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miss Akunina [59]
3 years ago
10

HELP PLS use the original price and markup to find the retail price. original price $68 markup 20%

Mathematics
1 answer:
insens350 [35]3 years ago
4 0

Answer: $81.60

Step-by-step explanation:

When we have an original price OP, and a markup of X% (this is how much increases the price with respect to the original price)

The retail price will be:

RP = OP + (X%/100%)*OP

RP = OP*(1 + X%/100%)

In this case, we have:

Original price = OP = $68

markup = X% = 20%

Replacing these in the above equation, we get:

RP= $68*(1 + 20%/100%) = $68*(1 + 0.20)

RP = $68*(1.20) = $81.60

Then the retail price of this particular item is $81.60

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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

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                       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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