Mathematics, 07.12.2019 03:31 tiwaribianca475

Aconical paper cup is 10 cm tall with a radius of 10 cm. the cup is being filled with water so that the water level rises at a rate of 2 cm/sec. at what rate is water being poured into the cup when the water level is 8 cm?

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Answer from: Vamoppppp

At the rate of $128\pi\ cm^{3}/sec$ water being poured into the cup when the water level is cm

The above figure shows the variables

Radius of conical cup() =10 cm

Height of the conical cup () cm

Height of water level in cup at time , cm

Radius of water at time cm

Volume of water at time $=V^{3}$

$\dfrac{\partial h}{\partial t}=2m/sec$

Triangle ABC is similar to triangle PBQ

$\frac{r}{h} = \frac{r}{h}$

$\frac{r}{h} = \frac{10}{10}$

$r=h\\r=8$

find $\dfrac{dV}{dt}$ which is rate of water being poured in cup.

The volume of a PBQ cone is,

$\begin{aligned}Volume&= Area\;of\;circle \times h\\&=\pi r^2 h\end{aligned}$

Differentiate with respect to we get:

$\dfrac{dV}{dt} = \pi r^{2}\frac{dh}{dt}$

$\dfrac{dV}{dt} = \pi (8)^{2}\times 2$

Therefore , $\dfrac{dV}{dt}=128\pi\ cm^{3}/sec$

Learn more about Volume related concepts here:

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Aconical paper cup is 10 cm tall with a radius of 10 cm. the cup is being filled with water so that

Answer from: onegirl435

Water is being pored in the cone with the rate of 401.92 cm³ per second.

Step-by-step explanation:

There is a cone having height h = 10 cm and radius r = 10 cm

That means h = r

Rate of change of water level $\frac{dh}{dt}=2$ cm per second

Since volume of a cone is represented by the formula,

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

Now we have to find the value of $\frac{dV}{dt}$ .

$\frac{dV}{dt}=\frac{d}{dt}(\frac{1}{3}\pi r^{2}h)$

Since r = h

$\frac{dV}{dt}=\frac{d}{dt}(\frac{1}{3}\pi h^{3} )$

$\frac{dV}{dt}=\frac{3}{3}\pi h^{2}.\frac{dh}{dt}$

For h = 8 cm and $\frac{dh}{dt}=2$ cm per second.

$\frac{dV}{dt}=\frac{3}{3}\pi (8)^{2}\times (2)$

$\frac{dV}{dt}=128\pi$ = 401.92 cubic cm per second

Therefore, water is being pored with the rate of 401.92 cm³ per sec.

Answer from: Quest

7.666270253x10 raised to the 18th power

Answer from: Quest

the system of inequity is

$y\leq -x+4$

$3y> x$

step-by-step explanation:

from the given graph it is noticed that we have two related lines. one is solid and another is dashed.

if a line passing through two points, then the equation of line is

$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

the solid line passing through the points (0,4) and (4,0).

$y-4=\frac{0-4}{4-0}(x-0)$

$y=-x+4$

the related equation of solid line is $y=-x+4$ .

similarly, find the related equation of dashed line. the dashed liune passing through the points (0,0) and (3,1).

$y-0=\frac{0-1}{3-0}(x-0)$

$3y=x$

the related equation of dashed line is $3y=x$

from the given graph it is noticed that the point (0,1) lies on the shaded region, therefore the point (0,1) must be satisfied by the related equations.

check the related equations by the point (0,1).

$1=-0+4$

$1=4$

this statement is true if the sign of inequality is < instead of equality. since it is an solid line therefore the sign of inequality will be $\leq$ . the required inequality is

$y\leq -x+4$

similarly, check the second related equation.

$3(1)=0$

it is a dashed line, so this statement is true if the sign of inequality is > . the required inequality is

$3y> x$

15 points write a system of inequalities to represent the shaded portion of the graph

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