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Evgesh-ka [11]
4 years ago
14

Determine whether the line Upper L 1L1 whose equation is yequals=5 x minus 95x−9 and the line Upper L 2L2 whose equation is yequ

als=one fifth x plus 6 1 5x+6 are​ parallel, perpendicular, or neither.
Mathematics
1 answer:
katovenus [111]4 years ago
5 0

Answer:

Neither

Step-by-step explanation:

Let m_{1} and m_{2} be the gradients of two lines.

  1. The two lines are parallel if their gradients m_{1} and m_{2} are equal.
  2. Two lines are perpendicular if the products of the gradients, m_{1}m_{2} =-1, i.e. gradient of one, m_{1}=-\frac{1}{m_{2}} m_{2}

Consider the given lines L_{1} whose equation is given as y=5x-9 and the line L_{2} whose equation is given as y=\frac{1}{5}x+6.

To make a comparison, ensure that the lines are in the slope-intercept form y=mx+c.

In L_{1} y=5x-9, m_{1} =5

In L_{2} , y=\frac{1}{5}x+6, m_{2} =\frac{1}{5}

Their product m_{1}m_{2} = 1,

Therefore the lines are neither parallel nor perpendicular.

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Let <em>s(t)</em> be the amount of salt in the tank at time <em>t</em>. Then <em>s(0)</em> = 50 lb.

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d<em>s</em>/d<em>t</em> = 12 - <em>s</em>/150

150 d<em>s</em>/d<em>t</em> = 1800 - <em>s</em>

150/(1800 - <em>s</em>) d<em>s</em> = d<em>t</em>

Integrate both sides to get

-150 ln|1800 - <em>s</em>| = <em>t</em> + <em>C</em>

Solve for <em>s</em> :

ln|1800 - <em>s</em>| = -<em>t</em>/150 + <em>C</em>

1800 - <em>s</em> = exp(-<em>t</em>/150 + <em>C </em>)

1800 - <em>s</em> = <em>C</em> exp(-<em>t</em>/150)

<em>s</em> = 1800 - <em>C</em> exp(-<em>t</em>/150)

Now given that <em>s(0)</em> = 50, we solve for <em>C</em> :

50 = 1800 - <em>C</em> exp(-0/150)

50 = 1800 - <em>C</em>

<em>C</em> = 1750

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<em>s(t)</em> = 1800 - 1750 exp(-<em>t</em>/150)

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By the way, we didn't have to solve for <em>s(t)</em>, we could have instead stopped with

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<em>t</em> = 150 ln(1750/1700)

<em>t</em> = 150 ln(35/34)

<em>t</em> ≈ 4.348

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