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stepladder [879]
3 years ago
10

Please so your work! 20pts and bainlyist for best answer! an = 2n + 11

Mathematics
1 answer:
miv72 [106K]3 years ago
5 0

Step-by-step explanation:

What's the value for a? This cannot be solved unless 'a' or 'n' is given

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Whats the solve for x?
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For the differential equations dydx=sqrt(y^2−36) does the existence/uniqueness theorem guarantee that there is a solution to thi
julsineya [31]

Answer:

1. (-4,6) there is no a solution to the equation through this point

2. (2,−6) there is no a solution to the equation through this point

3. (−5,39) there is a solution to the equation through this point

4. (−1,45)  there is a solution to the equation through this point

Step-by-step explanation:

Using the existence and uniqueness theorem:

Let:

F(x,y)=\sqrt{y^2-36} \\\\and\\\\\frac{\partial F}{\partial y} =\frac{y}{\sqrt{y^2-36} }

Now, let's find the domain of F(x,y), due to the square root:

y^2-36 \geq 0\\\\y^2\geq 36\\\\y \geq 6\hspace{12}or\hspace{12}y\leq-6

So the domain of the function is:

y \in R\hspace{12}y\geq6\hspace{12}or\hspace{12}y\leq-6

Now, due to the fraction \frac{\partial F}{\partial y} the denominator must be also different from 0, so:

y^2-36\neq0\\\\y \neq \pm6

So, the theorem  tells us that for each y_0\in R:\hspace{12}y_0>6\hspace{12}or\hspace{12}y_0 there exists a  unique solution defined in an open interval around x_0.

1. (-4,6)  there is no a solution to the equation through this point because y_0=6

2. (2,−6)  there is no a solution to the equation through this point because

y_0=-6

3. (−5,39) there is a solution to the equation through this point because

y_0>6

4. (−1,45)  there is a solution to the equation through this point because

y_0>6

6 0
3 years ago
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