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Ulleksa [173]
3 years ago
11

Three boys were running around the track. The first can run 1 lap in 15 min. The second can run 1 lap in 25 min. and the third c

an run a lap in 10 min. If the boys start running at 12:45, what time will all 3 boys pass each other again?​
Mathematics
1 answer:
Juliette [100K]3 years ago
3 0

Answer:

At 15:15

Step-by-step explanation:

Given,

the 1st boy can run a lap in 15 minutes

the 2nd boy can run a lap in 25 minutes

the 3rd boy can run a lap in 10 minutes

Now if they start running at 12:45,they will meet after the time which will be equal to the L.C.M. of their time need to run a lap.

L.C.M. of 15,25 & 10 = 150

So,they will pass each other after 150 minutes or 2 hours & 30 minutes.

If we add it with 12:45 we'll get 15:15 or 3:15 pm

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Answer: (A) 19.3%

<u>Step-by-step explanation:</u>

\text{Vodka: }2.5\text{ oz }\times\dfrac{\$35}{1\text{ liter }}\times\dfrac{1\text{ liter }}{33.814\text{ oz }}=\$2.59\\\\\text{Vermouth: }0.5\text{ oz }\times\dfrac{\$7.99}{1\text{ liter }}\times\dfrac{1\text{ liter }}{33.814\text{ oz }}=\$0.12\\\\\text{Olive: }\$0.15

Vodka + Vermouth + Olive = Total Cost

$2.59  +   $0.12      + $0.15 = $2.86

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Please see attachment
Dafna11 [192]

Answer:

a) The value of absolute minimum value = - 0.3536  

b) which is attained at   x = \frac{1}{\sqrt{2} }  

Step-by-step explanation:

<u>Step(i)</u>:-

Given function

                       f(x) = \frac{-x}{2x^{2} +1}     ...(i)

Differentiating equation (i) with respective to 'x'

                     f^{l} = \frac{2x^{2} +1(-1) - (-x) (4x)}{(2x^{2}+1)^{2}  }   ...(ii)

                    f^{l}(x) = \frac{2x^{2}-1}{(2x^{2}+1)^{2}  }

Equating Zero

                   f^{l}(x) = \frac{2x^{2}-1}{(2x^{2}+1)^{2}  } = 0

                 \frac{2x^{2}-1}{(2x^{2}+1)^{2}  } = 0

                2 x^{2}-1 = 0

               2 x^{2} = 1

             x^{2}  = \frac{1}{2}

             x = \frac{-1}{\sqrt{2} }  , x = \frac{1}{\sqrt{2} }

<u><em>Step(ii):</em></u>-

Again Differentiating equation (ii) with respective to 'x'

f^{ll}(x) = \frac{(2x^{2} +1)^{2} (4x) - 2(2x^{2} +1) (4x)(2x^{2}-1) }{(2x^{2}+1)^{4}  }

put

      x = \frac{1}{\sqrt{2} }

f^{ll} (x) > 0

The absolute minimum value at   x = \frac{1}{\sqrt{2} }

<u><em>Step(iii):</em></u>-

The value of absolute minimum value

                         f(x) = \frac{-x}{2x^{2} +1}

                       f(\frac{1}{\sqrt{2} } ) = \frac{-\frac{1}{\sqrt{2} } }{2(\frac{1}{\sqrt{2} } )^{2} +1}

         on calculation we get

The value of absolute minimum value = - 0.3536      

<u><em>Final answer</em></u>:-

a) The value of absolute minimum value = - 0.3536  

b) which is attained at   x = \frac{1}{\sqrt{2} }    

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