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ddd [48]
3 years ago
15

Please help?? Thank you if you help me

Mathematics
1 answer:
Luden [163]3 years ago
7 0
Y2-y1 over x2-x1 or you can go onto the calculator and push list and put  in the points and it will give you the answer of 68

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Identify the interval where y = x3 + 3x2 - 2x + 4 is decreasing.
VladimirAG [237]
Please be certain to use the symbol "^" to denote exponentiation.  Your function y should be written as

y = x^3 + 3x2 - 2x + 4.

Principle:  if the derivative of a function is negative on a certain x-interval, the function is decreasing on that interval.      Thus, you must differentiate the given function:  find dy/dx.  Set this dy/dx = to 0 and solve the resulting equation for x.  Set up intervals on the # line based upon your solutions.
For example, if x=-1 and x=2, then set up 3 intervals:  (-infinity,-1), (-1,2), and (2, infinity).  Looking at each interval separately, identify the interval or intervals on which the derivative dy/dx is negative.  

This is a critically important skill in calculus and well worth the time and effort required to learn it.
3 0
3 years ago
Which ratio forms a proportion with 25/35
Art [367]
5/7 is your answer

25/5 = 5
35/5 = 35

5:7 is proportionate to 25:35

hope this helps
4 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
Four of the seven students are from Middle Georgia State College. What is the probability that both of the interviewed students
Furkat [3]

The required value is 0.286 and \dfrac{2}{7}

Step-by-step explanation:

Since we have given that

Number of students = 7

Number of students are from Middle Georgia State College = 4

So, Probability that both of the interviewed students are from Middle Georgia State College is given by

\dfrac{4}{7}\times \dfrac{3}{6}\\\\=\dfrac{4}{7}\times \dfrac{1}{2}\\\\=\dfrac{2}{7}\\\\=0.286

Hence, the required value is 0.286 and \dfrac{2}{7}

3 0
3 years ago
Please help in confused
Darya [45]

Answer:

the first answer is correct and brainliest plsss (^_^)

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
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