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Licemer1 [7]
3 years ago
11

A 5-meter ladder is leaning against the side of a house. The foot of the ladder is pulled away from the house at a rate of 0.4 m

/sec. Determine how fast the top of the ladder is descending when the foot of the ladder is 3 meters from the house.
Mathematics
1 answer:
Flura [38]3 years ago
3 0
<h2>The top of the ladder is descending at 0.3 m/s.</h2>

Step-by-step explanation:

By Pythagoras theorem we know that

              Hypotenuse² = Base² + Perpendicular²

                   h² = b² + p²

We have for ladder

                        h = 5 m

                        b = 3 m

                        5² = 3² + p²

                        p = 4 m

                        \frac{db}{dt}=0.4m/s\\\\\frac{dh}{dt}=0

Differentiating h² = b² + p² with respect to time

                    2h\times \frac{dh}{dt}=2b\times \frac{db}{dt}+2p\times \frac{dp}{dt}\\\\5\times 0=3\times 0.4+4\times \frac{dp}{dt}\\\\\frac{dp}{dt}=-0.3m/s

The top of the ladder is descending at 0.3 m/s.

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Line segment NY has endpoints N(-11, 5) and Y(3,-3).
777dan777 [17]

Given:

Line segment NY has endpoints N(-11, 5) and Y(3,-3).

To find:

The equation of the perpendicular bisector of NY.

Solution:

Midpoint point of NY is

Midpoint=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)

Midpoint=\left(\dfrac{-11+3}{2},\dfrac{5-3}{2}\right)

Midpoint=\left(\dfrac{-8}{2},\dfrac{2}{2}\right)

Midpoint=\left(-4,1\right)

Slope of lines NY is

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m=\dfrac{-3-5}{3-(-11)}

m=\dfrac{-8}{14}

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Product of slopes of two perpendicular lines is -1. So,

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The perpendicular bisector of NY passes through (-4,1) with slope \dfrac{7}{4}. So, the equation of perpendicular bisector of NY is

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y-1=\dfrac{7}{4}(x-(-4))

y-1=\dfrac{7}{4}(x+4)

y-1=\dfrac{7}{4}x+7

Add 1 on both sides.

y=\dfrac{7}{4}x+8

Therefore, the equation of perpendicular bisector of NY is y=\dfrac{7}{4}x+8.

6 0
2 years ago
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