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3 years ago

12

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1 answer:

Pachacha [2.7K]3 years ago

4 0

Answer:

Yes, the triangles are similar by SAS

Step-by-step explanation:

Looking at the picture, we see that there are two triangles with side lengths of different sides. We can deduce that the scale factor is 1 2/3, because 8 divided by 4.8 is 1 2/3, and 10 divided by 6 is the same. Now that we have clarified that the triangles share the same scale factor, we notice that the angle is also the same, as mentioned in the picture. This leads us to say that the triangles are similar by the SAS similarity theorem (Side, Angle, Side). I hope this helped and please don't hesitate to reach out with more questions!

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2x+y=8<br> y+3z=5<br> z+2w=1<br> 5w+3x=9<br><br> what is x,y,z,w

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2 years ago

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3 years ago

Consider the differential equation y'' − y' − 20y = 0. Verify that the functions e−4x and e5x form a fundamental set of solution

KIM [24]

Answer:

Therefore $e^{-4x}$ and $e^{5x}$ are fundamental solution of the given differential equation.

Therefore $e^{-4x}$ and $e^{5x}$ are linearly independent, since $W(e^{-4x},e^{5x})=9e^x\neq 0$

The general solution of the differential equation is

$y=c_1e^{-4x}+c_2e^{5x}$

Step-by-step explanation:

Given differential equation is

y''-y'-20y =0

Here P(x)= -1, Q(x)= -20 and R(x)=0

Let trial solution be $y=e^{mx}$

Then, $y'=me^{mx}$ and $y''=m^2e^{mx}$

$\therefore m^2e^{mx}-m e^{mx}-20e^{mx}=0$

$\Rightarrow m^2-m-20=0$

$\Rightarrow m^2-5m+4m-20=0$

$\Rightarrow m(m-5)+4(m-5)=0$

$\Rightarrow (m-5)(m+4)=0$

$\Rightarrow m=-4,5$

Therefore the complementary function is = $c_1e^{-4x}+c_2e^{5x}$

Therefore $e^{-4x}$ and $e^{5x}$ are fundamental solution of the given differential equation.

If $y_1$ and $y_2$ are the fundamental solution of differential equation, then

$W(y_1,y_2)=\left|\begin{array}{cc}y_1&y_2\\y'_1&y'_2\end{array}\right|\neq 0$

Then $y_1$ and $y_2$ are linearly independent.

$W(e^{-4x},e^{5x})=\left|\begin{array}{cc}e^{-4x}&e^{5x}\\-4e^{-4x}&5e^{5x}\end{array}\right|$

$=e^{-4x}.5e^{5x}-e^{5x}.(-4e^{-4x})$

$=5e^x+4e^x$

$=9e^x\neq 0$

Therefore $e^{-4x}$ and $e^{5x}$ are linearly independent, since $W(e^{-4x},e^{5x})=9e^x\neq 0$

Let the the particular solution of the differential equation is

$y_p=v_1e^{-4x}+v_2e^{5x}$

$\therefore v_1=\int \frac{-y_2R(x)}{W(y_1,y_2)} dx$

and

$\therefore v_2=\int \frac{y_1R(x)}{W(y_1,y_2)} dx$

Here $y_1= e^{-4x}$ , $y_2=e^{5x}$ , $W(e^{-4x},e^{5x})=9e^x$ ,and $R(x)=0$

$\therefore v_1=\int \frac{-e^{5x}.0}{9e^x}dx$

=0

and

$\therefore v_2=\int \frac{e^{5x}.0}{9e^x}dx$

=0

The the P.I = 0

The general solution of the differential equation is

$y=c_1e^{-4x}+c_2e^{5x}$

7 0

3 years ago