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xxMikexx [17]
3 years ago
15

(x+7)/(x^2-49) find the domain. show work

Mathematics
1 answer:
harina [27]3 years ago
7 0
\large\begin{array}{l} \textsf{Find the domain of}\\\\ \mathsf{f(x)=\dfrac{x+7}{x^2-49}}\\\\ \mathsf{f(x)=\dfrac{x+7}{x^2-7^2}}\\\\\\ \textsf{Factor out the denominator using special products:}\\\\ \textsf{(a difference of squares)}\\\\ \mathsf{f(x)=\dfrac{x+7}{(x+7)(x-7)}} \end{array}


\large\begin{array}{l} \textsf{Restrictions for the domain:}\\\\ \bullet~~\textsf{Denominators must not be zero:}\\\\ \mathsf{(x+7)(x-7)\ne 0}\\\\ \begin{array}{rcl} \mathsf{x+7\ne 0}&~\textsf{ and }~&\mathsf{x-7\ne 0}\\\\ \mathsf{x\ne -7}&~\textsf{ and }~&\mathsf{x\ne 7} \end{array} \end{array}


\large\begin{array}{l} \textsf{Therefore, the domain of f is}\\\\ \mathsf{D_f=\{x\in\mathbb{R}:~~x\ne -7~~and~~x\ne 7\}}\\\\\\ \textsf{or using a more compact form}\\\\ \mathsf{D_f=\mathbb{R}\setminus\{-7,\,7\}}\\\\\\ \textsf{or using the interval notation}\\\\ \mathsf{D_f=\left]-\infty,\,-7\right[\,\cup\,\left]7,\,+\infty\right[.} \end{array}


<span>If you're having problems understanding this answer, try seeing it through your browser: brainly.com/question/2155752


\large\textsf{I hope it helps. :-)}



Tags: <em>function domain real rational factorizing special product interval</em>

</span>
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given y&gt;0 and dy/dx=(3x^2+4x)/yif the point (1,√10) is on the graph relating x and y, then what is y when x=0
jonny [76]
\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{3x^2+4x}y\iff y\,\mathrm dy=(3x^2+4x)\,\mathrm dx
\implies\displaystyle\int y\,\mathrm dy=\int(3x^2+4x)\,\mathrm dx
\implies \dfrac12y^2=x^3+2x^2+C

When x=1 you have y=\sqrt{10}, so

\dfrac12(\sqrt{10})^2=1^3+2(1)^2+C\implies 5=1+2+C\implies C=2

and so the particular solution to the ODE is

\dfrac12y^2=x^3+2x^2+2

Then when x=0, you get

\dfrac12y^2=0^3+2(0)^2+2=2
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where we omitted the negative root because it's given that y>0.
4 0
3 years ago
Solve the following systems by substitution
Paladinen [302]

Answer:

The answer to your question is below

Step-by-step explanation:

Question 1

                x = 5                       Equation l

                2x + y = 10              Equation ll

- Substitute Equation l in equation ll

               2(5) + y = 10

                y = 10 - 10

                y = 0

- Solution  (5, 0)

Question 2

                 x + 16y = 20          Equation l

                 x = 4y                    Equation ll

Substitute equation ll in equation l

                4y + 16y = 20

                        20y = 20

                           y = 20/20

                           y = 1

-Find x

                x = 4(1)

                x = 4

-Solution

                (4, 1)

Question 3

                   2x + 8y = 20            Equation l

                     x = 2                       Equation ll

-Substitute equation ll in equation l

                   2(2) + 8y = 20

                      4 + 8y = 20

                            8y = 20 - 4

                            8y = 16

                              y = 16/8

                              y = 2

- Solution

                   (2, 2)              

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fiasKO [112]
First of all, just to avoid being snookered by a trick question, we should verify that these are really right triangles:

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In the second one:
sin(one acute angle) = 8/17 = 0.4706...
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I'm sorry, but just now, I don't know how to do the 
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Ludmilka [50]

Answer:

A.

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ANEK [815]

Answer:

<h2><u>Answer</u><u> </u><u>:</u><u>-</u><u> </u></h2>

Option D :- (180 - 2x)°

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