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DanielleElmas [232]
3 years ago
7

(1 point) Find the general solution to the homogeneous differential equation. ????2y????????2−20????y????????+136y=0 Use c1 and

c2 in your answer to denote arbitrary constants, and enter them as c1 and c2. y(????)= ?
Mathematics
2 answers:
adell [148]3 years ago
5 0

Answer:

Question is not clear please post question clearly lots of question marks.

gregori [183]3 years ago
3 0

Your differential equation is not displayed well. It though looks like this:

2d²y/dx² - 20dy/dx + 136y = 0

If this is not the differential equation, the method of solving this would still be used in solving the correct one.

We first write an auxiliary equation to the differential equation.

The auxiliary equation is:

2m² - 20m + 136 = 0

Dividing by 2, we have

m² - 10m + 68 = 0

Next, we solve the auxiliary equation to obtain the values of m.

Solving using the quadratic formula

m = [-b ± √(b² - 4ac)]/2a

Where a = 1, b = -10, and c = 68

m = [10 ± √(100 - 272)]/2

= 5 ± (1/2)√(-172)

= 5 ± (1/2)i√172

= 5 ± 6.6i

For solutions of the form a ± ib, the complimentary solution is

y = e^(ax)[C1cosbx + C2sinbx]

Therefore, the complimentary solution is

y = e^(5x)[C1cos(6.6x) + C2sin(6.6x)]

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Rhombus ADEF is inscribed into a triangle ABC so that they share angle A and the vertex E lies on the side BC . What is the leng
Thepotemich [5.8K]

Answer:

Length of side of rhombus is x=\frac{ab}{a+b}  

Step-by-step explanation:

Given Rhombus ADEF is inscribed into a triangle ABC so that they share angle A and the vertex E lies on the side BC. We have to find the length of side of rhombus.

It is also given that AB=a and AC=b

Let side of rhombus is x.

In ΔCEF and ΔCBA

∠CEF=∠CBA        (∵Corresponding angles)

∠CFE=∠CAB        (∵Corresponding angles)

By AA similarity rule, ΔCEF~ΔCBA

∴ their sides are in proportion

\frac{EF}{AB}=\frac{CF}{AC}

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⇒ x=\frac{ab}{a+b}

Hence, length of side of rhombus is x=\frac{ab}{a+b}  

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What is the length side of a triangle that has vertices at (-5, -1), (-5, 5), and (3, -1)?
Rasek [7]

The side lengths of triangle are 6 units, 8 units and 10 units.

<u>SOLUTION: </u>

Given that, we have to find what is the length side of a triangle that has vertices at (-5, -1), (-5, 5), and (3, -1)  

We know that, distance between two points P\left(x_{1}, y_{1}\right) \text { and } Q\left(x_{2}, y_{2}\right) is given by  

P Q=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}

Now,  

\begin{array}{l}{\text { Distance between }(-5,-1) \text { and }(-5,5)=\sqrt{(-5-(-5))^{2}+(5-(-1))^{2}}} \\\\ {\qquad \begin{array}{l}{=\sqrt{(-5-(-5))^{2}+(5-(-1))^{2}}} \\\\ {=\sqrt{(0)^{2}+(5+1)^{2}}=\sqrt{(6)^{2}}=6} \\\\ {=\sqrt{(-5)^{2}+(5+1)^{2}+(5-(-1))^{2}}} \\\\ {=\sqrt{(-8)^{2}+(5+1)^{2}}=\sqrt{64+36}=\sqrt{100}=10} \\\\ {=\sqrt{(3-1)^{2}+(-1-(-1))^{2}}} \\\\ {=\sqrt{(5+3)^{2}+(0)^{2}}=\sqrt{(8)^{2}}=8}\end{array}}\end{array}

8 0
3 years ago
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