find the smallest number by which 29160 should be divided so that the quotient becomes a perfect cube

Mathematics

1 answer:

IgorLugansk [536]2 years ago

7 0

The smallest number by which 29160 should be divided so that the quotient becomes a perfect cube is 5

<h3>Perfect cube quotient of a division</h3>

The given number is 29160

Note that:

29160 can be factored into 5832

That is, 29160 = 5832 x 5

Take the cube of 5832:

$\sqrt[3]{5832}=18$

Since it has been confirmed that 5832 is a perfect cube, the smallest number by which 29160 should be divided so that the quotient becomes a perfect cube is 5

Learn more on perfect cube quotient here: brainly.com/question/17448146

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Find the x-coordinates of any relative extrema and inflection point(s) for the function f(x) = 6x(1/3) + 3x(4/3). You must justi

stealth61 [152]

Applying our power rule gets us our first derivative,

$\rm f'(x)=6\frac13x^{-2/3}+3\cdot\frac43x^{1/3}$

simplifying a little bit,

$\rm f'(x)=2x^{-2/3}+4x^{1/3}$

looking for critical points,

$\rm 0=2x^{-2/3}+4x^{1/3}$

We can apply more factoring.
I hope this next step isn't too confusing.
We want to factor out the smallest power of x from both terms,
and also the 2 from each.

$0=2x^{-2/3}\left(1+2x\right)$

When you divide x^(-2/3) out of x^(1/3),
it leaves you with x^(3/3) or simply x.

Then apply your Zero-Factor Property,

$\rm 0=2x^{-2/3}\qquad\qquad\qquad 0=(1+2x)$

and solve for x in each case to find your critical points.

Apply your First Derivative Test to further classify these points. You should end up finding that x=-1/2 is an relative extreme value, while x=0 is not.

Let's come back to this,

$\rm f'(x)=2x^{-2/3}+4x^{1/3}$

and take our second derivative.

$\rm f''(x)=-\frac43x^{-5/3}+\frac43x^{-2/3}$

Looking for inflection points,

$\rm 0=-\frac43x^{-5/3}+\frac43x^{-2/3}$

Again, pulling out the smaller power of x, and fractional part,

$\rm 0=-\frac43x^{-5/3}\left(1-x\right)$

And again, apply your Zero-Factor Property, setting each factor to zero and solving for x in each case. You should find that x=0 and x=1 are possible inflection points.

Applying your Second Derivative Test should verify that both points are in fact inflection points, locations where the function changes concavity.

8 0

4 years ago

In need help in this

Lyrx [107]

Answer:

Vertex form is

$a(x - h) {}^{2} + k = y$