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ElenaW [278]
4 years ago
13

Three Aztecs are dressing for a football game. They know there are 5 helmets in the locker, 3 black and 2 red. As they are about

to get their helmets, the lights go out. They all decide to just grab a helmet in the dark and head for the field. As they go in single file out into the tunnel, the one in the back says, "I can see both of the helmets in front of me, but I do not know the color of my helmet". The one in the middle says, "I can see the helmet in front of me, but I cannot determine the color of my own". The one leading the pack says, "I can see neither of the helmets behind me but I know the color of my own". How does he do it and what color was the helmet?
Mathematics
1 answer:
muminat4 years ago
4 0

We are given that there are:

Black helmets = 3

Red helments = 2

 

Let us start from the guy at the back. The guy at the back can see the two helmets in front of him. If the two helmets in front of him are red, then he can certainly say that his helmet is black since there are only 2 red helmets. Since that is not the case, then therefore he can either be black or red. Therefore this also means that at least 1 of the 2 guys in front of him has black helmet.

<span>Since the remaining two should have at least 1 black, therefore the middle guy can see that the helmet in front of him is not red (so 1st guy is black). Because if the guy in front of him is red, then he can know that he must be black since remember, at least one of the 2 front guys must be black, but the middle guy cannot say that. Therefore the 1st guy knows he has a black helmet.</span>

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The profile of the cables on a suspension bridge may be modeled by a parabola. The central span of the bridge is 1210 m long and
brilliants [131]

Answer:

The approximated length of the cables that stretch between the tops of the two towers is 1245.25 meters.

Step-by-step explanation:

The equation of the parabola is:

y=0.00035x^{2}

Compute the first order derivative of <em>y</em> as follows:

 y=0.00035x^{2}

\frac{\text{d}y}{\text{dx}}=\frac{\text{d}}{\text{dx}}[0.00035x^{2}]

    =2\cdot 0.00035x\\\\=0.0007x

Now, it is provided that |<em>x </em>| ≤ 605.

⇒ -605 ≤ <em>x</em> ≤ 605

Compute the arc length as follows:

\text{Arc Length}=\int\limits^{x}_{-x} {1+(\frac{\text{dy}}{\text{dx}})^{2}} \, dx

                  =\int\limits^{605}_{-605} {\sqrt{1+(0.0007x)^{2}}} \, dx \\\\={\displaystyle\int\limits^{605}_{-605}}\sqrt{\dfrac{49x^2}{100000000}+1}\,\mathrm{d}x\\\\={\dfrac{1}{10000}}}{\displaystyle\int\limits^{605}_{-605}}\sqrt{49x^2+100000000}\,\mathrm{d}x\\\\

Now, let

x=\dfrac{10000\tan\left(u\right)}{7}\\\\\Rightarrow u=\arctan\left(\dfrac{7x}{10000}\right)\\\\\Rightarrow \mathrm{d}x=\dfrac{10000\sec^2\left(u\right)}{7}\,\mathrm{d}u

\int dx={\displaystyle\int\limits}\dfrac{10000\sec^2\left(u\right)\sqrt{100000000\tan^2\left(u\right)+100000000}}{7}\,\mathrm{d}u

                  ={\dfrac{100000000}{7}}}{\displaystyle\int}\sec^3\left(u\right)\,\mathrm{d}u\\\\=\dfrac{50000000\ln\left(\tan\left(u\right)+\sec\left(u\right)\right)}{7}+\dfrac{50000000\sec\left(u\right)\tan\left(u\right)}{7}\\\\=\dfrac{50000000\ln\left(\sqrt{\frac{49x^2}{100000000}+1}+\frac{7x}{10000}\right)}{7}+5000x\sqrt{\dfrac{49x^2}{100000000}+1}

Plug in the solved integrals in Arc Length and solve as follows:

\text{Arc Length}=\dfrac{5000\ln\left(\sqrt{\frac{49x^2}{100000000}+1}+\frac{7x}{10000}\right)}{7}+\dfrac{x\sqrt{\frac{49x^2}{100000000}+1}}{2}|_{limits^{605}_{-605}}\\\\

                  =1245.253707795227\\\\\approx 1245.25

Thus, the approximated length of the cables that stretch between the tops of the two towers is 1245.25 meters.

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Answer:

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