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3 years ago

14

How do I sketch the graph y=-x?

1 answer:

pashok25 [27]3 years ago

6 0

x y
-1 1
-2 2
-3 3
---------------
x y
1 -1
2 -2
3 -3

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3 years ago

How do i do this? Circle Theorm

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2 years ago

Sand is leaking through a small hole at the bottom of a conical funnel at the rate of 12cm3/s . if the radius of the funnel is 8

Bezzdna [24]

Answer:

h ≅ 16.29 cm

$\mathbf{\dfrac{dh}{dt}= -0.1809 \ cm/s}$

Step-by-step explanation:

From the conical funnel;

Let the adjacent line at the base of the cone be radius (r) and the opposite ( vertical length) be the height (h)

Then;

$tan \theta= \dfrac{r}{h}$

$tan \theta= \dfrac{8}{16}$

$tan \theta= \dfrac{1}{2}$

∴

$\dfrac{r}{h} = \dfrac{1}{2}$

$r = \dfrac{h}{2}$

The volume (V) of the cone is expressed as:

$V = \dfrac{1}{3}\pi r ^2 h$

$V = \dfrac{1}{3}\pi (\dfrac{h}{2})^2 h$

$V = \dfrac{\pi h^3}{3\times 4}$

$\dfrac{dv}{dt} = \dfrac{3 \pi h^2}{3 \times 4} \dfrac{dh}{dt}$

$\dfrac{dv}{dt} = \dfrac{ \pi h^2}{4} \dfrac{dh}{dt}$

Given that:

$\dfrac{dv}{dt} = 12 \pi$

Then:

$-12 \pi = \dfrac{ \pi h^2}{4} \dfrac{dh}{dt}$

$- \int 48 \ dt = \int h^2 \ dh$

$\dfrac{h^3}{3}= -48 t + c$

$At \ t = 0 \ ; h = 0$

∴

$\dfrac{0^3}{3} = -48(0) + c$

c = 0

So;

$\dfrac{h^3}{3}= -48 t + 0$

$\dfrac{h^3}{3}= -48 t$

t = 30

$\implies h = (-48(30))^{1/3}$

h = 16.2865

h ≅ 16.29 cm

Thus;

from $-12 \pi = \dfrac{ \pi h^2}{4} \dfrac{dh}{dt}$

$\dfrac{-12 \pi} { \dfrac{ \pi h^2}{4} } =\dfrac{dh}{dt}$

$\dfrac{dh}{dt} = \dfrac{-12 \pi} { \dfrac{ \pi h^2}{4} }$

$\dfrac{dh}{dt}=\dfrac{-12 \times 4} { {16.29^2} }$

$\mathbf{\dfrac{dh}{dt}= -0.1809 \ cm/s}$

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