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olganol [36]
3 years ago
15

F(x)=4+5x+x2 how to identify vertex

Mathematics
2 answers:
MA_775_DIABLO [31]3 years ago
5 0
Attached the solution and step by step work.

chubhunter [2.5K]3 years ago
5 0
Please type "x^2" to indicate exponentiation.  Your function is

F(x) = 4 + 5x + x^2.  If written in "standard form," it's F(x) = x^2 + 5x + 4.

This has the configuration y = ax^2 + bx + c.  In your F(x), a=1, b=5 and c=4.

The x-coordinate of the vertex is x=-b/(2a).  Using a=1 and b=5, this x-coordinate is x= -5/(2[a]), or x = -5/2.

Find the y-coordinate of the vertex by plugging x=-5/2 into F(x)=x^2+5x+4:

F(-5/2) = (-5/2)^2 + 5(-5/2) + 4 = (25/4) - (25/2) + 4.

Simplifying this, using the LCD, we get (25/4) - (50/4) + (16/4).

Then F(-5/2) = -9/4.

The vertex is at (-5/2, -9/4).

There are other ways to find the vertex.  Let me know if you want to discuss this problem further.
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nevsk [136]

Answer:

10 i think because I added them i don't know though I hope it work sorry

6 0
3 years ago
Find \(\int \dfrac{x}{\sqrt{1-x^4}}\) Please, help
ki77a [65]
If you're using the app, try seeing this answer through your browser:  brainly.com/question/2867785

_______________


Evaluate the indefinite integral:

\mathsf{\displaystyle\int\! \frac{x}{\sqrt{1-x^4}}\,dx}\\\\\\ \mathsf{=\displaystyle\int\! \frac{1}{2}\cdot 2\cdot \frac{1}{\sqrt{1-(x^2)^2}}\,dx}\\\\\\ \mathsf{=\displaystyle \frac{1}{2}\int\! \frac{1}{\sqrt{1-(x^2)^2}}\cdot 2x\,dx\qquad\quad(i)}


Make a trigonometric substitution:

\begin{array}{lcl}
\mathsf{x^2=sin\,t}&\quad\Rightarrow\quad&\mathsf{2x\,dx=cos\,t\,dt}\\\\
&&\mathsf{t=arcsin(x^2)\,,\qquad 0\ \textless \ x\ \textless \ \frac{\pi}{2}}\end{array}


so the integral (i) becomes

\mathsf{=\displaystyle\frac{1}{2}\int\!\frac{1}{\sqrt{1-sin^2\,t}}\cdot cos\,t\,dt\qquad\quad (but~1-sin^2\,t=cos^2\,t)}\\\\\\
\mathsf{=\displaystyle\frac{1}{2}\int\!\frac{1}{\sqrt{cos^2\,t}}\cdot cos\,t\,dt}

\mathsf{=\displaystyle\frac{1}{2}\int\!\frac{1}{cos\,t}\cdot cos\,t\,dt}\\\\\\
\mathsf{=\displaystyle\frac{1}{2}\int\!\f dt}\\\\\\
\mathsf{=\displaystyle\frac{1}{2}\,t+C}


Now, substitute back for t = arcsin(x²), and you finally get the result:

\mathsf{\displaystyle\int\! \frac{x}{\sqrt{1-(x^2)^2}}\,dx=\frac{1}{2}\,arcsin(x^2)+C}          ✔

________


You could also make

x² = cos t

and you would get this expression for the integral:

\mathsf{\displaystyle\int\! \frac{x}{\sqrt{1-(x^2)^2}}\,dx=-\,\frac{1}{2}\,arccos(x^2)+C_2}          ✔


which is fine, because those two functions have the same derivative, as the difference between them is a constant:

\mathsf{\dfrac{1}{2}\,arcsin(x^2)-\left(-\dfrac{1}{2}\,arccos(x^2)\right)}\\\\\\
=\mathsf{\dfrac{1}{2}\,arcsin(x^2)+\dfrac{1}{2}\,arccos(x^2)}\\\\\\
=\mathsf{\dfrac{1}{2}\cdot \left[\,arcsin(x^2)+arccos(x^2)\right]}\\\\\\
=\mathsf{\dfrac{1}{2}\cdot \dfrac{\pi}{2}}

\mathsf{=\dfrac{\pi}{4}}         ✔


and that constant does not interfer in the differentiation process, because the derivative of a constant is zero.


I hope this helps. =)

6 0
3 years ago
Hey can you please help me posted picture of question
zavuch27 [327]
A relation can be a function if an x value is not paired with more than y-values.
The x and y values are flipped when finding an inverse.

So, if in the inverse function a y value is paired with more than one x values, this will mean that in the function an x-value was paired with more than one y-values.

Looking at the graph we can say that no y-value in the inverse is paired with more than one x values, so the original relation would be a function.

So the answer is TRUE
4 0
3 years ago
PLS HELP ASAP I SUCK AT THIS PLSS
Ilia_Sergeevich [38]

Answer:

60

Step-by-step explanation:

60*0.45=27

5 0
3 years ago
Geometry question to verify it​
Oksana_A [137]

Step-by-step explanation:

here's the solution,

in the given figure , sum of all angles formed with O measures 360°

because, it forms a complete angle

so,

=》mPOQ + mQOR + mROS + mSOT + mTOP = 360°

=》mPOQ + mQOR + mROS + mSOT + mTOP = (90° × 4)

=》mPOQ + mQOR + mROS + mSOT + mTOP = 4 × right angle

(cuz.. right angle = 90°)

4 0
3 years ago
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