the constant outside:

$\displaystyle B^\prime(t)=24.6\frac{d}{dt}[\sin(\frac{\pi t}{10})(8-t)]$

Now, we will utilize the product rule. The product rule is:

$(uv)^\prime=u^\prime v+u v^\prime$

We will let:

$\displaystyle u=\sin(\frac{\pi t}{10})\text{ and } \\ \\ v=8-t$

Then:

$\displaystyle u^\prime=\frac{\pi}{10}\cos(\frac{\pi t}{10})\text{ and } \\ \\ v^\prime= -1$

(The derivative of u was determined using the chain rule.)

Then it follows that:

$\displaystyle \begin{aligned} B^\prime(t)&=24.6\frac{d}{dt}[\sin(\frac{\pi t}{10})(8-t)] \\ \\ &=24.6[(\frac{\pi}{10}\cos(\frac{\pi t}{10}))(8-t) - \sin(\frac{\pi t}{10})] \end{aligned}$