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Inessa05 [86]
2 years ago
15

A construction company built a scale model of a building. The model was built using a scale of 3 inches = 32 feet. If the buildi

ng is expected to be 200 feet tall, how tall will the model be
Mathematics
1 answer:
Vlad1618 [11]2 years ago
7 0

Let x represent the height of the model.

We have been given that a construction company built a scale model of a building. The model was built using a scale of 3 inches = 32 feet. We are asked to find the height of the model, if  the building is expected to be 200 feet tall.

We will use proportions to solve our given problem as:

\frac{\text{Model height}}{\text{Actual height}}=\frac{\text{Model length}}{\text{Actual length}}

Upon substituting our given values, we will get:

\frac{x}{200\text{ ft}}=\frac{3\text{ in}}{\text{32 ft}}

\frac{x}{200\text{ ft}}\times 200\text{ ft}=\frac{3\text{ in}}{\text{32 ft}}\times 200\text{ ft}

x=3\text{ in}\times 6.25

x=18.75\text{ in}

Therefore, the model will be 18.75 inches tall.

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We have the given differential equation: y′′+4y=5xcos(2x)

We use the Method of Undetermined Coefficients.

We first solve the homogeneous differential equation y′′+4y=0.

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Now, we finding a particular solution.

y_p(t)=A5x\cos 2x\\\\y'_p(t)=A5\cos 2x-A10x\sin 2x\\\\y''_p(t)=-A20\sin 2x-A20x\cos 2x\\\\\\\implies y''+4y=5x\cos 2x\\\\-A20\sin 2x-A20x\cos 2x+4\cdot A5x\cos 2x=5x\cos 2x\\\\-A20\sin 2x=5x\cos 2x\\\\A=-\frac{x}{4} \cot 2x\\

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So, solution of  the differential equation is

y(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos  2x+c_1e^{-2it}+c_2e^{2it}\\

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