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FinnZ
1 month ago
12

Una avenida está siendo asfaltada por etapas. En la primera etapa se asfaltó la mitad; en la segunda, la quinta parte, y en la t

ercera, la cuarta parte del total. ¿Cuál es la longitud de la avenida si aún faltan 200 m por asfaltar?
Mathematics
1 answer:
babunello [11.8K]1 month ago
7 0

Respuesta:

Sabemos que:

L representa la longitud de la avenida.

En la primera fase se asfaltó la mitad, es decir, L/2, quedando lo siguiente por asfaltar:

L - L/2 = L/2.

En la segunda fase se asfaltó un quinto, o L/5, por lo que el restante es:

L/2 - L/5 = 5*L/10 - 2*L/10 = (3/10)*L

En la tercera fase se asfaltó un cuarto del total, es decir, L/4, quedando:

(3/10)*L - L/4 = 12*L/40 - 10L/40 = (2/40)*L

Y sabemos que la parte que queda por asfaltar mide 200m:

(2/40)*L = 200m

L = 200m*(40/2) = 4,000m

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1.4in

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Length of Photo = 4in

Width of Photo = 3in

Unknown:

Value of X =?

Solution:

Following these steps:

  Area of a rectangle  = l x w

Considering the photo is rectangular; area of photo:

  Area of photo  = 4in x 3in = 12in²

For the advertisement's area;

   Length of ad  = 4 + x

   Width of ad = 3 + x

Based on the relation that

                  the area of the photo  = \frac{1}{2} area of ad

                           12in²  = \frac{1}{2} area of ad

                   Area of ad  = 24in²

Advertising Area:

             (4 + x) (3 + x)  = 24

               12 + 4x + 3x + x² = 24

                12 + 7x + x²  = 24

                       x² + 7x = 24 - 12

                       x² + 7x = 12

                     x² + 7x - 12  = 0

                    Using the quadratic formula where

     a  = 1, b = 7 and c = -12

          x  = \frac{-b (+/-) \sqrt{b^{2} } - 4ac}{2a}

     

          x = \frac{-7 + \sqrt[]{-7^{2} - 4x1x-12 } }{2x1}   or \frac{-7 - \sqrt[]{-7^{2} - 4x1x-12 } }{2x1}

          x  = 1.4  or  -8.4

therefore the answer is 1.4in

x is 1.4in

4 0
17 days ago
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