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svetlana [45]
3 years ago
6

The quadratic function modeling the height of a ball over time is symmetric about the line , where t is time in seconds. Which s

tatement is true about this situation?
A.

The height of the ball is the same after 1 second and 3 seconds.

B.

The height of the ball is the same after 0 seconds and 4 seconds.

C.

The height of the ball is the same after 0.5 second and 5.5 seconds.

D.

The height of the ball is the same after 1.5 seconds and 3.5 seconds.
Mathematics
1 answer:
Rashid [163]3 years ago
3 0

Answer:

A. The height of the ball is the same after 1 second and 3 seconds.

Step-by-step explanation:

The quadratic equation function modeling is used to analyze the relationship between variables in the equation. If the ball is at height and t is time then the ball is on symmetric line which will move for initial 1 second then stabilizes at 3 seconds.

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Write the following expression using a single exponent. 3-5 × 36 A. 30 B. 3-11 C. 31 D. 3-30
fiasKO [112]

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3^1

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3/8 divided by -0.25​
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Step-by-step explanation:

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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
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