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dlinn [17]
3 years ago
13

Gina looks at the architectural plan of a four-walled room in which the walls meet each other at right angles. The length of one

wall in the plan is 17 inches. The length of the diagonal of the floor of the room in the plan is approximately 18.79 inches.
Is the room in the shape of a square? Explain how you determined your answer. Show all your work.
Mathematics
2 answers:
Alja [10]3 years ago
8 0
No. A square has equal sides that are perpendicular and parallel. I'm guessing this room resembles some other figure.
disa [49]3 years ago
6 0
Your shape is a cube.
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poizon [28]

Answer:

See below

Step-by-step explanation:

It has something to do with the<em> </em><u><em>Weierstrass substitution</em></u>, where we have

$\int\, f(\sin(x), \cos(x))dx = \int\, \dfrac{2}{1+t^2}f\left(\dfrac{2t}{1+t^2}, \dfrac{1-t^2}{1+t^2} \right)dt$

First, consider the double angle formula for tangent:

\tan(2x)= \dfrac{2\tan(x)}{1-\tan^2(x)}

Therefore,

\tan\left(2 \cdot\dfrac{x}{2}\right)= \dfrac{2\tan(x/2)}{1-\tan^2(x/2)} = \tan(x)=\dfrac{2t}{1-t^2}

Once the double angle identity for sine is

\sin(2x)= \dfrac{2\tan(x)}{1+\tan^2(x)}

we know \sin(x)=\dfrac{2t}{1+t^2}, but sure,  we can derive this formula considering the double angle identity

\sin(x)= 2\sin\left(\dfrac{x}{2}\right)\cos\left(\dfrac{x}{2}\right)

Recall

\sin \arctan t = \dfrac{t}{\sqrt{1 + t^2}} \text{ and } \cos \arctan t = \dfrac{1}{\sqrt{1 + t^2}}

Thus,

\sin(x)= 2 \left(\dfrac{t}{\sqrt{1 + t^2}}\right) \left(\dfrac{1}{\sqrt{1 + t^2}}\right) = \dfrac{2t}{1 + t^2}

Similarly for cosine, consider the double angle identity

Thus,

\cos(x)=  \left(\dfrac{1}{\sqrt{1 + t^2}}\right)^2- \left(\dfrac{t}{\sqrt{1 + t^2}}\right)^2 = \dfrac{1}{t^2+1}-\dfrac{t^2}{t^2+1} =\dfrac{1-t^2}{1+t^2}

Hence, we showed \sin(x) \text { and } \cos(x)

======================================================

5\cos(x) =12\sin(x) +3, x \in [0, 2\pi ]

Solving

5\,\overbrace{\frac{1-t^2}{1+t^2}}^{\cos(x)} = 12\,\overbrace{\frac{2t}{1+t^2}}^{\sin(x)}+3

\implies \dfrac{5-5t^2}{1+t^2}= \dfrac{24t}{1+t^2}+3 \implies  \dfrac{5-5t^2 -24t}{1+t^2}= 3

\implies 5-5t^2-24t=3\left(1+t^2\right) \implies -8t^2-24t+2=0

t = \dfrac{-(-24)\pm \sqrt{(-24)^2-4(-8)\cdot 2}}{2(-8)} = \dfrac{24\pm 8\sqrt{10}}{-16} =  \dfrac{3\pm \sqrt{10}}{-2}

t=-\dfrac{3+\sqrt{10}}{2}\\t=\dfrac{\sqrt{10}-3}{2}

Just note that

\tan\left(\dfrac{x}{2}\right) =  \dfrac{3\pm 8\sqrt{10}}{-2}

and  \tan\left(\dfrac{x}{2}\right) is not defined for x=k\pi , k\in\mathbb{Z}

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What is the best approximation for the input value when f(x)=g(x)?
Lostsunrise [7]

Answer:

x=0 and x=1.

Step-by-step explanation:

If we have to different functions like the ones attached, one is a parabolic function and the other is a radical function. To know where f(x)=g(x), we just have to equalize them and find the solution for that equation:

x^{2}=\sqrt{x} \\(x^{2} )^{2}=(\sqrt{x} )^{2}\\x^{4}=x\\x^{4}-x=0\\x(x^{3}-1)=0\\

So, applying the zero product property, we have:

x=0\\x^{3}-1=0\\x^{3}=1\\x=\sqrt[3]{1}=1

Therefore, these two solutions mean that there are two points where both functions are equal, that is, when x=0 and x=1.

So, the input values are  x=0 and x=1.

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3 years ago
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Answer:

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Step-by-step explanation:

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