In the

ASTRONOMY!!!

Helen [10]

Answer:

I think it's

B. UV radiation from the Sun causes overheating of transformers.

Sorry if i'm wrong.

4 0

3 years ago

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

Triangle EFG has coordinates E (-4,4), F (-4,-2), and G (2,-2). Triangle JKL will be a dilation of triangle EFG with a scale fac

denis-greek [22]

The point L is plotted at A. (1, -1).

Step-by-step explanation:

Step 1:

First, we plot the coordinates of both triangles.

The coordinates of the triangle EFG are as follows;

E = (-4, 4), F = (-4, -2), and G = (2, -2).

The coordinates of triangle JKL are;

J = (-2, 2), K = (-2, -1), and L = (x, y).

Step 2;

Next, we need to find the distances between at least two coordinates for both triangles.

The distances between vertical lines are found by calculating the difference in y values while for horizontal lines, the distances are found by calculating the difference between the x values.

The distance of EF $= 4-(-2) =6.$