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Alinara [238K]
3 years ago
10

HELP ME PLEASE FAST NOW PLEASE!!! ASAP

Mathematics
2 answers:
Juliette [100K]3 years ago
6 0
Range is greater for the 13-14 year olds.
Yuliya22 [10]3 years ago
4 0
Range is greater for the children 13 -14
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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
3 years ago
Is the distribution a discrete probability​ distribution?
Tanzania [10]
N fknvnab;nkfanlvkbdsljnv.jrehv;iavdsirhkebnkljarwnfbnkvbw<em>n;kjnv;kjnrk</em>
3 0
3 years ago
What is the answer of square root of -24 in complex numbers
Sergio039 [100]

With \sqrt{-1}=i, we have

\sqrt{-24}=\sqrt{(-1)(24)}=\sqrt{-1}\sqrt{24}=i\sqrt{24}

Since 24=2^3\cdot3, we can further write this as

i\sqrt{24}=i\sqrt{2^2\cdot6}=2i\sqrt6

3 0
3 years ago
Simplify.<br><br> 5 - {-(-2)}<br><br> A)-3<br> B) 3<br> C)-7<br> D) 7
Fed [463]
<span>5 - {-(-2)}
</span>5 - {+2} (negative + negative = positive)
<span>5 - 2 (negative + positive = negative)
</span>3
5 0
3 years ago
Solve the inequality to - 1/3k &gt;= 2
Xelga [282]

Answer;k≤−6

Step-by-step explanation:(

3

−1

)*(

−1

3

k)≥(

3

−1

)*(2)=

k≤−6

5 0
2 years ago
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