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Lyrx [107]
3 years ago
12

What is the equation of a line that passes through the point (8, 1) and is perpendicular to the line whose equation is y=−2/3x+5

y=−2/3x+5 ?
Mathematics
2 answers:
Kazeer [188]3 years ago
5 0
Y = -2/3x + 5....the slope here is -2/3. A perpendicular line will have a negative reciprocal slope. To find the negative reciprocal, we flip the slope and change the sign. So our perpendicular line will have a slope of 3/2 (see how I flipped -2/3 making it -3/2....and then changed the sign, making it 3/2)

y = mx + b
slope(m) = 3/2
(8,1)....x = 8 and y = 1
now we sub and find b, the y int
1 = 3/2(8) + b
1 = 12 + b
1 - 12 = b
-11 = b

so ur perpendicular equation is : y = 3/2x - 11 <==
VikaD [51]3 years ago
4 0
Your equation oinitial is y=-2/3x+5

perpendicular lines have slopes that mulitply to get -1

y=mx+b
m=slope

given
y=-2/3x+5
slope=-2/3

perpendicular so
-2/3 times what=-1
times -3/2 both sides
what=3/2
the slope is 3/2

let's use point slope form

equation of a line that passes through (x1,y1) and has a slope of m is
y-y1=m(x-x1)
so given point (8,1) and slope 3/2

y-1=3/2(x-8)
is yo equation
if yo wanted slope intercept
y-1=3/2x-12
y=3/2x-11 is slope intercept
if wanted standard form

11=3/2x-y
3x-2y=22


equaiton is
y-1=\frac{3}{2}(x-8) for point slope form
y=\frac{3}{2}x-11 in slope intercept form
3x-2y=22 in standard form

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Hello people ~
Luden [163]

Cone details:

  • height: h cm
  • radius: r cm

Sphere details:

  • radius: 10 cm

================

From the endpoints (EO, UO) of the circle to the center of the circle (O), the radius is will be always the same.

<u>Using Pythagoras Theorem</u>

(a)

TO² + TU² = OU²

(h-10)² + r² = 10²                                   [insert values]

r² = 10² - (h-10)²                                     [change sides]

r² = 100 - (h² -20h + 100)                       [expand]

r² = 100 - h² + 20h -100                        [simplify]

r² = 20h - h²                                          [shown]

r = √20h - h²                                       ["r" in terms of "h"]

(b)

volume of cone = 1/3 * π * r² * h

===========================

\longrightarrow \sf V = \dfrac{1}{3}  * \pi  * (\sqrt{20h - h^2})^2  \  ( h)

\longrightarrow \sf V = \dfrac{1}{3}  * \pi  * (20h - h^2)  (h)

\longrightarrow \sf V = \dfrac{1}{3}  * \pi  * (20 - h) (h) ( h)

\longrightarrow \sf V = \dfrac{1}{3} \pi h^2(20-h)

To find maximum/minimum, we have to find first derivative.

(c)

<u>First derivative</u>

\Longrightarrow \sf V' =\dfrac{d}{dx} ( \dfrac{1}{3} \pi h^2(20-h) )

<u>apply chain rule</u>

\sf \Longrightarrow V'=\dfrac{\pi \left(40h-3h^2\right)}{3}

<u>Equate the first derivative to zero, that is V'(x) = 0</u>

\Longrightarrow \sf \dfrac{\pi \left(40h-3h^2\right)}{3}=0

\Longrightarrow \sf 40h-3h^2=0

\Longrightarrow \sf h(40-3h)=0

\Longrightarrow \sf h=0, \ 40-3h=0

\Longrightarrow \sf  h=0,\:h=\dfrac{40}{3}<u />

<u>maximum volume:</u>                <u>when h = 40/3</u>

\sf \Longrightarrow max=  \dfrac{1}{3} \pi (\dfrac{40}{3} )^2(20-\dfrac{40}{3} )

\sf \Longrightarrow maximum= 1241.123 \ cm^3

<u>minimum volume:</u>                 <u>when h = 0</u>

\sf \Longrightarrow min=  \dfrac{1}{3} \pi (0)^2(20-0)

\sf \Longrightarrow minimum=0 \ cm^3

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