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3 years ago

6

Hans leaned a 17 ft ladder against his house. He placed the base of the ladder 8 ft from his house, and the top of the ladder re

ached the base of his window.
How high is the base of his window from the ground?
$a^{2}$ + $b^{2}$ = $c^{2}$
A.25
B.15
C.23
D.17

1 answer:

Fynjy0 [20]3 years ago

7 0

Answer:

b

Step-by-step explanation:

so 17 squared minus 8 squared is 225. the square root of 225 is 15

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3 years ago

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For the nonhomogeneous part of the ODE, let $y_p=(a_2x^2+a_1x+a_0)\sin2x+(b_2x^2+b_1x+b_0)\cos2x$ . Then

${y_p}''=(-4b_2x^2+(8a_2-b_1)x+4a_1-4b_0+2b_2)\cos2x+(-4a_2x^2+(-4a_1-8b_2)x-4a_0+2a_2-4b_1)\sin2x$

Substituting into the ODE gives

$(-8b_2x^2+(8a_2-b_1)x+4a_1-8b_0+2b_2)\cos2x+(-8a_2x^2+(-8a_1-8b_2)x-8a_0+2a_2-4b_1)\sin2x=(x^2-3)\sin2x$

It follows that

$\begin{cases}-8b_2=0\\8a_2-8b_1=0\\4a_1-8b_0+2b_2=0\\-8a_2=1\\-8a_1-8b_2=0\\-8a_0+2a_2-4b_1=-3\end{cases}\implies\begin{cases}a_2=-\dfrac18\\\\a_1=0\\\\a_0=\dfrac{13}{32}\\\\b_2=0\\\\b_1=-\dfrac18\\\\b_0=0\end{cases}$

which yields the particular solution

$y_p=-\dfrac18x^2\sin2x+\dfrac{13}{32}\sin2x-\dfrac18x\cos2x$

So the general solution is

$y=y_c+y_p$
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