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Svetach [21]
2 years ago
7

During the Lantern Festival, the Chinese store sells lanterns for $ 8.88 each, including tax. The shop received a total of $106.

56 from the sales of lantern one night. Write an equation that can be used to determine the number of lanterns, x, sold by the Chinese store that night. How many lanterns was sold?
Mathematics
1 answer:
torisob [31]2 years ago
4 0

Answer:

12 Lanterns.

Step-by-step explanation:

To find how many they sold in one night, divide 106.56 by 8.88 and you get 12.

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Consider the following. (A computer algebra system is recommended.) y'' + 3y' = 2t4 + t2e−3t + sin 3t (a) Determine a suitable f
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First look for the fundamental solutions by solving the homogeneous version of the ODE:

y''+3y'=0

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For the non-homogeneous version, you can exploit the superposition principle and consider one term from the right side at a time.

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Assume the ansatz solution,

{y_p}=at^5+bt^4+ct^3+dt^2+et

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\implies {y_p}''=20at^3+12bt^2+6ct+2d

(You could include a constant term <em>f</em> here, but it would get absorbed by the first solution C_1 anyway.)

Substitute these into the ODE:

(20at^3+12bt^2+6ct+2d)+3(5at^4+4bt^3+3ct^2+2dt+e)=2t^4

15at^4+(20a+12b)t^3+(12b+9c)t^2+(6c+6d)t+(2d+e)=2t^4

\implies\begin{cases}15a=2\\20a+12b=0\\12b+9c=0\\6c+6d=0\\2d+e=0\end{cases}\implies a=\dfrac2{15},b=-\dfrac29,c=\dfrac8{27},d=-\dfrac8{27},e=\dfrac{16}{81}

y''+3y'=t^2e^{-3t}

e^{-3t} is already accounted for, so assume an ansatz of the form

y_p=(at^3+bt^2+ct)e^{-3t}

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\implies {y_p}''=(9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c)e^{-3t}

Substitute into the ODE:

(9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c)e^{-3t}+3(-3at^3+(3a-3b)t^2+(2b-3c)t+c)e^{-3t}=t^2e^{-3t}

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