The coordinates of the vertices of quadrilateral HIJK are H (-2, -1), I (2, 1), J (5, 0), and K (-1, -3). Isabella states that q
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Answer:
Step-by-step explanation:
The original equation is . We propose that the solution of this equations is of the form
. Then, by replacing the derivatives we get the following
Since we want a non trival solution, it must happen that A is different from zero. Also, the exponential function is always positive, then it must happen that
Recall that the roots of a polynomial of the form are given by the formula
In our case a = 121, b = 110 and c = -24. Using the formula we get the solutions
So, in this case, the general solution is
a) In the first case, we are given that y(0) = 1 and y'(0) = 0. By differentiating the general solution and replacing t by 0 we get the equations
(or equivalently
By replacing the second equation in the first one, we get which implies that
.
So
b) By using y(0) =0 and y'(0)=1 we get the equations
(or equivalently
By solving this system, the solution is
Then
c)
The Wronskian of the solutions is calculated as the determinant of the following matrix
By plugging the values of and
We can check this by using Abel's theorem. Given a second degree differential equation of the form y''+p(x)y'+q(x)y the wronskian is given by
In this case, by dividing the equation by 121 we get that p(x) = 10/11. So the wronskian is
Note that this function is always positive, and thus, never zero. So is a fundamental set of solutions.
Answer:
Step-by-step explanation:
<u>Angles in a Circle</u>
An exterior angle of a circle is an angle whose vertex is outside a circle and the sides of the angle are secants or tangents of the circle.
Segments AE and DE are secants of the given circle. They form an exterior angle called AED.
The measure of an exterior angle is equal to half the difference of the measure of their intercepted arcs.
Intercepted arcs in the given circle are AD=113° and BC=48°. The exterior angle is: