$\text{The vertex form of a quadratic equation:}\\\\y=a(x-h)^2+k\\\\(h,\ k)-vertex\\---------------\\\text{We have:}\\\\y=-\dfrac{1}{3}(x+2)^2+5=-\dfrac{1}{3}(x-(-2))^2+5\Rightarrow h=-2\ \text{and}\ k=5\\\\\text{therefore the vertex is:}\ (-2,\ 5)$

6 0

3 years ago

MODEL HOUSE Baron built a square pyramid block to use as the roof of a model house he was making. The square base had sides of l

ololo11 [35]

Answer:

I think the answer is 14, but rounded to the nearest tenth will be 10.......

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

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

Tell whether -2 is a solution of the inequality n-5<8

ruslelena [56]

Answer:

-2 is a solution

Step-by-step explanation:

-2-5<8

-7<8

6 0

3 years ago