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3 years ago

Parallel Lines A line is parallel to y = 2x - 8 and intersects the point (-4,-1). What is the equation of this parallel line?

Mathematics

1 answer:

Assoli18 [71]3 years ago

4 0

Answer:

y= 2x+7

Step-by-step explanation:

Since the line is parallel to y = 2x - 8 it has slope m= 2

The equation of the line passing through (-4,-1) is given by,

$\frac{y - ( - 1)}{x - ( - 4)} = m$

$\frac{y + 1}{x + 4} = 2$

$y + 1 = 2(x + 4)$

$y + 1 = 2x + 8$

$y = 2x + 8 - 1$

$y = 2x + 7$

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Gravel is being dumped from a conveyor belt at a rate of 20 ft3 /min and its coarseness is such that it forms a pile in the shap

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The height of the pile is increasing at the rate of $\mathbf{ \dfrac{20}{56.25 \pi} \ \ \ \ \ ft/min}$

Step-by-step explanation:

Given that :

Gravel is being dumped from a conveyor belt at a rate of 20 ft³ /min

i.e $\dfrac{dV}{dt}= 20 \ ft^3/min$

we know that radius r is always twice the diameter d

i.e d = 2r

Given that :

the shape of a cone whose base diameter and height are always equal.

then d = h = 2r

h = 2r

r = h/2

The volume of a cone can be given by the formula:

$V = \dfrac{\pi r^2 h}{3}$

$V = \dfrac{\pi (h/2)^2 h}{3}$

$V = \dfrac{1}{12} \pi h^3$

$V = \dfrac{ \pi h^3}{12}$

Taking the differentiation of volume V with respect to time t; we have:

$\dfrac{dV}{dt }= (\dfrac{d}{dh}(\dfrac{\pi h^3}{12})) \times \dfrac{dh}{dt}$

$\dfrac{dV}{dt }= (\dfrac{\pi h^2}{4} ) \times \dfrac{dh}{dt}$

we know that:

$\dfrac{dV}{dt}= 20 \ ft^3/min$

So;we have:

$20= (\dfrac{\pi (15)^2}{4} ) \times \dfrac{dh}{dt}$

$20= 56.25 \pi \times \dfrac{dh}{dt}$

$\mathbf{\dfrac{dh}{dt}= \dfrac{20}{56.25 \pi} \ \ \ \ \ ft/min}$

The height of the pile is increasing at the rate of $\mathbf{ \dfrac{20}{56.25 \pi} \ \ \ \ \ ft/min}$

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